TITLE: Estimate $\int_0^{\pi}\sin(x)dx$ using monte carlo
QUESTION [1 upvotes]: So I'm reading about monte carlo simulation and stumbled upon a way to simulate $\int_0^{\pi}\sin(x)dx$: Generate independent standard uniforms $U^1$ and $U^2$ and find the proportion of pairs such that $\sin(\pi U^1)>U^2$ and multiply by $\pi$. It is then stated that it will converge to $\int_0^{\pi}\sin(x)dx=2$. However how would one prove that? Do we want to show that $N^{-1} \pi\sum_{i=1}^N1_{(\sin(\pi U^1_i)>U_i^2)}\to 2$ almost surely? If so I guess we can say that $1_{(\sin(\pi U^1_i)>U^2_i)}$ is Bernoulli with success probability $p=\mathbb{P}(\sin(\pi U^1_i)>U^2_i)$. However I'm not sure how to continue.

REPLY [0 votes]: As you pointed out, the result follows from the law of large numbers. For $p$ (in the OP's notation), suppose $U$ and $V$ are independent uniform $0-1$ distributed random variables. An application of Fubini's theorem gives
$$\begin{align}
\pi\mathbb{P}[\sin(\pi U)>V]&=\pi\int_{[0,1\times[0,1]}\mathbb{1}_{\sin(\pi u)>v}(u,v)\,dudv\\
&=\pi\int^1_0\Big(\int^{\sin \pi u}_0\,dv\Big)\,du\\
&=\pi\int^1_0\sin\pi u\,du=\int^\pi_0\sin x\,dx
\end{align}$$


TITLE: Proof Verification: Prove $\sqrt{x}$ is uniformly continuous on [0, $\infty$)
QUESTION [4 upvotes]: Proof:
Fix $\epsilon \gt 0$. We want to find $\delta \gt 0$ such that:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}| \lt \epsilon$$
$$\phantom{2000i11111}\Rightarrow |\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \epsilon$$
If $|x-a|\lt \delta$ then $|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}$ for $\delta \lt 1$
Thus, choose $\delta = min\{1, (\sqrt{a+1}+\sqrt{a})\epsilon\}$. Then we have:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}|=|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}=\frac{(\sqrt{a+1}+\sqrt{a})\epsilon}{\sqrt{a+1}+\sqrt{a}}=\epsilon$$

REPLY [2 votes]: Simply, if $|x - a| < \delta(\epsilon) = \epsilon^2$, then, since $|\sqrt{x} - \sqrt{a}| \leqslant |\sqrt{x} + \sqrt{a}|$, we have
$$|\sqrt{x} - \sqrt{a}|^2 = |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}-\sqrt{a}|\leqslant |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}+\sqrt{a}| = |x - a|< \epsilon^2\\ \implies|\sqrt{x} - \sqrt{a}|< \epsilon$$


TITLE: Why does $\langle W(t) W(s) \rangle = \min (t,s)$ for Wiener process $W$?
QUESTION [1 upvotes]: In section 4.2 of Gardiner's Stochastic Methods, we find
\begin{align}
& \left\langle \sum_{i=1}^n W(\tau_i)[W(t_i) - W(t_{i-1})]\right\rangle \\[10pt]
= & \sum_{i=1}^n [ \min(\tau_i,t_i) - \min(\tau_i,t_{i-1})]
\end{align}
and I do not understand this step, although I do understand that $\langle W(t)^2 \rangle = t$.

REPLY [3 votes]: Suppose for simplicity that $s<t$. Then using independence of increments, we get
$$ \mathbb{E}[W_tW_s]=\mathbb{E}[(W_t-W_s)W_s]+\mathbb{E}[W_s^2]=\mathbb{E}[W_t-W_s]\mathbb{E}[W_s]+\mathbb{E}[W_s^2]=s$$


TITLE: Counter example in Sobolev spaces: Is there a standard example for showing that $H^1(\mathbb{R}^2)\not\subset L^\infty(\mathbb R^2)$?
QUESTION [1 upvotes]: One way to define the Sobolev space $H^s(\mathbb{R}^n)$ is as the completion of the $C^\infty$ functions on $\mathbb{R}^n$ under the norm
$$\lVert f \rVert_{H^s(\mathbb{R}^n)}:= \left( \sum_{|\alpha|\leq s} \lVert  \partial^\alpha f \rVert^2_{L^2(\mathbb{R^n})}\right)^{1/2}$$
where $\alpha=(\alpha_1,...,\alpha_n)$ with $|\alpha|:=\sum_{i=1}^n\alpha_i$ and $\partial^\alpha$ is the standard multi-index notation for partial derivatives. The Sobolev embedding theorems give $H^s(\mathbb{R}^n)\subset L^\infty(\mathbb{R}^n)$ whenever $n < s$.
What would be a standard example of $H^1(\mathbb{R}^2)\not\subset L^\infty(\mathbb{R}^2)$ (that is, an $L^2$ function in the plane having a weak first derivative in $L^2$ but the function itself unbounded in a set of positive measure).
I have come up with some involved example myself (defining it piecewise in an unelegant way), so I wonder if there is a simple enlightening example of this fact.

REPLY [1 votes]: Let
$$
j(x,y)=\left\{\begin{array}{lll}\exp\big((1-x^2-y^2)^{-1}\big) & \text{if} & x^2+y^2<1,\\ 0 & \text{if} & x^2+y^2\ge 1. \end{array}\right.
$$
Then $j\in C^\infty(\mathbb R^2)$.
A function in $H^1(\mathbb R^2)\setminus L^\infty(\mathbb R^2)$ is the following
$$
f(x,y)=j(x,y)\log\big(\log(1+x^2+y^2)\big).
$$


TITLE: Why is $|\alpha\rangle$ not eigenstate of $a^{\dagger}$ for $\alpha^*$
QUESTION [1 upvotes]: I know that even if we have:
$$a |\alpha \rangle = \alpha |\alpha\rangle$$
We don't have:
$$a^{\dagger} |\alpha \rangle = \alpha^* |\alpha\rangle$$
Actually as explained in the second answer here Eigenvalue for the creation operator for a coherent state $a^{\dagger}$ doesn't have eigenvalues.
But what disturbs me is the "matrix" vision.
If I choose a basis composed of $|\alpha\rangle$ and then some other states such that I have an orthogonal basis in the end, I can write down the (infinite) matrix $a$ in this basis.
The first diagonal coefficient will be $\alpha$. Then to go to $a^{\dagger}$ I conjugate transpose the matrix which will give me $\alpha^*$ on the first diagonal element.
Then through this vision it seems that $|\alpha\rangle$ is eigenstate of $a^{\dagger}$ with $\alpha^*$ as eigenvalue.
Where is the mistake in this reasoning ? Is it because we work in infinite dimension space so the matrix vision can induce errors ?

REPLY [4 votes]: Your mistake - while $\alpha$ will be on the diagonal (and $\alpha^*$ on the diagonal of the hermitian conjugate) it is not guaranteed that all the other terms in the row (and the column in the Hermitian conjugate representation of $a^{\dagger}$) will be zero. Therefore it will not be an eigenstate. It will have the property $\langle \alpha | a^{\dagger} |\alpha\rangle = \alpha^*$ as this is the value on the diagonal.
Note, for example, that the coherent states do not form a normal orthogonal basis, but an over-complete basis. Such that $\langle \alpha | \beta \rangle \propto \exp(|\alpha-\beta|^2)$ if I remember correctly.


text:
\section{Polynomial Addition is Associative}
Tags: Polynomial Theory

\begin{theorem}
[[Definition:Addition of Polynomial Forms|Addition of polynomials]] is an [[Definition:Associative|associative]] operation.
\end{theorem}

\begin{proof}
Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\left\{{X_j: j \in J}\right\}$ be a set of [[Definition:Indeterminate (Polynomial Theory)|indeterminates]].
Let $Z$ be the set of all [[Definition:Multiindex|multiindices]] indexed by $\left\{{X_j: j \in J}\right\}$.
Let:
:$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
:$\displaystyle g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
:$\displaystyle h = \sum_{k \mathop \in Z} c_k \mathbf X^k$
be arbitrary [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\left\{{X_j: j \in J}\right\}$ over $R$.
Then:
{{begin-eqn}}
{{eqn|l = \left({ f + g }\right) + h
     |r = \sum_{k \mathop \in Z} \left({ \left({ a_k + b_k }\right) + c_k }\right) \mathbf X^k
     |c = applying the definition of [[Definition:Addition of Polynomial Forms|polynomial addition]] twice
}}
{{eqn|r = \sum_{k \mathop \in Z} \left({ a_k + \left({ b_k + c_k }\right) }\right) \mathbf X^k
     |c = because $+$ in $R$ is [[Definition:Associative|associative]]
}}
{{eqn|r = f + \left({ g + h }\right)
     |c = applying the definition of [[Definition:Addition of Polynomial Forms|polynomial addition]] twice
}}
{{end-eqn}}
{{qed}}
[[Category:Polynomial Theory]]
khbinhw56oq156ek6gvmmv21h544jhg
\end{proof}

config: wiki


TITLE: How to add a potential to a wave equation?
QUESTION [2 upvotes]: Imagine you have the wave equation:
$$
\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}= \nabla^2
u$$
And you have your solution wave is "confined in a box" (or with the extremes of the ropes attached). Or maybe, your rope is confined to a paraboloid well. How would you add this information to the wave equation (in a similar way to the  potential on the Schrödinger Equation).

REPLY [2 votes]: You can try to add a source term. First, I'm going to define:
$$\partial^2 \equiv \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \nabla^2$$
and so the wave equation becomes:
$$\partial^2 u = 0$$
you can add a source term like the following:
$$\partial ^2 u = J(\vec{x})$$.
In fact, in electromagnetism, the source term is the current and charge densities, which affect the electromagnetic waves.
In quantum field theory, the wave equation (for spin 0 particle) is:
$$\partial^2\psi - m^2\psi = 0$$
and it's common to add source terms which corresponds to particles:
$$\partial^2 \psi - m^2 \psi = J(\vec{x})$$

I'm not sure if that's what you want though. If you are looking at a classical situation and let's say want to add gravity to a suspended waving rope, it's better to find the lagrangian.
Each piece of a rope has elastic energy:
$$\tau(\sqrt{dx^2 +du^2} - dx)=\tau (\sqrt{1+(\frac{du}{dx})^2}-1)dx$$
(where $\tau$ is elastic constant) and gravitational potential energy:
$$dx \sigma g u$$
(where $\sigma$ is mass per unit length, and assuming u is pointing upwards.)
the total energy is:
$$E=\int dx (\sigma g u + \tau(\sqrt{1+(\frac{du}{dx})^2)} -1))$$
and if you want the rope to be moving through time, you can add the kinetic energy term:
$$dx \frac{1}{2} \sigma (\frac{du}{dt})^2 $$
So the action becomes:
$$S = \int dt T-E = \int dtdx (\frac{1}{2} \sigma (\frac{du}{dt})^2
-\sigma g u - \tau(\sqrt{1+(\frac{du}{dx})^2)} -1))$$
You can use Euler Lagrange equations to find how the wavy rope evolves with time and you've successfully added potentials to a wave. I'm not sure which do you want so I provided both.

REPLY [0 votes]: Suppose that the potential energy per unit length associated with a displacement $u$ at a point $x$, is $V(u,x)$, then the wave equation will have an extra term of the form $-\frac{\partial V}{\partial u}$ on the RHS. To see why that is, it is easiest to look at the wave-action, taking the form
$$
S=\int dtdx \frac{1}{2c^2}(\partial_tu)^2-\frac{1}{2}( \nabla u)^2
$$
In general, the action has form $S=\int \mathrm{d}t\, T-V$, where $T$ is kinetic and $V$ is potential energy. Now we simply add our extra potential, and derive the equations of motion from the Euler Lagrange equations. This leads us to
$$
0=-\frac{1}{c^2}\partial_t^2u+\nabla^2u-\partial_uV.
$$


TITLE: Elliptic functions as inverses of Elliptic integrals
QUESTION [15 upvotes]: Let us begin with some (standard, I think) definitions.
Def: An elliptic function is a doubly periodic meromorphic function on $\mathbb{C}$.
Def: An elliptic integral is an integral of the form
$$f(x) = \int_{a}^x R\left(t,\sqrt{P(t)}\right)\ dt,$$
where $R$ is a rational fucntion of its arguments and where $P(t)$ is a third or fourth degree polynomial with simple roots.
I have often heard the claim that an elliptic function is (or can be) defined as the inverse of an elliptic integral. However, I have never seen a proof of this statement. As someone who is largely unfamiliar with the subject, most of the references I could dig up seem to refer to the special case of the Jacobi elliptic functions, which appear as inverse functions of the elliptic integrals of the first kind. Maybe the claim I'm referring to is simply talking about the special case of Jacobi elliptic functions, but I believe the statement holds in generality (I could be wrong).
So, can anyone provide a proof or reference (or counter-example) to something akin to the following?
Claim: The elliptic functions are precisely the inverses of the elliptic integrals, as I've defined them above. That is, every elliptic function arises as the inverse of some elliptic integral, and conversely every elliptic integral arises as the inverse of some elliptic function.

REPLY [4 votes]: The claim as stated is not true. (E.g., if $R$ has only even powers of the second variable, the resulting function $f$ is the integral of a rational function.) What is true is that every general elliptic integral of this form can be expressed as a linear combination of integrals of rational functions and the three Legendre canonical forms (elliptic integrals of the first, second, and third kind). This is a classical result, and there are several different algorithms to reduce a general elliptic integral to this form, some of them implemented in common computer algebra systems.
A modern (freely available) reference with a list of classical references is here: B.C. Carlson, Toward Symbolic Integration of Elliptic Integrals, Journal of Symbolic Computation, 28 (6), 1999, 739–753


TITLE: Mean of samples from Exponential distribution
QUESTION [1 upvotes]: I please ask someone to check if my calculations are right.
I have $X_1, ..., X_n$ from a $\mathcal{E}(\lambda): f(x, \lambda) = \lambda e^{-\lambda x}$,
$\mathcal n = 50, {\lambda} = 3$.
I have to find $\mathcal \alpha $ such that $P(\bar{X} \le 1) = \alpha$, where $\bar{X}$ is the sample mean; i did:
$$Y=\sum_{i = 1}^{n} X_i$$
$$Y \sim \Gamma (n, 1/\lambda)$$
$$\bar{X} = \frac{1}{n} Y \sim \Gamma(n, \frac{1/\lambda}{n})$$
I do not know how to proceed further. Any help would be appreciated.

REPLY [0 votes]: I think you forgot that they tell you that $n = 50$, $\lambda = 3$. Also, I will be a little more explicit.
$$P(\bar X\leq  1) = P(Y\leq n) = P(Y\leq 50)$$
where $Y$ is defined like you did. To me $Y\sim \text{Gamma}(n, \lambda)$, which implies a pdf
$$f_Y(y) = \frac1{\Gamma(n)}\lambda^ny^{n-1}e^{-\lambda y}.$$
The cdf of the gamma distribution is well-known,
$$P(Y\leq y) = 1-e^{-\lambda y}\sum_{k=0}^{n-1}\frac{(\lambda y)^k}{k!}.$$
So you can plug in the values and get
$$\alpha = 1-e^{-3\cdot 50}\sum_{k=0}^{49}\frac{(3\cdot 50)^k}{k!}.$$
W|A gave something like $0.99999999999999999999925878991$.
We can also try a normal approximation: $\mu = 50/3, \sigma^2 = 50/9$, and so
$$P(Y\leq 50)\approx P\left(Z\leq\frac{50-\mu}{\sigma}\right) = \Phi(14.14213562)$$
where $\Phi$ is the standard normal cdf.


TITLE: Please explain this step in proving the square root of 3 is irrational
QUESTION [4 upvotes]: Assume that
$$3 = \frac{p^2}{q^2}$$
So,
$$ 3 q^2 = p^2$$
So $p^2$ is divisible by $3$. How we can conclude this?

REPLY [1 votes]: $p^2$ is divisible by 3 because $p^2 = 3 q^2$, and $3 q^2$ is divisible by 3.


TITLE: Mirrors/optics and physical adaptation of the eye
QUESTION [0 upvotes]: Is it possible to use a clever combination of lenses and mirrors placed between your eye and a screen 1m away from you to make the eye react to the screen as if it were 20m away from you?
Note that I'm not asking whether it's possible to make the light rays travel a total distance of 20m, but whether the eyes will react as if the screen were 20m away.
(Not sure if this is the right StackExchange for this. If it isn't, please let me know where else I should post this, thanks!)

REPLY [0 votes]: You mean aside from using lots of mirrors so that the light actually travels 20 m before it hits your eye?
Yes you can do this, from a lens point of view you just want to change the divergence of the rays from a single point. If it is 1 m away they will be more divergent and require more power to focus by your eye. If it's further away they will be less divergent. From an image size point of view, think of looking through a set of binoculars the wrong way around.


text:
\begin{document}

\title{Degree One Milnor $K$-invariants of Groups of Multiplicative Type}

\author[A. Wertheim]{Alexander Wertheim}

\email{awertheim@math.ucla.edu}

\begin{abstract} Let $G$ be a commutative affine algebraic group over a field $F$, and let $H \colon \Fields_{F} \to \AbGrps$ be a functor. A (homomorphic) $H$-invariant of $G$ is a natural transformation $\Tors(-, G) \to H$, where $\Tors(-, G)$ is the functor $\Fields_{F} \to \AbGrps$ taking a field extension $L/F$ to the group of isomorphism classes of $G_{L}$-torsors over $\Spec(L)$. The goal of this paper is to compute the group $\Inv_{\hom}^{1}(G, H)$ of $H$-invariants of $G$ when $G$ is a group of multiplicative type, and $H$ is the functor taking a field extension $L/F$ to $L^{\times} \otimes_{\ZZ} \QQ/\ZZ$.
\end{abstract}

\maketitle

\section{Introduction}
Let $G$ be an affine algebraic group over a field $F$ (of arbitrary characteristic), and let $\Fields_{F}$ denote the category of field extensions of $F$. Let
\[H \colon \Fields_{F} \longrightarrow \AbGrps\]
be a functor. In \cite{GMS}, an {\bf $H$-invariant of $G$} is defined to be a natural transformation of set-valued functors
\[I \colon \Tors(-, G) \longrightarrow H\]
where $\Tors(-, G)$ is the functor from $\Fields_{F}$ to $\Set$ taking a field extension $L/F$ to $\Tors(L, G_{L})$, the set of isomorphism classes of $G_{L}$-torsors over $\Spec(L)$. Invariants were first introduced by Serre in \cite[Section 6]{SerreInvariants}, where he defined invariants in the case when $H$ is a (Galois) cohomological functor.

There is another type of invariant that one may consider, however. Namely, since any affine group scheme over $F$ may be viewed as a functor from $F$-algebras to groups, we define a {\bf type-zero} $H$-invariant of $G$ to be a natural transformation of set-valued functors $G \to H$, where by $G$ we mean the restriction of $G$ to $\Fields_{F}$. We denote the group of type-zero $H$-invariants of $G$ by $\Inv^{0}(G, H)$. To distinguish the invariants introduced in the previous paragraph from type-zero invariants, we will call them {\bf type-one} invariants, and we denote the group of type-one $H$-invariants of $G$ by $\Inv^{1}(G, H)$.

In this paper, we study type-one invariants when $G$ is an algebraic group of multiplicative type, i.e. when $G$ is a twisted form of a diagonalizable group. We note that every torus is a group of multiplicative type; in general, groups of multiplicative type need not be smooth or connected. We consider a slightly more restrictive class of invariants than those introduced above, however. If $G$ is \emph{commutative}, then for any affine $F$-scheme $X$, the pointed set $\Tors(X, G)$ can be given the structure of an abelian group. Since groups of multiplicative type are commutative, we may view the functor $\Tors(-, G)$ as a functor from $\Fields_{F}$ to $\AbGrps$. Accordingly, we will focus our attention on invariants which are morphisms of \emph{group-valued} functors. We will call such invariants {\bf homomorphic}, and we denote the subgroup of homomorphic type-one $H$-invariants of $G$ by $\Inv^{1}_{\hom}(G, H)$. Likewise, one may consider homomorphic type-zero invariants, which we will similarly denote $\Inv^{0}_{\hom}(G, H)$.

The goal of this paper is to determine $\Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)$, the group of (type-one) {\bf degree one Milnor $K$-invariants of $G$}, where $K^{M}_{i}$ denotes the functor sending a field extension $L/F$ to the $i^{\thh}$ Milnor $K$-group of $L$ (see \cite{Milnor}); we recall that $K^{M}_{0}(L) = \ZZ, K^{M}_{1}(L) = L^{\times}$. For any $n \in \NN$, let $K^{M}_{1}/n$ denote the functor $K^{M}_{1} \otimes_{\ZZ} \ZZ/n\ZZ$. The embedding $\ZZ/n\ZZ \hookrightarrow \QQ/\ZZ$ induces a morphism of functors $\iota_{n} \colon K^{M}_{1}/n \to K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ$; likewise, if $n$ and $m$ are positive integers such that $n$ divides $m$, then the embedding $\ZZ/n\ZZ \to \ZZ/m\ZZ$ sending $[1]_{n}$ to $[m/n]_{m}$ induces a morphism of functors $\beta_{n, m} \colon K^{M}_{1}/n \to K^{M}_{1}/m$. One may check that the collection of functors $\{K^{M}_{n} \otimes_{\ZZ} \QQ/\ZZ\}_{n \in \NN}$ defines the data of a cycle module in the sense of Rost (see \cite{Rost}), as does $\{K^{M}_{n} \otimes_{\ZZ} \ZZ/m\ZZ\}_{n \in \NN}$ for any $m \in \NN$. The functors $K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ$ and $K^{M}_{1}/m$ respectively form the first graded components of these cycle modules.

As we will explain in Section \ref{BundlesAreTorsors}, a classic Kummer theory argument shows that there is an isomorphism $\Sigma_{n} \colon K^{M}_{1}/n \to \Tors(-, \bmu_{n, F})$ of group valued functors. On the other hand, any $\chi \in \Hom(G, \bmu_{n, F}) = G^{\ast}[n]$ gives rise to a morphism of group-valued functors $\Tors_{\ast}(\chi) \colon \Tors(-, G) \to \Tors(-, \bmu_{n, F})$. Thus, we may associate to any element $\chi \in G^{\ast}[n]$ a homomorphic invariant $I_{\chi} \in \Inv^{1}_{\hom}(G, K^{M}_{1}/n)$ which is the composition of $\Tors_{\ast}(\chi)$ with $\Sigma_{n}^{-1}$. This leads us to our first main theorem.

\begin{theorem}[\ref{TypeOneGMTModN}] \label{BigA}
	The map $\Phi(G, n) \colon G^{\ast}[n] \to \Inv^{1}_{\hom}(G, K^{M}_{1}/n)$ sending $\chi$ to $I_{\chi}$ is a group isomorphism.
\end{theorem}

The composition of any such $I_{\chi}$ with $\iota_{n}$ produces an element of $\Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \mathbb{Q}/\mathbb{Z})$, and so defines a group homomorphism $\tilde{\Phi}(G, n) \colon G^{\ast}[n] \to \Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \mathbb{Q}/\mathbb{Z})$ for each $n \in \NN$. Passing to the colimit as $n$ varies, we obtain a universally induced group morphism $\Phi(G) \colon G^{\ast}_{\tors} \to \Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \mathbb{Q}/\mathbb{Z})$.

\begin{theorem}[\ref{TypeOneGMTQZ}] \label{BigB}
The map $\Phi(G) \colon G^{\ast}_{\tors} \to \Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \mathbb{Q}/\mathbb{Z})$ is a group isomorphism.
\end{theorem}

Our proofs of Theorems \ref{TypeOneGMTModN} and \ref{TypeOneGMTQZ} depend critically on the determination of homomorphic type-zero invariants for tori with values in $K^{M}_{1}/n$ for each $n \in \NN$. The following result was proven by Merkurjev (cf. \cite[Corollary 3.7]{MerkurjevIAG}) in the case when the characteristic of $F$ does not divide $n$; we give a proof in this paper which holds independent of the characteristic of $F$.

\begin{theorem}[\ref{TypeZeroInvariantsGaloisModN}] \label{BigC}
	If $T$ is an algebraic torus, then $\Inv^{0}_{\hom}(T, K^{M}_{1}/n) \cong H^{0}(F, T_{\sep}^{\ast}/(T_{\sep}^{\ast})^{n})$.
\end{theorem}

The results we have obtained above follow a rich history of work on cohomological invariants: here are a few related recent examples. In \cite{Totaro}, Totaro computed all mod $p$ cohomological invariants for many important affine group schemes in characteristic $p$; in particular, under the assumption that $\chara(F) = p > 0$, Totaro independently computed $\Inv^{1}(G, K^{M}_{1}/p)$ for \emph{any} affine group scheme (\cite[Theorem 12.2]{Totaro}). The computation of invariants for smooth linear algebraic groups with values in $H^{2}(-, \QQ/\ZZ(1))$ was carried out by Alexandre Lourdeaux in \cite{Lourdeaux}.

\subsection{Acknowledgments} I would like to express my gratitude to my advisor, Alexander Merkurjev, for his advice, encouragement, and many helpful meetings. I am also grateful to Bar Roytman, David Hemminger, Will Baker, and Burt Totaro for helpful conversations.

\subsection{Notation and Conventions} Throughout, $F$ denotes a fixed base field of arbitrary characteristic, and $F_{\sep}$ denotes a fixed separable closure. We put $\Gamma = \Gal(F_{\sep}/F)$. If $G$ is a group scheme over $F$, we write $G_{\sep}$ to denote the base change of $G$ to $F_{\sep}$, and $G^{\ast}$ to denote the character group of $G$. For an abelian group $A$ and a positive integer $n$, we write $A[n]$ to denote the subgroup of $n$-torsion elements of $A$. All group schemes are affine unless otherwise indicated. For any group scheme $G$ over $F$ and any $F$-algebra $R$, we write $\varepsilon_{R}$ to denote the identity element of $G(R)$. If $\varphi \colon Q \to Q'$ is a morphism of commutative $F$-group schemes, we write $Q^{\varphi}$ to denote the image of the embedding $Q \to Q \times Q'$ induced by $\Id_{Q}$ and the composition of $\varphi$ with the inversion map $Q' \to Q'$. For an $F$-scheme $X$, we write $\Tors_{\ast}(\varphi)(X)$ to denote the morphism $\Tors(X, Q) \to \Tors(X, Q')$ induced by $\varphi$. Likewise, if $f \colon Y \to X$ is a morphism of $F$ schemes, we write $\Tors^{\ast}(f)(Q)$ to denote the pullback morphism $\Tors(X, Q) \to \Tors(Y, Q)$.

\section{An Outline of the Argument} \label{Outline}

In this section, we give a structural overview of our argument.

\subsection{Resolution by Tori} \label{Versal}

Recall that a group scheme $G$ over $F$ is said to be {\bf diagonalizable} if the natural embedding $G^{\ast} \to F[G]^{\times}$ induces an isomorphism of Hopf $F$-algebras $F\langle G^{\ast} \rangle \to F[G]$, where $F\langle G^{\ast} \rangle$ denotes the group algebra of $G^{\ast}$ over $F$. As noted in the introduction, a group scheme $G$ over $F$ is a {\bf group of multiplicative type} if $G_{\sep}$ is diagonalizable over $F_{\sep}$. The functors
\[G \longmapsto G_{\sep}^{\ast}, \qquad M \longmapsto (F_{\sep}\langle M \rangle)^{\Gamma}\]
define a short exact sequence-preserving equivalence between the category of (algebraic) groups of multiplicative type over $F$ and the category of (finitely generated) $\Gamma$-modules (\cite[Theorem 12.23]{MilneAG}). Under this equivalence, the full subcategory of diagonlizable $F$-group schemes is equivalent to the subcategory of $\Gamma$-modules with trivial $\Gamma$-action.

When $G$ is an algebraic group of multiplicative type, $G$ may be embedded in a quasisplit torus $P$ such that every $G_{L}$ torsor over a field $L/F$ is the pullback of the $G$-torsor $P \to P/G$ along an $L$-point of $P/G$. Indeed, since $G_{\sep}^{\ast}$ is finitely generated, it admits a surjective morphism of $\Gamma$-modules $W \to G_{\sep}^{\ast}$ from a permutation $\Gamma$-module $W$. If $S$ denotes the kernel of this map, then let $P, T$ be the groups of multiplicative type respectively associated to $W, S$. Note that $P$ is a quasisplit torus, $T$ is a torus, and the exact sequence
\[1 \to S \to W \to G_{\sep}^{\ast} \to 1\]
of $\Gamma$-modules yields an exact sequence
\begin{equation} \label{ResolutionByTori}
1 \to G \xrightarrow{~f~} P \xrightarrow{~g~} T \to 1
\end{equation}

of $F$-group schemes. We will call such an exact sequence \ref{ResolutionByTori} a {\bf resolution of $G$ by tori}.

For every field extension $L/F$, the exact sequence on points $1 \to G(L) \to P(L) \to T(L)$ may be continued as follows. Let $\rho(L) \colon T(L) \to \Tors(L, G_{L})$ be the group homomorphism sending a point $\alpha \in T(L)$ to the pullback of the $G$-torsor $P \to T$ along $\alpha$. One may check that the sequence
\begin{equation} \label{ExactSeqTors}
1 \to G(L) \xrightarrow{~f(L)~} P(L) \xrightarrow{~g(L)~} T(L) \xrightarrow{~\rho(L)~} \Tors(L, G_{L}) \xrightarrow{~\Tors_{\ast}(f_{L})~} \Tors(L, P_{L})
\end{equation}
is exact; we note that this does not depend on the fact that $G, P, T$ are of multiplicative type, and can be proven for any exact sequence of commutative group schemes. Since $P_{L}$ is a quasisplit torus, every $P_{L}$-torsor over $\Spec(L)$ is trivial. Therefore, the map $\rho(L) \colon T(L) \to \Tors(L, G_{L})$ is surjective.

The surjectivity of $\rho(L)$ allows us to relate type-one invariants for $G$ to type-zero invariants for tori, which are well understood for certain functors $H$. As $L$ varies over all field extensions of $F$, the morphisms $\rho(L)$ define a morphism of functors $\rho \colon T \to \Tors(-, G)$, which gives rise to a map $\Inv(\rho, H) \colon \Inv^{1}_{\hom}(G, H) \to \Inv^{0}_{\hom}(T, H)$ given by composition with $\rho$. Likewise, the group homomorphism $g \colon P \to T$ is a natural transformation of group-valued functors, and so induces a map $\Inv(g, H) \colon \Inv^{0}_{\hom}(T, H) \to \Inv^{0}_{\hom}(P, H)$ given by composition with $g$. The exactness of \ref{ExactSeqTors} shows that the resulting sequence
\begin{equation} \label{ExactSeqInvariants}
1 \to \Inv^{1}_{\hom}(G, H) \xrightarrow{~\Inv(\rho, H)~} \Inv^{0}_{\hom}(T, H) \xrightarrow{~\Inv(g, H)~} \Inv^{0}_{\hom}(P, H)
\end{equation}
is exact. To describe $\Inv^{1}_{\hom}(G, H)$, it therefore suffices to determine the image of $\Inv(\rho, H)$ in $\Inv^{0}_{\hom}(T, H)$.

\subsection{The Argument} \label{Pushforward}

Fix a positive integer $n$, let $G, P, T$ be as in the exact sequence \ref{ResolutionByTori}, and let $H = K^{M}_{1}/n$. For any group scheme $Q$ over $F$, we say that a class $V \in \Tors(Q, G)$ is {\bf normalized} if the pullback of $V$ along $\varepsilon_{F}\in Q(F)$ represents the trivial class in $\Tors(F, G)$. Let $\Tors_{\nm}(Q, G)$ denote the subgroup of normalized $G$-torsors over $Q$.

Consider the map $\upsilon_{n}(G) \colon G^{\ast}[n] \to \Tors_{\nm}(T, \bmu_{n, F})$ which sends a character $\chi \in G^{\ast}[n]$ to $(\Tors_{\ast}(\chi)(T))(P \to T)$. We note that $\upsilon_{n}(G)$ is a group homomorphism. Indeed, for any $F$-scheme $X$, the map $\Tors(X, G) \times \Tors(X, G) \to \Tors(X, G \times G)$ sending a pair of representatives $E_{1} \to X, E_{2} \to X$ to the universal map $E_{1} \times E_{2} \to X$ is a group isomorphism. If $\Delta_{G} \colon G \to G \times G$ denotes the diagonal map, and $m_{G} \colon G \times G \to G$ denotes the group multiplication, then up to the preceding identification, $\Tors_{\ast}(m_{G})(X)$ is the group operation, and $\Tors_{\ast}(\Delta_{G})(X)$ is the diagonal embedding. Hence, if $\chi, \chi' \in G^{\ast}[n]$, then we have $\Tors_{\ast}(\chi\chi')(X) = \Tors_{\ast}(\chi)(X) + \Tors_{\ast}(\chi')(X)$, since $\chi\chi'$ factors as $m_{\bmu_{n, F}} \circ (\chi \times \chi') \circ \Delta_{G}$. This argument also explains why $\Phi(G, n)$ is a group homomorphism.

Suppose we were armed with the following facts:

\begin{enumerate}[(1)]
\item The sequence
\begin{equation} \label{CharactersToTorsors}
1 \to G^{\ast}[n] \xrightarrow{~\upsilon_{n}(G)~} \Tors_{\nm}(T, \bmu_{n, F}) \xrightarrow{~\Tors^{\ast}(g)(\bmu_{n, F})~} \Tors_{\nm}(P, \bmu_{n, F})
\end{equation}
is exact.
\item For any smooth, connected, reductive group $R$ over $F$, there is a group isomorphism $\tilde{\Lambda}_{n}(R) \colon \Tors_{\nm}(R, \bmu_{n, F}) \to \Inv^{0}_{\hom}(R, K^{M}_{1}/n)$.
\item The diagram
\begin{center}
	\begin{tikzcd}
	{G^{\ast}[n]} \arrow[rr, "\upsilon_{n}(G)"] \arrow[dd, "{\Phi(G, n)}"] &  & {\Tors_{\nm}(T, \bmu_{n, F})} \arrow[rr, "{\Tors^{\ast}(g)(\bmu_{n, F})}"] \arrow[dd, "\tilde{\Lambda}_{n}(T)"] &  & {\Tors_{\nm}(P, \bmu_{n, F})} \arrow[dd, "\tilde{\Lambda}_{n}(P)"] \\
	&  &                                                                        &  &                                                                                &  &                                                                    \\
	{\Inv^{1}_{\hom}(G, K^{M}_{1}/n)} \arrow[rr, "{\Inv(\rho, K^{M}_{1}/n)}"']            &  & {\Inv^{0}_{\hom}(T, K^{M}_{1}/n)} \arrow[rr, "{\Inv(g, K^{M}_{1}/n)} "']                                                        &  & {\Inv^{0}_{\hom}(P, K^{M}_{1}/n)}
	\end{tikzcd}
\end{center}
commutes.
\end{enumerate}

If these three statements hold, then an easy diagram chase using the exactness of \ref{CharactersToTorsors} and \ref{ExactSeqInvariants} shows that $\Phi(G, n)$ is an isomorphism. The remainder of this paper is dedicated to proving these three facts, and carefully explaining why the induced map $\Phi(G)$ is an isomorphism. The remaining sections are organized as follows.

Section \ref{MuNTorsors} gives a thorough treatment of $\bmu_{n, F}$-torsors, laying the ground work for facts $(1)$ and $(2)$. We will provide a Galois theoretic-interpretation of the group $\Tors_{\nm}(T, \bmu_{n, F})$ which will allow us to interpret sequence \ref{CharactersToTorsors} as an exact sequence arising in Galois cohomology in section \ref{TypeOneInvariants}. We will also prove a pullback formula for $\bmu_{n, F}$-torsors over a smooth, connected, reductive group $R$ which shows that normalized $\bmu_{n, F}$-torsors over $R$ give rise to homomorphic type-zero invariants of $R$.

Section \ref{TypeZeroInvariants} is devoted to constructing the map $\tilde{\Lambda}_{n}(R)$ for any smooth, connected, reductive group $R$, and proving it is a group isomorphism. We also give a description of type-zero invariants for $R$ with values in $K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ$.

The final section (\ref{TypeOneInvariants}) will prove facts (1) and (3), yielding Theorem \ref{BigA}. As noted, we will then deduce Theorem \ref{BigB} from Theorem \ref{BigA} via a detailed examination of $\Phi(G)$.

\section{$\mu_{n, F}$-Torsors} \label{MuNTorsors}

Throughout this section, let $n$ denote a fixed positive integer. As indicated in the previous section, an essential ingredient in the proof of Theorem \ref{TypeOneGMTModN} is a robust understanding $\bmu_{n, F}$-torsors over an $F$-scheme $X$. In this section, we recall several well-known characterizations of $\bmu_{n, F}$-torsors. Our main results are Theorems \ref{TorsKerDivMap}, \ref{PointsMultTorsors} and \ref{GalFixedTorsors}. Theorem \ref{TorsKerDivMap} explains that when $G$ is a smooth, connnected group, $\Tors(G, \bmu_{n, F})$ may be identified with the kernel of the divisor map $\partial_{n}(G) \colon F(G)^{\times}/(F(G)^{\times})^{n} \to \Div(G)/n\Div(G)$. Under the further assumption that $G$ is reductive, Theorem \ref{PointsMultTorsors} proves a formula relating the pullbacks of a class in $\Tors(G, \bmu_{n, F})$ along points $\alpha, \beta \in G(M)$ to its pullback along the product $\alpha\beta \in G(M)$, where $M$ is a field extension of $F$. Theorem \ref{GalFixedTorsors} computes the Galois fixed points of $\Tors_{\nm}(G_{\sep}, \bmu_{n, F_{\sep}})$ when $G$ is geometrically integral, $G_{\sep}^{\ast}$ is torsion-free, and $G_{\sep}$ has trivial divisor class group.

\subsection{The Group $\Psi(A, n)$} \label{BundlesAreTorsors}

\begin{defn}
	For any commutative ring $A$, let $\Psi(A, n)$ denote the set of equivalence classes of pairs $(\mathcal{L}, \varphi)$, where $\mathcal{L} \in  \Pic(A)[n]$, $\varphi$ is an $A$-module isomorphism $\mathcal{L}^{\otimes n} \to A$, and two pairs $(\mathcal{L}, \varphi), (\mathcal{L}', \varphi')$ are equivalent if and only if there is an isomorphism of $A$-modules $\rho \colon \mathcal{L} \to \mathcal{L}'$ such that $\varphi' \circ \rho^{\otimes n} = \varphi$.
\end{defn}

We record the following observations about $\Psi(A, n)$, which are straightforward to check:

\begin{enumerate}[(1)]
	\item The tensor product induces a group operation on $\Psi(A, n)$: one defines the product of classes $[(\mathcal{L}, \varphi)], [(\mathcal{L}', \varphi')] \in \Psi(A, n)$ to be $[(\mathcal{L} \otimes_{A} \mathcal{L}', \varphi \otimes_{A} \varphi')]$, where $\varphi \otimes_{A} \varphi'$ really refers to the composition
	\[(\mathcal{L} \otimes_{A} \mathcal{L}')^{\otimes n} \xrightarrow{~\sim~} (\mathcal{L}^{\otimes n}) \otimes_{A} (\mathcal{L}')^{\otimes n} \xrightarrow{~\varphi \otimes_{A} \varphi'~} A \otimes_{A} A \xrightarrow{~\sim~} A.\]
	The identity class is represented by the pair $(A, \Id_{A})$, and the inverse of a class $[(\mathcal{L}, \varphi)]$ is given by $[(\mathcal{L}^{\ast}, (\varphi^{-1})^{\ast})]$, where $\mathcal{L}^{\ast}$ is the dual bundle to $\mathcal{L}$, and $(\varphi^{-1})^{\ast}$ is the composition
	\[(\mathcal{L}^{\ast})^{\otimes n} \xrightarrow{~\sim~} (\mathcal{L}^{\otimes n})^{\ast} \xrightarrow{~(\varphi^{-1})^{\ast}~} A^{\ast} \xrightarrow{~\sim~} A. \]
	\item For any ring morphism $f \colon A \to B$, extension of scalars induces a group morphism $\Psi(-, n)(f) \colon \Psi(A, n) \to \Psi(B, n)$ sending $[(\mathcal{L}, \varphi)]$ to $[(\mathcal{L} \otimes_{A} B, \varphi \otimes_{A} \Id_{B})]$, where $\varphi \otimes_{A} \Id_{B}$ really denotes the composition
	\[(\mathcal{L} \otimes_{A} B)^{\otimes n} \xrightarrow{~\sim~} \mathcal{L}^{\otimes n} \otimes_{A} B \xrightarrow{\varphi \otimes_{A} \Id_{B}} A \otimes_{A} B \xrightarrow{~\sim~} B.\]
	In this way, the association $A \mapsto \Psi(A, n)$ defines a functor $\Psi(-, n)$ from $\Rings$ to $\AbGrps$.
	\item For any positive integer $m$ with $n$ dividing $m$, there is a morphism of functors $\omega_{n, m} \colon \Psi(-, n) \to \Psi(-, m)$ defined for a commutative ring $A$ by $\omega_{n, m}(A)[(\mathcal{L}, \varphi)] = [(\mathcal{L}, \varphi^{\otimes m/n})]$, where by $\varphi^{m/n}$ we mean the composition of isomorphisms
	\[\mathcal{L}^{\otimes m} \xrightarrow{~\sim~} (\mathcal{L}^{\otimes n})^{\otimes m/n}\xrightarrow{~\varphi^{\otimes m/n}~} A^{\otimes m/n} \xrightarrow{~\sim~} A.\]
\end{enumerate}

There is a convenient way to produce elements of $\Psi(A, n)$ which can be described as follows. Fix an element $y \in A^{\times}$, and consider the $A$-algebra $R_{y} := A[X]/\langle X^{n} - y \rangle$; we denote the residue class of $X$ in $R_{y}$ by $y^{1/n}$. If $\mathcal{L}_{y}$ denotes the free $A$-submodule of $R_{y}$ generated by $y^{1/n}$, one immediately sees that the ``multiplication'' map $\varphi_{y} \colon \mathcal{L}_{y}^{\otimes n} \to A$ sending $x_{1}y^{1/n} \otimes \cdots \otimes x_{n}y^{1/n}$ to $yx_{1}x_{2} \cdots x_{n}$ is an isomorphism of $A$-modules, and the pair $(\mathcal{L}_{y}, \varphi_{y})$ represents a class in $\Psi(A, n)$.

One readily checks that the map $A^{\times} \to \Psi(A, n)$ sending $y \in A^{\times}$ to $[(\mathcal{L}_{y}, \varphi_{y})]$ is a group homomorphism whose kernel is exactly $(A^{\times})^{n}$, and so we obtain a well-defined injective group morphism $\Delta_{n}(A) \colon A^{\times}/(A^{\times})^{n} \to \Psi(A, n)$. Moreover, this collection of maps is functorial in $A$: in other words, if $\mathcal{K}^{n}$ denotes the functor from $\Rings$ to $\AbGrps$ sending a commutative ring $A$ to $A^{\times}/(A^{\times})^{n}$, the collection of maps $\Delta_{n}(A)$ as $A$ varies defines a natural transformation $\Delta_{n} \colon \mathcal{K}^{n} \to \Psi(-, n)$.

On the other hand, for any commutative ring $A$, there is a well-defined surjective group homomorphism $\Theta_{n}(A) \colon \Psi(A, n) \to \Pic(A)[n]$ which sends a class $[(\mathcal{L}, \varphi)] \in \Psi(A, n)$ to $[\mathcal{L}]$, and the collection of such $\Theta_{n}(A)$ as $A$ varies likewise determines a natural transformation $\Theta_{n} \colon \Psi(-, n) \to \Pic(-)[n]$. The relationship between $\Delta_{n}$ and $\Theta_{n}$ is explained by the following proposition.

\begin{prop} \label{MuTorsExactSeq}
	For any commutative ring $A$, the sequence
	\[1 \to A^{\times}/(A^{\times})^{n} \xrightarrow{~\Delta_{n}(A)~} \Psi(A, n) \xrightarrow{~\Theta_{n}(A)~} \Pic(A)[n] \to 0\]
	is exact.
\end{prop}

\begin{proof}
	The inclusion $\Imm(\Delta_{n}(A)) \subset \ker(\Theta_{n}(A))$ is immediate, since $\mathcal{L}_{y}$ is a free $A$-module for any $y \in A^{\times}$ by construction. Suppose that $(\mathcal{L}, \varphi) \in \ker(\Theta_{n}(A))$, i.e. that $\mathcal{L}$ is free. Let $\psi \colon \mathcal{L} \to A$ be an isomorphism of $A$-modules. Consider the composition of isomorphisms
	\[A \xrightarrow{~\varphi^{-1}~} \mathcal{L}^{\otimes n} \xrightarrow{~\psi^{\otimes n}} A^{\otimes n} \xrightarrow{~\sim~} A. \]
	Every $A$-module isomorphism $A \to A$ is given by multiplication by some invertible element of $A$, so the composition above is multiplication by $x$ for some $x \in A^{\times}$. Put $y = x^{-1}$, and let $\alpha \colon A \to \mathcal{L}_{y}$ be the isomorphism sending $a$ to $ay^{1/n}$. Then one easily checks that $\alpha \circ \psi$ is an isomorphism between $(\mathcal{L}, \varphi)$ and $(\mathcal{L}_{y}, \varphi_{y})$.
\end{proof}

\begin{cor}\label{KMuNTorsors}
	If $\Pic(A)[n] = 0$, then $\Delta_{n}(A)$ is an isomorphism. $\qed$
\end{cor}

Suppose now that $A$ is an $F$-algebra. To any element $[(\mathcal{L}, \varphi)]$ of $\Psi(A, n)$, one may associate a $\ZZ/n\ZZ$-graded $A$-algebra $\Tw(\mathcal{L}, \varphi)$. As an $A$-module, we set
\[\Tw(\mathcal{L}, \varphi) := A \oplus \mathcal{L} \oplus \mathcal{L}^{\otimes 2} \oplus \cdots \oplus \mathcal{L}^{\otimes n-1}.\]
The multiplicative structure on $\Tw(\mathcal{L}, \varphi)$ is induced by the isomorphisms $\mathcal{L}^{\otimes i} \otimes_{A} \mathcal{L}^{\otimes j} \to \mathcal{L}^{\otimes i+j}$ for $i+j < n$, and $\mathcal{L}^{\otimes i} \otimes_{A} \mathcal{L}^{\otimes j} \to \mathcal{L}^{\otimes n} \otimes_{A} \mathcal{L}^{\otimes (n-(i+j))} \xrightarrow{\varphi \otimes \Id} A \otimes_{A} \mathcal{L}^{\otimes (n-(i+j))} \to \mathcal{L}^{n-(i+j)}$ for $i+j \geqslant n$. Note that the inclusion morphism $A \to \Tw(\mathcal{L}, \varphi)$ is faithfully flat, because $\Tw(\mathcal{L}, \varphi)$ is finitely generated and projective as an $A$-module. In fact, the dual morphism $\Spec(\Tw(\mathcal{L}, \varphi)) \to \Spec(A)$ is a $\bmu_{n, F}$-torsor over $\Spec(A)$, and we can say yet more, as the next theorem explains. Let $\lambda_{n}(A) \colon \Psi(A, n) \to \Tors(\Spec(A), \bmu_{n, F})$ be the set map sending $[(\mathcal{L}, \varphi)]$ to the $\bmu_{n, F}$-torsor class represented by the map $\Spec(\Tw(\mathcal{L}, \varphi)) \to \Spec(A)$.

\begin{thm} \label{PsiTorsors}
	The map $\lambda_{n}(A)$ is a well-defined {\bf group} isomorphism. Moreover, as $A$ varies over all $F$-algebras, the collection of maps $\lambda_{n}(A)$ defines a natural isomorphism $\lambda_{n} \colon \Psi(-, n) \to \Tors(-, \bmu_{n, F})$.
\end{thm}

\begin{proof}
	See \cite[\href{https://stacks.math.columbia.edu/tag/03PK}{Tag 03PK}]{Stacks}. Alternatively, see \cite[page 125]{EC}.
\end{proof}

We note that for any $y \in A^{\times}$, the universal map $A[X]/\langle X^{n} -y \rangle \to \Tw(\mathcal{L}_{y}, \varphi_{y})$ sending $\overline{X}$ to $y^{1/n} \in \mathcal{L}_{y}$ is an isomorphism of $(\ZZ/n\ZZ)$-graded $A$-algebras. Hence, the composition $\lambda_{n}(A) \circ \Delta_{n}(A)$ takes $y \in A^{\times}$ to the class of the $\bmu_{n, F}$-torsor $\Spec(A[X]/\langle X^{n}- y \rangle) \to \Spec(A)$. We put $\mathbf{ \Sigma_{n} := \lambda_{n} \circ \Delta_{n}}$.

\subsection{Divisors}

When $A$ is a domain, there is another description of $\Psi(A, n)$ in terms of divisors. Let $K$ denote the field of fractions of $A$, and let $\Cart(A)$ denote the group of invertible fractional ideals of $A$. Likewise, if $A$ is a Krull domain, let $\Div(A)$ be the free abelian group generated by the codimension $1$ points of $\Spec(A)$. We write $\dv(A) \colon \Cart(A) \to \Div(A)$ to denote the usual valuation homomorphism which sends a fractional ideal $I$ to the formal sum of its valuations at each height one prime of $A$. We let $\partial(A) \colon K^{\times} \to \Div(A)$ denote the group morphism sending $x \in K^{\times}$ to $\dv(xA)$.

Consider the set $C(A, n)$ consisting of pairs $(I, f)$ where $I \in \Cart(A)$, and $f \in K^{\times}$ such that $I^{n} = fA$. The binary operation on $C(A, n)$ defined by $(I, f) \cdot (I', f') = (II', ff')$ gives $C(A, n)$ the structure of a group with identity element $(A, 1)$. There is a group homomorphism $K^{\times} \to C(A, n)$ sending $x \in K^{\times}$ to $(xA, x^{n})$, and we set $\Cart(A, n)$ to be the cokernel of this morphism.

If we further assume that $A$ is a Krull domain, then there is an analagous construction $\Div(A, n)$. If $D(A, n)$ denotes the set of pairs $(D, g)$ where $D \in \Div(A)$ and $g \in K^{\times}$ such that $\partial(A)(g) = nD$, then $\Div(A, n)$ is defined to be the cokernel of group homomorphism $K^{\times} \to D(A, n)$ which sends $x \in K^{\times}$ to the pair $(\partial(A)(x), x^{n})$. One may check that the map $\dv(A) \colon \Cart(A) \to \Div(A)$ described above descends to a group morphism $\dv_{n}(A) \colon \Cart(A, n) \to \Div(A, n)$, and this map is an isomorphism if $A$ is regular.

For any element $I \in \Cart(A)$, the multiplication map $I^{\otimes n} \to I^{n}$ is an isomorphism of $A$-modules, since $I$ is projective of rank $1$. Given a pair $(I, f)$ in $C(A, n)$, we may produce a pair $(I, m_{f})$ which represents a class in $\Psi(A, n)$, where $m_{f}$ is the composition of $A$-isomorphisms $I^{\otimes n} \xrightarrow{~\sim~} I^{n} \xrightarrow{~\cdot f^{-1}~} A$. One may check that the resulting set map $\Omega(A) \colon C(A, n) \to \Psi(A, n)$ sending a pair $(I, f)$ to $[(I, m_{f})]$ is a group homomorphism.

\begin{prop}
	The morphism $\Omega(A) \colon C(A, n) \to \Psi(A, n)$ is surjective, and the kernel is precisely the image of the group morphism $K^{\times} \to C(A, n)$ sending $x \in K^{\times}$ to $(xA, x^{n})$. Therefore, $\Omega(A)$ descends to a well-defined group isomorphism $\Omega_{n}(A) \colon \Cart(A, n) \to \Psi(A, n)$.
\end{prop}

\begin{proof}
	Let $m \colon A^{\otimes n} \to A$ denote the multiplication map. If $(I, f) \in C(A, n)$ belongs to $\ker(\Omega(A))$, then there is an isomorphism $\rho \colon A \to I$ such that $m_{f} \circ \rho^{\otimes n} = m$. Then $I$ is principal, generated by $\rho(1) =: x \in K^{\times}$, and
	\[1 = m(1 \otimes \cdots \otimes 1) = m_{f}(x \otimes \cdots \otimes x) = x^{n}/f,\]
	so $x^{n} = f$, and $(I, f) = (xA, x^{n})$. On the other hand, for any $x \in K^{\times}$, the class $[(xA, m_{x^{n}})]$ in $\Psi(A, n)$ is trivial, via the isomorphism $A \to xA$ sending $1$ to $x$. \\
	Now, let the pair $(\mathcal{L}, \varphi)$ represent a class in $\Psi(A, n)$. Let $I \subset K$ denote the image of $\mathcal{L}$ under the composition of the $A$-embedding $\mathcal{L} \to \mathcal{L} \otimes_{A} K$ with a fixed $K$-module isomorphism $\mathcal{L} \otimes_{A} K \to K$. After clearing denominators, we may assume that $I \subset A \subset K$, so that $I$ is an ideal of $A$. Since $\mathcal{L}$ is projective of rank $1$, $I$ is an invertible ideal of $A$. \\
	Let $\alpha \colon \mathcal{L} \to I$ denote our $A$-module isomorphism of $\mathcal{L}$ onto $I$. If $f$ denotes the image of $1$ under the sequence of isomorphisms $A \xrightarrow{~\varphi^{-1}~} \mathcal{L}^{\otimes n} \xrightarrow{~\alpha^{\otimes n}~} I^{\otimes n} \xrightarrow{~\sim~} I^{n}$, then one sees that $I^{n} = fA$, and $\alpha$ is an isomorphism between $(\mathcal{L}, \varphi)$ and $(I, m_{f})$.
\end{proof}

The above proof shows that every element of $\Psi(A, n)$ admits a representative of the form $(I, m_{f})$ where $I \subset A$ is an invertible fractional ideal of $A$ satisfying $I^{n} = fA$ for some nonzero $f \in A$. We will call such a representative an {\bf ideal representative} of a class in $\Psi(A, n)$.

\begin{cor}\label{NonzeroIdealTors}
	Let $A$ be a normal domain with field of fractions $K$, and let $X$ be a class in $\Psi(A, n)$. Let $M$ be a domain, and let $\alpha_{1}, \ldots, \alpha_{n} \colon A \to M$ be ring morphisms. Then one can choose an ideal representative $(\tilde{I}, m_{\tilde{f}})$ for $X$ such that $\alpha_{i}(\tilde{f}) \in M\setminus \{0\}$ for each $1 \leqslant i \leqslant n$.
\end{cor}

\begin{proof}
	For each $1 \leqslant i \leqslant n$, put $\mfp_{i} = \ker(\alpha_{i}) \in \Spec(A)$. Let $S$ be the multiplicative subset of $A$ defined by $S = A\setminus\bigcup_{i=1}^{n} \mfp_{i}$; then $B := S^{-1}A$ is a semi-local ring whose maximal ideals are a subset of $\{\mfp_{i}B\}_{i = 1}^{n}$. Let $(I, m_{f})$ be an ideal representative for $X$, and put $J = S^{-1}I$. Since $J$ is a $B$-module of constant rank $1$ and $B$ is semi-local, $J$ is a free $B$-module of rank $1$, hence principal. Say $J$ is generated by $0 \neq y/z \in B$. Since $I^{n} = fA$, we have $J^{n} = fB$, whence
	\[\left(\frac{y}{z}\right)^{n} = f \cdot u \]
	for some unit $u \in B^{\times}$. If $u = v/w$ for $v \in A, w \in S$, we must have $v \in S$ as well. Put $g = vz/y \in K^{\times}$, so that
	\[v^{n-1}w = g^{n}f, \]
	and let $\tilde{I} = gI$, $\tilde{f} = v^{n-1}w \in S \subset A$. Then $\tilde{I}^{n} = g^{n}(I^{n}) = g^{n}fA = v^{n-1}wA$, and the map $I \to \tilde{I}$ given by multiplication by $g$ is an isomorphism between $(I, m_{f})$ and $(\tilde{I}, m_{\tilde{f}})$ in $\Psi(A, n)$. Moreover, $\tilde{f} \in S$, and so $\alpha_{i}(\tilde{f}) \in M\setminus \{0\}$ for each $i$. It remains to show that $\tilde{I} \subset A$. Let $x \in I$; then $(gx)^{n} \in \tilde{I}^{n} \subset A$. But then $gx$ is a root of $X^{n}-(gx)^{n} \in A[X]$, and so is integral over $A$, and therefore belongs to $A$.
\end{proof}

Notice that if $M$ is a field, then $\Pic(M)$ is trivial, so $\Delta_{n}(M)$ is an isomorphism by Corollary \ref{KMuNTorsors}.

\begin{prop}\label{FieldMorphismTorsImage}
	Let $A$ be a normal domain, and let $M$ be a field. Let $\alpha \colon A \to M$ be a ring morphism. Let $X \in \Psi(A, n)$, and let $(I, m_{f}) \in \Psi(A, n)$ be an ideal representative for $X$ satisfying $\alpha(f) \neq 0$. Then $(\Delta_{n}(M)^{-1} \circ \Psi(-, n)(\alpha))(X) = [\alpha(f)^{-1}].$
\end{prop}

\begin{proof}
	Consider the morphism of $M$-vector spaces $\tau \colon I \otimes_{A} M \to \mathcal{L}_{\alpha(f)^{-1}}$ given on simple tensors by $\tau(x \otimes z) = \alpha(x)z\alpha(f)^{-1/n}$. Since $I \otimes_{A}M$ and $L_{\alpha(f)^{-1}}$ are both $M$-vector spaces of dimension $1$, the map $\tau$ is an isomorphism provided it is nonzero. Indeed, this is the case, since $f \in I$, and so $\tau(f \otimes 1) = \alpha(f)\alpha(f)^{-1/n}$ is nonzero. It is straightforward to check that $\tau$ is an isomorphism between $\Psi(-, n)(\alpha)(X)$ and $\mathcal{L}_{\alpha(f)^{-1}}$.
\end{proof}

\begin{cor} \label{IdealRepDelta}
	Let $A$ be a normal domain with field of fractions $K$, let $M$ be a field, and let $\alpha \colon A \to M$ be a morphism of rings. Let $x \in A^{\times}$, and let $(I, m_{f})$ be an ideal representative for $\Delta_{n}(A)(x)$. Then there is a nonzero element $y \in A$ such that $I = yA$ and $y^{n}/f = x$ in $K$, and $[\alpha(x)] = [\alpha(f)^{-1}]$ in $M^{\times}/(M^{\times})^{n}$. We deduce $\mathcal{K}^{n}(\alpha) = \Delta_{n}(M)^{-1} \circ \Psi(-, n)(\alpha) \circ \Delta_{n}(A)$.
\end{cor}
\begin{proof}
	Since $[(I, m_{f})]$ and $[(\mathcal{L}_{x}, \varphi_{x})]$ are equal as classes in $\Psi(A, n)$, there is an isomorphism of $A$-modules $\omega \colon \mathcal{L}_{x} \to I$ such that $m_{f} \circ \omega^{\otimes n} = \varphi_{x}$. As $\mathcal{L}_{x}$ is free, $I$ is a (nonzero) principal ideal, generated by $y := \omega(1 \cdot x^{1/n}) \in A$. We thus have
	\[x = \varphi_{x}(x^{1/n} \otimes \cdots \otimes x^{1/n}) = m_{f}(\omega^{n}(x^{1/n} \otimes \cdots \otimes x^{1/n})) = m_{f}(y \otimes \cdots \otimes y) = y^{n}/f \]
	as claimed. Moreover, since $xf = y^{n}$ and $\alpha(x), \alpha(f) \in M^{\times}$, this forces $\alpha(y) \in M^{\times}$, and so $[\alpha(x)] \cdot [\alpha(f)] = [\alpha(y)^{n}] = [1] \in M^{\times}/(M^{\times})^{n}$.
\end{proof}

Suppose $A$ is a normal domain, let $K$ be its field of fractions, and let $\xi \colon A \to K$ be the canonical localization map. Fix $X \in \Psi(A, n)$, and let $(I, m_{f})$ be an ideal representative for $X$. By Proposition \ref{FieldMorphismTorsImage}, $\Delta_{n}(K)^{-1}(\Psi(-, n)(\xi)(X)) = [\xi(f)^{-1}] = [1/f]$. If  $\partial_{n}(A) \colon K^{\times}/(K^{\times})^{n} \to \Div(A)/n\Div(A)$ denotes the map induced by $\partial(A)$, then
\[\partial_{n}(A)([1/f]) = -[\partial(A)(f)] = -[\dv(fA)] = -[n\dv(I)] = 0 \in \Div(A)/n\Div(A) \]
Hence, the map $\Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi)$ takes image in $\ker(\partial_{n}(A)) \subset K^{\times}/(K^{\times})^{n}$. If we further assume $A$ is regular, then Theorem \ref{TorsKerDivMap} shows $\Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi)$ (viewed by abuse of notation as a map $\Psi(A, n) \to \ker(\partial_{n}(A))$) is an isomorphism.

\begin{thm} \label{TorsKerDivMap}
	Let $A$ be a regular domain, let $K$ be its field of fractions, and let $\xi \colon A \to K$ be the canonical localization map. Then the map $\Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi) \colon \Psi(A, n) \to \ker(\partial_{n}(A))$ is an isomorphism.
\end{thm}
\begin{proof}
	First, consider the morphism $\zeta(A) \colon \ker(\partial_{n}(A)) \to \Cl(A)[n]$ defined as follows: if $[x] \in K^{\times}/(K^{\times})^{n}$ belongs to the kernel of $\partial_{n}(A)$, then $\partial(A)(x) \in n\Div(A)$. Define $\zeta(A)$ by sending $[x]$ to $[D]$, where $D$ satisfies $nD = \partial(A)(x)$; note that $D$ must be unique, since $\Div(A)$ is free. This map is well-defined, because if $x' = xy^{n}$ for $y \in K^{\times}$, then $\partial(A)(xy^{n}) = n(D+\partial(A)(y))$, and $[D] = [D+\partial(A)(y)]$ in $\Cl(A)$. We claim that the sequence
	\[1 \to A^{\times}/(A^{\times})^{n} \xrightarrow{~\mathcal{K}^{n}(\xi)~} \ker(\partial_{n}(A)) \xrightarrow{~\zeta(A)~} \Cl(A)[n] \to 0 \]
	is exact. Indeed, $\mathcal{K}^{n}(\xi)$ is an injection because $A$ is integrally closed in $K$. Moreover, if $[D] \in \Cl(A)[n]$, then $nD = \partial(A)(x)$ for some $x \in K^{\times}$, and so $\zeta(A)([x]) = [D]$. Hence, $\zeta(A)$ is surjective. \\
	It remains to check exactness at $\ker(\partial_{n}(A))$. Clearly, $\Imm(\mathcal{K}^{n}(\xi)) \subset \ker(\zeta(A))$, so suppose that $[x] \in \ker(\zeta(A))$. Then $\partial(A)(x) = n\partial(A)(y) = \partial(A)(y^{n})$ for some $y \in K^{\times}$, whence $x = y^{n} \cdot x'$ for some $x' \in A^{\times}$, and so $[x] = [x']$ in $K^{\times}/(K^{\times})^{n}$. \\
	Now, since $A$ is regular, $\dv(A) \colon \Cart(A) \to \Div(A)$ is an isomorphism, and so induces an isomorphism $\Pic(A)[n] \to \Cl(A)[n]$. Let $\nu(A) \colon \Pic(A)[n] \to \Cl(A)[n]$ be the composition of this isomorphism with the inversion automorphism $\Cl(A)[n] \to \Cl(A)[n]$. I claim that $\zeta(A) \circ \Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi) = \nu(A) \circ \Theta_{n}(A)$. Indeed, let $X$ be a class in $\Psi(A, n)$, and let $(I, m_{f})$ be an ideal representative for $X$. Then
	\[(\nu(A) \circ \Theta_{n}(A))(X) = \nu(A)([I]) =  -[\dv(I)].\]
	On the other hand,
	\[(\zeta(A) \circ \Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi))(X) = \zeta(A)([1/f])\]
	by Proposition \ref{FieldMorphismTorsImage}. But $I^{n} = fA$, so $\partial(A)(1/f) = -\dv(I^{n}) = -n\dv(I)$, and thus $\zeta(A)(1/f) = -[\dv(I)]$. By Proposition \ref{IdealRepDelta}, $\mathcal{K}^{n}(\xi) = \Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi) \circ \Delta_{n}(A)$, so we have a commutative diagram of abelian groups
	\begin{center}
		\begin{tikzcd}[column sep = .5 in, row sep = .7 in]
		1 \arrow{r} & A^{\times}/(A^{\times})^{n} \arrow{d}{\Id_{A^{\times}/(A^{\times})^{n}}} \arrow{r}{\Delta_{n}(A)} & \arrow{d}{\Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi)} \Psi(A, n) \arrow{r}{\Theta_{n}(A)} & \arrow{d}{\nu(A)} \Pic(A)[n] \arrow{r} & 0 \\
		1 \arrow{r} & A^{\times}/(A^{\times})^{n} \arrow{r}{\mathcal{K}^{n}(\xi)} & \ker(D(A)_{n}) \arrow{r}{\zeta(A)} & \Cl(A)[n] \arrow{r} & 0
		\end{tikzcd}
	\end{center}
	whose rows are exact. Since $\Id_{A^{\times}/(A^{\times})^{n}}$ and $\nu(A)$ are isomorphisms, $\Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi)$ must be an isomorphism as well.

\end{proof}

\subsection{Pulling Back Torsors Along Products of Points}

\begin{defn}
	Let $G$ be an algebraic group over a field $F$, and let $A = F[G]$. Let $\varepsilon_{F} \in G(F)$ denote the identity element. We say a class $X \in \Psi(A, n)$ is {\bf normalized} if $X \in \ker(\Psi(-, n)(\varepsilon_{F}))$. We denote the subgroup of $\Psi(A, n)$ consisting of normalized elements by $\Psi_{\nm}(A, n)$. Likewise, we set $\Cart_{\nm}(A, n) = \Omega_{n}(A)^{-1}(\Psi_{\nm}(A, n)))$, and if $G$ is smooth, then we set $\Div_{\nm}(A, n) = \dv_{n}(A)(\Cart_{\nm}(A, n))$.
\end{defn}

We note the following properties of the subgroup $\Psi_{\nm}(A, n)$:

\begin{enumerate}[(1)]
	\item By Theorem \ref{PsiTorsors}, one sees that $\Psi_{\nm}(A, n) = \lambda_{n}(A)^{-1}(\Tors_{\nm}(G, \bmu_{n, F}))$.
	\item The assignment $A \mapsto \Psi_{\nm}(A, n)$ defines a functor from the category of Hopf $F$-algebras to $\AbGrps$. Moreover, if $M/F$ is a field extension, and $\alpha \colon A \to A_{M}$ denotes the canonical base change morphism, then the restriction of $\Psi(-, n)(\alpha)$ to $\Psi_{\nm}(A, n)$ takes image in $\Psi_{\nm}(A_{M}, n)$.
	\item If $G$ is a smooth, connected group, then $\Delta_{n}(A)^{-1}(\Psi_{\nm}(A, n)) = G^{\ast}/(G^{\ast})^{n} \subset A^{\times}/(A^{\times})^{n}$. This follows from Rosenlicht's theorem (\cite[Theorem 3]{Rosenlicht}).
\end{enumerate}

Normalized elements play a key role in the following situation. Let $M/F$ be a field extension, and fix a class $X \in \Psi(A, n)$. Consider the map $G(M) \to \Psi(M, n)$ which sends $\alpha \in G(M)$ to $\Psi(-, n)(\alpha)(X)$. Under what conditions is this map a group homomorphism? As the following theorem shows, this is the case precisely when $X$ is normalized, provided that $G$ is smooth, connected, and reductive.

\begin{thm}\label{PointsMultTorsors}
	Let $G$ be a smooth, connected, reductive, algebraic group over a field $F$. Put $A = F[G]$, and let $M$ be a field extension of $F$. For any $\alpha, \beta \in G(M)$, and any class $X \in \Psi(A, n)$, we have
	\[\Psi(-, n)(\alpha)(X) \cdot \Psi(-, n)(\beta)(X) = \Psi(-, n)(\alpha\beta)(X) \cdot \Psi(-, n)(\varepsilon_{M})(X) \]
\end{thm}

\begin{proof}
	Let $(I, m_{f})$ be an ideal representative for $X$ such that $\alpha(f), \beta(f), (\alpha\beta)(f), \varepsilon_{M}(f) \in M^{\times}$; this is possible by Corollary \ref{NonzeroIdealTors}. By Proposition \ref{FieldMorphismTorsImage}, it suffices to show that
	\[[(\alpha\beta)(f)\varepsilon_{M}(f)] = [\alpha(f)\beta(f)]\]
	as classes in $M^{\times}/(M^{\times})^{n}$. Let $B = F[G \times_{F} G] = A \otimes_{F} A$, and let $E$ be the field of fractions of $B$. Let $c, p_{1}, p_{2} \colon A \to B$ be the $F$-algebra morphisms corresponding respectively to the morphisms $G \times G \to G$ given by multiplication and projection onto each component. We note that $c, p_{1}, p_{2}$ are each flat, hence injective. \\
	By \cite[Theorems 16.56 and 21.84]{MilneAG}, any connected, reductive algebraic group over a separably closed field is rational, and so the natural map $\Pic(B) \to \Pic(A) \oplus \Pic(A)$ is an isomorphism by \cite[Lemma 6.6]{Sansuc}. Moreover, up to this identification, $\Pic(c) \colon \Pic(A) \to \Pic(B)$ is the diagonal embedding, and  $\Pic(p_{i}) \colon \Pic(A) \to \Pic(B)$ is the embedding onto the $i^{\thh}$ component. Let $J_{c} = c(I)B, J_{i} = p_{i}(I)B$; since $c, p_{1}, p_{2}$ are flat, $J_{c}, J_{1}, J_{2} \in \Cart(B)$, and we have $[J_{c}] = [J_{1}] + [J_{2}]$ as classes in $\Pic(B)$. In light of the classical exact sequence
	\begin{equation} \label{ExactSeqCart}
	1 \to B^{\times} \to E^{\times} \to \Cart(B) \to \Pic(B) \to 0
	\end{equation}
	there exists $h \in E^{\times}$ such that $J_{c} = hB \cdot J_{1} \cdot J_{2}$. Raising each side of this equation to the $n^{\thh}$ power and using the relation $I^{n} = fA$ gives the equation $c(f)B = h^{n}B \cdot (f \otimes f)B$. Appealing again to \ref{ExactSeqCart}, there exists $b \in B^{\times}$ such that $b c(f) = h^{n} (f \otimes f)$. Let $x, y \in B$ such that $h = x/y$, so that our equation reads $b c(f) y^{n} = x^{n} (f \otimes f)$. \\
	Let $\omega \colon B \to M$ be the composition of $\alpha \otimes_{F} \beta \colon B \to M \otimes_{F} M$ and the multiplication map $M \otimes_{F} M \xrightarrow{\sim} M$. Then $\omega(c(f)) = (\alpha\beta)(f)$, and $\omega(f \otimes f) = \alpha(f)\beta(f)$, so applying $\omega$ to the equation above gives
	\[(\alpha\beta)(f) \omega(b)\omega(y)^{n} = \alpha(f)\beta(f) \omega(x)^{n}\]
	Since $M$ is a field, $\mfp := \ker(\omega)$ is a prime ideal of $B$. We know that $\alpha(f), \beta(f) \in M^{\times}$, so $f \otimes f$ belongs to $B \setminus \mfp$. Hence, $h^{n} = bc(f)/(f\otimes f) \in B_{\mfp}$. Since $B$ is regular, it follows that $B_{\mfp}$ is integrally closed in $E$, so $h^{n} \in B_{\mfp}$ implies $h \in B_{\mfp}$; in particular, we have $\omega(y) \neq 0$. This also forces $\omega(x) \neq 0$, since $(\alpha\beta)(f), \omega(b) \in M^{\times}$, so we have
	\[[(\alpha\beta)(f)\omega(b)] = [\alpha(f)\beta(f)]\]
	as classes in $M^{\times}/(M^{\times})^{n}$. It remains to show that $\omega(b)$ and $\varepsilon_{M}(f)$ belong to the same class in $M^{\times}/(M^{\times})^{n}$. By Rosenlicht's theorem (\cite[Theorem 3]{Rosenlicht}), the map $F^{\times} \oplus G^{\ast} \oplus G^{\ast} \to B^{\times}$ sending $(z, \chi, \rho)$ to $z(\chi^{\sharp}(t) \otimes \rho^{\sharp}(t) )$ is an isomorphism, so $b$ can be written as $z(g \otimes g')$ for $g, g' \in A^{\times}$ group-like elements, $z \in F^{\times}$. Then $\omega(b) = z\alpha(g)\beta(g')$, and our equation in $M^{\times}/(M^{\times})^{n}$ therefore reads
	\[[(\alpha\beta)(f) \cdot z \cdot \alpha(g)\beta(g')] = [\alpha(f)\beta(f)]\]
	Our derivation of this equation did not depend on our choice of $\alpha, \beta \in G(M)$, only on the fact that $\alpha(f), \beta(f), (\alpha\beta)(f) \in M^{\times}$. In particular, since we arranged that $\varepsilon_{M}(f) \neq 0$, we can substitute $\varepsilon_{M}$ for $\alpha$ or $\beta$ in our equation. Plugging in $\alpha = \varepsilon_{M}$ and using $\varepsilon_{M}(g) = 1$ gives $[\varepsilon_{M}(f)] = [z\beta(g')]$, and likewise, plugging in $\beta = \varepsilon_{M}$ yields $[\varepsilon_{M}(f)] = [z\alpha(g)]$. Substituting both $\alpha = \varepsilon_{M}, \beta = \varepsilon_{M}$ simultaneously gives us $[z] = [\varepsilon_{M}(f)]$, whence $[\alpha(g)] = [\beta(g')] = 1$, and so $[z\alpha(g)\beta(g')] = [\varepsilon_{M}(f)]$, completing the proof.

\end{proof}

\begin{cor} \label{NormalizedGroupOp}
	Let $G, A, M$ be as in the statement of Theorem \ref{NormalizedGroupOp}. If $X \in \Psi_{\nm}(A, n)$, then the map $G(M) \to \Psi(M, n)$ sending $\alpha \in G(M)$ to $\Psi(-, n)(\alpha)(X)$ is a group homomorphism. \qed
\end{cor}

\subsection{The Galois Action on Torsors}

Suppose that our group $G$ is smooth and connected, and $\Pic(G_{\sep})[n] = 0$. Then putting $A = F[G]$, $\Delta_{n}(A_{\sep})$ is an isomorphism by Corollary \ref{KMuNTorsors}, and the subgroup of $\Psi_{\nm}(A_{\sep}, n)$ of $\Psi(A_{\sep}, n)$ is the image of $G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}$. Via the embedding $\Gamma \to \Aut_{F-\alg}(A_{\sep})$, $\Gamma$ acts functorially on $A_{\sep}^{\times}/(A_{\sep}^{\times})^{n}$ and $\Psi(A_{\sep}, n)$, and the map $\Delta_{n}(A_{\sep})$ is $\Gamma$-equivariant. Since the action of $\Gamma$ on $A_{\sep}^{\times}/(A_{\sep}^{\times})^{n}$ preserves the summand $G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n} \subset A_{\sep}^{\times}/(A_{\sep}^{\times})^{n}$, this shows that the action of $\Gamma$ on $\Psi(A_{\sep}, n)$ restricts to an action on $\Psi_{\nm}(A_{\sep}, n)$.

Throughout this section, let $\alpha \colon A \to A_{\sep}$ denote the canonical base change morphism. The associated map $\Psi(-, n)(\alpha) \colon \Psi_{\nm}(A, n) \to \Psi_{\nm}(A_{\sep}, n)$ has image in $H^{0}(F, \Psi_{\nm}(A_{\sep}, n))$. If we assume that $G$ is geometrically integral, then $\Psi(-, n)(\alpha)$ is an embedding with image $H^{0}(F, \Psi_{\nm}(A_{\sep}, n))$; this is the content of Theorem \ref{GalFixedTorsors}. First, we require a lemma.

\begin{lem} \label{ClassGroupCohomology}
	Let $G$ be a smooth, geometrically integral group variety over $F$, and let $A = F[G]$. If $\Cl(A_{\sep}) = 0$, then there is an isomorphism $Z(A) \colon H^{1}(F, G_{\sep}^{\ast}) \to \Cl(A)$.
\end{lem}

\begin{proof}
	Let $K_{s} = \Frac(A_{\sep})$; since $G$ is geometrically integral, $K_{s} = KF_{\sep}$. Since $\Cl(A_{\sep}) = 0$, we have an exact sequence of $\Gamma$-modules
	\[1 \to (A_{\sep})^{\times} \to (K_{s})^{\times} \xrightarrow{\partial(A_{\sep})} \Div(A_{\sep}) \to 0 \]
	and therefore obtain the following long exact sequence in Galois cohomology:
	\[H^{0}(F, (A_{\sep})^{\times}) \to H^{0}(F, K_{s}^{\times}) \to H^{0}(F, \Div(A_{\sep})) \xrightarrow{~\delta~} H^{1}(F, (A_{\sep})^{\times}) \to H^{1}(F, K_{s}^{\times}) \to \cdots \]
	Since $\Gamma \cong \Gal(KF_{\sep}/K) = \Gal(K_{s}/K)$, we have $H^{1}(F, K_{s}^{\times})$ by Hilbert Theorem 90. As $A$ is regular and geometrically integral, $\Div(\alpha)$ embeds $\Div(A)$ onto $H^{0}(F, \Div(A_{\sep}))$. By Rosenlicht's Theorem (\cite[Theorem 3]{Rosenlicht}), the map $F_{\sep}^{\times} \oplus G_{\sep}^{\ast} \to A_{\sep}^{\times}$ sending $(z, \chi)$ to $z\chi^{\sharp}(t)$ is an isomorphism of $\Gamma$-modules. Hence, $H^{1}(F, (A_{\sep})^{\times}) \cong H^{1}(F, G_{\sep}^{\ast}) \oplus H^{1}(F, (F_{\sep})^{\times}) = H^{1}(F, G_{\sep}^{\ast})$. We thus have a commutative diagram
	\begin{center}
		\begin{tikzcd}[column sep = .4 in, row sep = .5 in]
		A^{\times} \arrow{d}{} \arrow{r} & K^{\times} \arrow{r}{\partial(A)} \arrow{d}{} & \Div(A)  \arrow{d}{\Div(\alpha)} \arrow{r}{\dv(A)} & \Cl(A) \arrow{r} & 0  \\
		H^{0}(F, (A_{\sep})^{\times}) \arrow{r} & H^{0}(F,K_{s}^{\times}) \arrow{r}{\partial(A_{\sep})} & H^{0}(F,\Div(A_{\sep})) \arrow{r}{\delta} & H^{1}(F, G_{\sep}^{\ast}) \arrow{r} & 0
		\end{tikzcd}
	\end{center}
	with exact rows and vertical arrows isomorphisms. By the universal property of the cokernel, $\Div(\alpha)$ descends to a well-defined map $Z(A) \colon \Cl(A) \to H^{1}(F, G_{\sep}^{\ast})$ which sends the $[D] \in \Cl(A)$ to $\delta(\Div(\alpha)(D))$. By (e.g.) the Five Lemma, $Z(A)$ is an isomorphism.

\end{proof}

Note that we can be more explicit in describing $Z(A)$. Let $[D] \in \Cl(A)$, and set $D' = \Div(\alpha)(D)$. Since the map $\partial(A_{\sep}) \colon K_{s}^{\times} \to \Div(A_{\sep})$ is surjective, there exists $x \in K_{s}^{\times}$ such that $D' = \partial(A_{\sep})(x)$. One can accordingly define a cocycle $\sigma_{x} \colon \Gamma \to A_{\sep}^{\times}$ by setting $\sigma_{x}(\gamma) = \gamma(x)/x$ for $\gamma \in \Gamma$, and the class of $\sigma_{x}$ in $H^{1}(F, A_{\sep}^{\times}) = H^{1}(F, G_{\sep}^{\ast})$ does not depend on the choice of $x$. The map $Z(A)$ then takes $[D]$ to $[\sigma_{x}]$ in $H^{1}(F, G_{\sep}^{\ast})$.

\begin{thm}\label{GalFixedTorsors}
	Let $G$ be a geometrically integral, smooth group scheme over $F$. Put $A = F[G], K = \Frac(A)$, and $K_{s} = \Frac(A_{\sep})$. Suppose that $\Cl(A_{\sep}) = 0$, and $G_{\sep}^{\ast}[n] = 0$. Then the natural map $\Psi(-, n)(\alpha) \colon \Psi_{\nm}(A, n) \to \Psi_{\nm}(A_{\sep}, n)$ is an embedding of $\Psi_{\nm}(A, n)$ onto $H^{0}(F, \Psi_{\nm}(A_{\sep}, n))$.
\end{thm}

\begin{proof}
	Put $\eta(A) = \dv_{n}(A) \circ \Omega_{n}(A)^{-1} \circ \Delta_{n}(A)$. By Proposition \ref{MuTorsExactSeq}, we have an exact sequence
	\[1 \to G^{\ast}/(G^{\ast})^{n} \xrightarrow{~\eta(A)~} \Div_{\nm}(A, n) \xrightarrow{~~} \Cl(A) \xrightarrow{~\cdot n ~} \Cl(A) \to 0.\]
	Because $G_{\sep}^{\ast}[n] = 0$, there is an exact sequence of $\Gamma$-modules
	\[1 \to G_{\sep}^{\ast} \xrightarrow{~ \cdot n ~} G_{\sep}^{\ast} \to G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n} \to 1 \]
	which yields the following long exact sequence in Galois cohomology:
	\[1 \to H^{0}(F, G_{\sep}^{\ast}) \xrightarrow{~ \cdot n ~} H^{0}(F, G_{\sep}^{\ast}) \to H^{0}(F, G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}) \xrightarrow{~ \delta ~} H^{1}(F, G_{\sep}^{\ast}) \xrightarrow{~ \cdot n ~} H^{1}(F, G_{\sep}^{\ast}) \to \cdots \]
	We can rewrite the above (truncated) long exact sequence as
	\[1 \to G^{\ast}/(G^{\ast})^{n} \to H^{0}(F, G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}) \xrightarrow{~ \delta ~} H^{1}(F, G_{\sep}^{\ast}) \xrightarrow{~ \cdot n ~} H^{1}(F, G_{\sep}^{\ast}) .\]
	The boundary map $\delta$ can be described as follows: let $[u] \in H^{0}(F, G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n})$. Then $\gamma(u)/u \in (G_{\sep}^{\ast})^{n}$ for any $\gamma \in \Gamma$, so let $x_{\gamma}$ be the {\bf unique} element of $G_{\sep}^{\ast}$ such that $x_{\gamma}^{n} = \gamma(u)/u$. Then $\delta([u])$ is the class of the cocycle $\sigma_{u} \colon \Gamma \to G_{\sep}^{\ast}$ which sends $\gamma$ to $x_{\gamma}$. \\
	Note that $\Cl(A_{\sep}) \cong \Pic(A_{\sep}) = 0$, and so $\Delta_{n}(A_{\sep})$ is an isomorphism by Corollary \ref{KMuNTorsors}. Let $\tau(A)$ denote the composition
	\[\Div_{\nm}(A, n) \xrightarrow{~ \Omega_{n}(A) \circ \dv_{n}(A)^{-1} ~} \Psi_{\nm}(A, n) \xrightarrow{~ \Psi(-, n)(\alpha)~} \Psi_{\nm}(A_{\sep}, n) \xrightarrow{~ \Delta_{n}(A_{\sep})^{-1}~} G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}. \]
	Explicitly, given the class of a pair $(D, g)$ in $\Div_{\nm}(A, n)$, $D' := \Div(\alpha)(D)$ is principal, since $\Cl(A_{\sep}) = 0$, so there exists $x \in K_{s}^{\times}$ such that $\partial(A_{\sep})(x) = D'$; $\tau(A)$ sends $[(D, g)]$ to $[x^{n}/g] \in G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}$. If the diagram
	\begin{center}
		\begin{tikzcd}[column sep = .4 in, row sep = .5 in]
		1 \arrow{r} &  G^{\ast}/(G^{\ast})^{n} \arrow{d}{\Id_{G^{\ast}/(G^{\ast})^{n}}} \arrow{r}{\eta(A)} & \Div_{\nm}(A, n) \arrow{r}{} \arrow{d}{\tau(A)} & \Cl(A) \arrow{d}{Z(A)} \arrow{r}{\cdot n} & \Cl(A) \arrow{d}{Z(A)} \arrow{r} & 0  \\
		1 \arrow{r} & G^{\ast}/(G^{\ast})^{n} \arrow{r} & H^{0}(F,G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}) \arrow{r}{\delta} & H^{1}(F,G_{\sep}^{\ast}) \arrow{r}{\cdot n} & H^{1}(F, G_{\sep}^{\ast}) \arrow{r} & 0
		\end{tikzcd}
	\end{center}
	commutes, then $\tau(A)$ must be an isomorphism, so $\Psi(-, n)(\alpha)$ must be one as well. The last square is manifestly commutative. We have $\tau(A) \circ \eta(A) = \Delta_{n}(A_{\sep})^{-1} \circ \Psi(-, n)(\alpha) \circ \Delta_{n}(A)$, which is easily seen to be the inclusion $G^{\ast}/(G^{\ast})^{n} \to G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}$. It remains to show that the middle square commutes. \\
	Let the pair $(D, g)$ represent a class in $\Div_{\nm}(A, n)$, and put $D' = \Div(\alpha)(D)$. Let $x \in K_{s}^{\times}$ such that $D' = \partial(A_{\sep})(x)$, so that $x^{n} = gu$ for some $u \in A_{\sep}^{\times}$. As explained above, $\tau(A)([(D, g)]) = [u] \in G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}$. For any $\gamma \in \Gamma$,
	\[\frac{\gamma(u)}{u} = \frac{\gamma(x^{n}g^{-1})}{x^{n}g^{-1}} = \frac{\gamma(x^{n})}{x^{n}} = \left(\frac{\gamma(x)}{x}\right)^{n}\]
	because $g^{-1}$ is $\Gamma$-invariant. Therefore, $\delta$ takes $[u]$ to the class of the cocycle $\sigma_{u} \colon \Gamma \to G_{\sep}^{\ast}$ defined by $\gamma \mapsto \gamma(x)/x$. On the other hand, as explained in paragraph immediately following Theorem \ref{ClassGroupCohomology}, $Z(A)$ takes $[D]$ to the very same cocycle class, so we're done.

\end{proof}

\section{Type-Zero Invariants for Connected Reductive Groups} \label{TypeZeroInvariants}

Throughout this section, let $G$ be a smooth, connected, reductive algebraic group over $F$. Let $A = F[G], K = F(G)$, and let $\xi \colon A \to K$ denote the generic point of $G$.

We are now equipped to determine the groups $\Inv_{\hom}^{0}(G, H)$ for $H = K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ$ and $H = K^{M}_{1}/n$ for all $n \in \NN$; this is the content of Theorems \ref{TypeZeroInvariantsGaloisQZ} and \ref{TypeZeroInvariantsReductiveModN} respectively. A key step is the observation that, under suitable conditions, a type-zero $H$-invariant of $G$ is determined by its value at $\xi$; precisely, the evaluation homomorphism $\ev_{\xi}(H) \colon \Inv^{0}_{\hom}(G, H) \to H(K)$ sending an invariant $I$ to $I(K)(\xi)$ is injective. Before proving this in Proposition \ref{TypeZeroGenericSet}, we need a technical lemma. For any positive integer $n$, let $p_{n} \colon G \to G$ denote the $n^{\thh}$ power map, which sends $x$ to $x^{n}$ for any $F$-algebra $R$ and any $x \in G(R)$.

\begin{lem} \label{NthPowerDominant}
	The map $p_{n}$ is dominant.
\end{lem}

\begin{proof}
	Since the property of dominance descends under faithfully flat base change, we may assume that our base field $F$ is algebraically closed. By (e.g.) \cite[Theorem 17.44]{MilneAG}, the union of the Cartan subgroups of $G$ contains a dense open subset of $G$. Since $G$ is reductive, the Cartan subgroups of $G$ are precisely the maximal tori in $G$. But the restriction of $p_{n}$ to any torus in $G$ is surjective, and so the image of $p_{n}$ contains every torus in $G$.

\end{proof}

\begin{prop}\label{TypeZeroGenericSet}
	Suppose $H$ is the $d^{\thh}$ graded component of a torsion cycle module. Let $I \in \Inv^{0}(G, H)$, and suppose $I(K)(\xi) = 1_{H(K)}$. Suppose that for any field extension $L/F$ and any $\alpha, \beta \in G(L)$, $I$ satisfies
	\[I(L)(\alpha)I(L)(\beta) = I(L)(\alpha\beta)I(L)(\varepsilon_{L}).\]
	Then $I$ is trivial.
\end{prop}

\begin{proof}
	Let $L/F$ be a field extension, and fix $t \in G(L)$. Put $S := G_{L}$, and let $g \colon S \to G$ be the canonical base change morphism, with comorphism $f \colon A \to A_{L}$.  Let $E = L(S)$, and let $\xi' \colon A_{L} \to E$ be the generic point of $S$. Since $f$ is injective, the composition $\xi' \circ f$ extends to a morphism $u \colon K \to E$ of $F$-algebras such that $u \circ \xi = \xi' \circ f$. Put $\xi_{E} := u \circ \xi$, and let $n$ be a positive integer such that $I(E)(\xi_{E})^{n} = I(E)(\varepsilon_{E})^{n} = 1$. \\
	Suppose that there exist morphisms $i \colon K \to E, j \colon L \to E$ satisfying the following two properties:
	\begin{enumerate}[(a)]
		\item $H(j) \colon H(L) \to H(E)$ is injective;
		\item $G(i)(\xi) = (\xi_{E})^{n} \cdot t_{E}$, where $t_{E} := j \circ t$.

	\end{enumerate}
	Then we have

	\[H(j)(I(L)(t)) = I(E)(t_{E}) = I(E)(\xi_{E})^{n}I(E)(t_{E})I(E)(\varepsilon_{E})^{-n} = I(E)(({\xi}_{E})^{n} \cdot t_{E}), \]
	whence we conclude
	\[H(j)(I(L)(t)) = I(E)(G(i)(\xi)) = H(i)(I(K)(\xi)) = 1_{H(E)}.\]
	We therefore devote the remainder of the proof to constructing such a pair $(i, j)$. Let $j \colon L \to E$ denote the composition of the structural map $L \to A_{L}$ with $\xi'$. Since $S$ is a smooth algebraic $L$-variety such that $S(L) \neq \emptyset$, $H(j)$ is injective by \cite[Lemma 1.3]{MerkurjevIAG}. \\
	To construct $i$, let $s \colon A_{L} \to L$ be the unique $L$-algebra morphism such that $t = s \circ f = g(L)(t)$, and put $s_{E} = j \circ s$, so that $t_{E} = s_{E} \circ f$. Let $p_{n, s} \colon S \to S$ be the morphism of $L$-schemes given by the composition of the $n^{\thh}$ power map $p_{n}$ with right translation by $s$. By Corollary \ref{NthPowerDominant}, $p_{n, s}$ is dominant, and so the associated comorphism $h \colon A_{L} \to A_{L}$ is injective. In particular, the composition $\xi' \circ h$ extends to a morphism $v \colon E \to E$ of $L$-algebras such that $v \circ \xi' = \xi' \circ h$. Putting $i = v \circ u$,  we claim that $i$ satisfies (b). \\
	On the one hand, we have $p_{n, s}(E)(\xi') = \xi' \circ h$, but by definition of $p_{n, s}$, we also have $p_{n, s}(E)(\xi') = (\xi')^{n} \cdot s_{E}$. Accordingly, this yields
	\[G(i)(\xi) = v \circ u \circ \xi = \xi' \circ h \circ f = g(E)(p_{n, s}(E)(\xi')).\]
	But we compute
	\[g(E)(p_{n, s}(\xi')) = g(E)((\xi')^{n} \cdot s_{E}) = g(E)(\xi')^{n} \cdot g(E)(s_{E}) = (\xi_{E})^{n} \cdot t_{E},\]
	which establishes (b).

\end{proof}

\begin{cor}\label{TypeZeroGenericGroup}
	The morphism $\ev_{\xi}(H) \colon \Inv^{0}_{\hom}(G, H) \to H(K)$ is injective. \qed
\end{cor}

For any fixed field extension $L/F$, there is a map $\Psi(A, n) \times G(L) \to L^{\times}/(L^{\times})^{n}$ which sends the pair $(X, y)$ to $\Delta_{n}(L)^{-1}(\Psi(-, n)(y)(X))$. If we fix a class $X \in \Psi(A, n)$ in the first argument, we obtain a set map $I_{X}(L) \colon G(L) \to L^{\times}/(L^{\times})^{n}$. As $L$ varies, the collection of maps $I_{X}$ determines an invariant in $\Inv^{0}(G, K^{M}_{1}/n)$. If $X$ is \emph{normalized}, then Corollary \ref{NormalizedGroupOp} shows that $I_{X}$ is homomorphic. We thus obtain a group homomorphism $\Lambda_{n}(G) \colon \Psi_{\nm}(A, n) \to \Inv^{0}_{\hom}(G, K^{M}_{1}/n)$. As the next theorem shows, $\Lambda_{n}(G)$ is in fact an isomorphism.

\begin{thm} \label{TypeZeroInvariantsReductiveModN}
	The map
	\[\Lambda_{n}(G) \colon \Psi_{\nm}(A, n) \to \Inv^{0}_{\hom}(G, K^{M}_{1}/n)\]
	sending a class $X \in \Psi_{\nm}(A, n)$ to the invariant $I_{X}$ is an isomorphism.
\end{thm}

\begin{proof}
	By Lemma \ref{TorsKerDivMap}, the map $\Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi) \colon \Psi(A, n) \to \ker(\partial_{n}(A))$ is an isomorphism. Thus, since $\ev_{\xi}(K^{M}_{1}/n) \circ \Lambda_{n}(G)$ coincides with the restriction of $\Delta_{n}(K)^{-1} \circ \Psi(-, n)(\xi)$ to $\Psi_{\nm}(A, n)$, $\Lambda_{n}(G)$ must be injective. \\
	Now, fix an invariant $I \in \Inv^{0}_{\hom}(G, K^{M}_{1}/n)$. By Corollary \ref{TypeZeroGenericGroup}, $\ev_{\xi}(K^{M}_{1}/n)$ is injective. The sequence
	\[\Inv^{0}_{\hom}(G, K^{M}_{1}/n) \xrightarrow{~\ev_{\xi}(K^{M}_{1}/n)~} K^{\times}/(K^{\times})^{n} \xrightarrow{~\partial_{n}(A)~} \Div(A)/n\Div(A) \]
	is a complex by \cite[Lemma 2.1]{MerkurjevIAG}, so $\ev_{\xi}(K^{M}_{1}/n)$ has image contained in $\ker(\partial_{n}(A))$. Letting $X \in \Psi(A, n)$ be a class such that $\Delta_{n}(K)^{-1}(\Psi(-, n)(\xi)(X)) = I(K)(\xi)$, we have $I_{X}(K)(\xi) = I(K)(\xi)$ by construction. We must therefore have $I_{X} = I$ by Theorem \ref{PointsMultTorsors} and Proposition \ref{TypeZeroGenericSet}. But as $I$ is homomorphic, it must be the case that $I_{X}(F)(\varepsilon_{F}) = I(F)(\varepsilon_{F})$ is the trivial class in $F^{\times}/(F^{\times})^{n}$, whence $X$ is normalized, and $\Lambda_{n}(G)(X) = I$.
\end{proof}

\begin{cor} \label{TypeZeroInvariantsGaloisModN}
	Suppose that $G$ is a torus, and let $\alpha \colon A \to A_{\sep}$ be the canonical base change morphism. Then the map
	\[(\Delta(A_{\sep}) \circ \Psi(-, n)(\alpha))^{-1} \circ \Lambda_{n}(G) \colon H^{0}(F, G_{\sep}^{\ast}/(G_{\sep}^{\ast})^{n}) \to \Inv^{0}_{\hom}(G, K^{M}_{1}/n)\]
	is an isomorphism. \qed
\end{cor}

For any natural number $n$, let $\Inv^{0}(G, \iota_{n})$ denote the group morphism $\Inv^{0}_{\hom}(G, K^{M}_{1}/n) \to \Inv^{0}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)$ given by composition with $\iota_{n}$. Likewise, if $n$ and $m$ are positive integers such that $n$ divides $m$, let $\Inv^{0}(G, \beta_{n, m})$ denote the group morphism $\Inv^{0}_{\hom}(G, K^{M}_{1}/n) \to \Inv^{0}_{\hom}(G, K^{M}_{1}/m)$ given by composition with $\beta_{n, m}$. Since $\iota_{n} = \iota_{m} \circ \beta_{n, m}$, we obtain a universal induced map
\[\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(G, \iota_{n}) \colon \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}_{\hom}(G, K^{M}_{1}/n) \to \Inv^{0}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ).\]

\begin{prop} \label{ColimTypeZero}
	The map $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(G, \iota_{n})$ is an isomorphism.
\end{prop}

\begin{proof}
	We have the following commutative diagram:
	\begin{center}
		\begin{tikzcd}
		{\displaystyle\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}_{\hom}(G, K^{M}_{1}/n) } \arrow[rrr, "\operatornamewithlimits{colim}\limits_{n \in \NN} \ev_{\xi}(K^{M}_{1}/n)"] \arrow[ddd, "{\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(G, \iota_{n})}"'] &  &  & \displaystyle\operatornamewithlimits{colim}\limits_{n \in \NN} \ker(\partial_{n}(A)) \arrow[ddd, "\operatornamewithlimits{colim}\limits_{n \in \NN} \iota_{n}(K)"] \\
		&  &  &                                                                                \\
		&  &  &                                                                                \\
		{\Inv^{0}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)} \arrow[rrr, "\ev_{\xi}(K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)"']                                                         &  &  & \ker(\partial(A) \otimes_{\ZZ} \Id_{\QQ/\ZZ})
		\end{tikzcd}
	\end{center}
	The rightmost arrow is an isomorphism, and the lower and upper horizontal arrows are injective by Corollary \ref{TypeZeroGenericGroup}, so it follows that $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(G, \iota_{n})$ is injective. To see that $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(G, \iota_{n})$ is surjective, fix an invariant $I \in \Inv^{0}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)$ and let $x = I(K)(\xi) \in \ker(\partial(A) \otimes_{\ZZ} \Id_{\QQ/\ZZ})$. There exists some positive integer $n$ and $y \in \ker(\partial_{n}(A))$ such that $\iota_{n}(K)(y) = x$. Let $Y \in \Psi(A, n)$ with $\Psi(-, n)(\xi)(Y) = \Delta_{n}(K)(y)$. Then the associated invariant $I_{Y} \in \Inv^{0}(G, K^{M}_{1}/n)$ satisfies $(\iota_{n} \circ I_{Y})(K)(\xi) = x = I(K)(\xi)$, and so $\iota_{n} \circ I_{Y} = I$ by Theorem \ref{PointsMultTorsors} and Proposition \ref{TypeZeroGenericSet}. In particular, $((\iota_{n} \circ I_{Y})(F))(\varepsilon_{F})$ is the trivial class in $F^{\times} \otimes_{\ZZ} \QQ/\ZZ$, which means that $z := I_{Y}(F)(\varepsilon_{F})$ belongs to the kernel of $\iota_{n}(F) \colon F^{\times}/(F^{\times})^{n} \to F^{\times} \otimes_{\ZZ} \QQ/\ZZ$. \\
	This can only be the case if $z \in \ker(\beta_{n, nd}(F))$ for some $d \in \NN$, so fix such a $d$. For any field extension $M/F$, the diagram
	\begin{center}
		\begin{tikzcd}
		M^{\times}/(M^{\times})^{n} \arrow[ddd, "{\beta_{n, nd}(M)}"'] \arrow[rrr, "\Delta_{n}(M)"] &  &  & {\Psi(M, n)} \arrow[ddd, "{\omega_{n, nd}(M)}"] \\
		&  &  &                                                 \\
		&  &  &                                                 \\
		M^{\times}/(M^{\times})^{nd} \arrow[rrr, "\Delta_{nd}(M)"']                                &  &  & {\Psi(M, nd)}
		\end{tikzcd}
	\end{center}
	commutes, and so putting $Y' = \omega_{n, nd}(A)(Y)$, $\Psi(-, nd)(\varepsilon_{F})(Y') = \Delta_{nd}(F)(\beta_{n, nd}(F)(z))$, whence $Y'$ is normalized. Thus, $I_{Y'} = \Lambda_{nd}(G)(Y')$ is homomorphic, and
	\[((\iota_{nd} \circ I_{Y'})(K))(\xi)) = \iota_{nd}(K)(\beta_{n, nd}(K)(y)) = \iota_{n}(K)(y) = x,\]
	so $\iota_{nd} \circ I_{Y'} = I$ by Corollary \ref{TypeZeroGenericGroup}.

\end{proof}

\begin{cor} \label{TypeZeroInvariantsGaloisQZ}
	If $G$ is a torus, then $\Inv^{0}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ) \cong H^{0}(F, G_{\sep}^{\ast} \otimes \QQ/\ZZ)$.
\end{cor}

\begin{proof}
	If $\alpha \colon A \to A_{\sep}$ denotes the canonical base change morphism, this follows from Proposition \ref{ColimTypeZero}, Theorem \ref{GalFixedTorsors}, and the fact that the diagram
	\begin{center}
		\begin{tikzcd}
		{H^{0}(F, G^{\ast}_{\sep}/(G^{\ast}_{\sep})^{n})} \arrow[ddd, "\cdot {m/n}"] &  &  & {\Psi_{\nm}(A, n)} \arrow[lll, "{\Delta_{n}(A_{\sep})^{-1} \circ \Psi(-, n)(\alpha)}"'] \arrow[rrr, "\Lambda_{n}(G)"] \arrow[ddd, "{\omega_{n, m}(A)}"] &  &  & {\Inv^{0}_{\hom}(G, K^{M}_{1}/n)} \arrow[ddd, "{\Inv^{0}(G, \beta_{n, m})}"] \\
		&  &  &                                                                                &  &  &                                                                       \\
		&  &  &                                                                                &  &  &                                                                       \\
		{H^{0}(F, G^{\ast}_{\sep}/(G_{\sep}^{\ast})^{m})}                            &  &  & {\Psi_{\nm}(A, m)} \arrow[lll, "{\Delta_{m}(A_{\sep})^{-1} \circ \Psi(-, m)(\alpha)}"] \arrow[rrr, "\Lambda_{m}(G)"']                                   &  &  & {\Inv^{0}_{\hom}(G, K^{M}_{1}/m)}
		\end{tikzcd}
	\end{center}
	commutes for all $n, m \in \NN$ with $n$ dividing $m$.
\end{proof}

\section{Computation of Degree One Milnor $K$-invariants of Groups of Multiplicative Type} \label{TypeOneInvariants}

In this section, we determine the degree one Milnor $K$-invariants of an algebraic group $G$ of multiplicative type. To begin, fix a resolution \ref{ResolutionByTori} of $G$ by tori. Applying the snake lemma to the diagram
\begin{center}
	\begin{tikzcd}[column sep = .5 in, row sep = .5 in]
	1 \arrow{r} & T_{\sep}^{\ast} \arrow{d}{\cdot n}  \arrow{r}{g_{\sep}^{\ast}} & P_{\sep}^{\ast} \arrow{d}{\cdot n} \arrow{r}{f_{\sep}^{\ast}} & G_{\sep}^{\ast} \arrow{d}{\cdot n} \arrow{r} & 1 \\
	1 \arrow{r} & T_{\sep}^{\ast} \arrow{r}{g_{\sep}^{\ast}} & P_{\sep}^{\ast} \arrow{r}{f_{\sep}^{\ast}} & G_{\sep}^{\ast} \arrow{r} & 1
	\end{tikzcd}
\end{center}
yields the exact sequence of $\Gamma$-modules
\[1 \to G_{\sep}^{\ast}[n] \xrightarrow{~~} T_{\sep}^{\ast}/(T_{\sep}^{\ast})^{n} \xrightarrow{~~} P_{\sep}^{\ast}/(P_{\sep}^{\ast})^{n}, \]
and after taking $\Gamma$-fixed points we obtain the exact sequence
\[1 \to H^{0}(F, G_{\sep}^{\ast}[n]) \xrightarrow{~~} H^{0}(F, T_{\sep}^{\ast}/(T_{\sep}^{\ast})^{n}) \xrightarrow{~~} H^{0}(F, P_{\sep}^{\ast}/(P_{\sep}^{\ast})^{n}) \]
of abelian groups. Let $A = F[G], B = F[P], C = F[T]$, let $g^{\sharp} \colon C \to B, f^{\sharp} \colon B \to A$ be the associated comorphisms, and let $\alpha_{X} \colon X \to X_{\sep}$ denote the canonical base change morphism for $X = A, B, C$. For $Y = B, C$, let $\ell_{n}(Y) := \Delta_{n}(Y_{\sep})^{-1} \circ \Psi(-, n)(\alpha_{Y}) \circ \lambda_{n}(Y)^{-1}$.

\begin{prop} \label{ExactSeqCharsTors}
	The diagram
	\begin{center}
		\begin{tikzcd}
		{G^{\ast}[n]} \arrow[rrr, "\upsilon_{n}(G)"] \arrow[ddd] &  &  & {\Tors_{\nm}(T, \bmu_{n, F}}) \arrow[rrr, "{\Tors^{\ast}(g)(\bmu_{n, F})}"] \arrow[ddd, "{\ell_{n}(C)}"'] &  &  & {\Tors_{\nm}(P, \bmu_{n, F})} \arrow[ddd, "{\ell_{n}(B)}"'] \\
		&  &  &                                                                                &  &  &                                                                                \\
		&  &  &                                                                                &  &  &                                                                                \\
		{H^{0}(F, G_{\sep}^{\ast}[n])} \arrow[rrr]               &  &  & {H^{0}(F, T_{\sep}^{\ast}/(T_{\sep}^{\ast})^{n})} \arrow[rrr, "\mathcal{K}^{n}(g_{\sep}^{\sharp})"']                                                                            &  &  & {H^{0}(F, P_{\sep}^{\ast}/(P_{\sep}^{\ast})^{n})}
		\end{tikzcd}
	\end{center}
	commutes.
\end{prop}

\begin{proof}
	The right square commutes because $\Delta_{n}$ and $\lambda_{n}$ are natural transformations. To see that the left square commutes, fix $\chi \in G^{\ast}[n]$.
	Consider the commutative diagram
	\begin{center}
		\begin{tikzcd}
		1 \arrow[rr] &  & T_{\sep}^{\ast} \arrow[rr, "j_{\sep}^{\ast}"] \arrow[dd, "\Id_{T_{\sep}^{\ast}}"'] &  & P_{\sep}^{\ast} \times_{G_{\sep}^{\ast}} \mathbb{Z}/n\mathbb{Z} \arrow[rr, "(\pi_{n})_{\sep}^{\ast}"] \arrow[dd, "(\pi_{P})_{\sep}^{\ast}"] &  & \mathbb{Z}/n\mathbb{Z} \arrow[rr] \arrow[dd, "\chi_{\sep}^{\ast}"] &  & 1 \\
		&  &                                                                      &  &                                                                                &  &                                                             &  &   \\
		1 \arrow[rr] &  & T_{\sep}^{\ast} \arrow[rr, "g_{\sep}^{\ast}"']                          &  & P_{\sep}^{\ast} \arrow[rr, "f_{\sep}^{\ast}"']                                                                &  & G_{\sep}^{\ast} \arrow[rr]                                  &  & 1
		\end{tikzcd}
	\end{center}
	of $\Gamma$-modules with exact rows, and let $H$ denote the group of multiplicative type dual to $P_{\sep}^{\ast} \times_{G_{\sep}^{\ast}} \mathbb{Z}/n\mathbb{Z}$. The $F$-group morphism $j \colon H \to T$ dual to $j_{\sep}^{\ast}$ is a $\mu_{n, F}$-torsor over $T$, and we claim that $j$ represents the class $\upsilon_{n}(G)(\chi)$. Indeed, let $\pi_{n} \colon \bmu_{n, F} \to H, \pi_{P} \colon P \to H$ be the morphisms dual to $(\pi_{n})_{\sep}^{\ast}$ and $(\pi_{P})_{\sep}^{\ast}$ respectively. The morphism of $T$-schemes $P \times \bmu_{n, F} \to H$ defined on $R$-points by $(x, y) \mapsto \pi_{P}(R)(x)\pi_{n}(R)(y)$ for any $F$-algebra $R$ and any $x \in P(R), y \in \bmu_{m, F}(R)$ is constant on $G^{\chi}$-orbits. It therefore descends to a universal map $(P \times \bmu_{n, F})/G^{\chi} \to H$ over $T$, which one may check is $\bmu_{n, F}$-equivariant. \\
	Now, let $y \in P_{\sep}^{\ast}$ be such that $f_{\sep}^{\ast}(y) = \chi$, and let $z \in T_{\sep}^{\ast}$ be such that $g_{\sep}^{\ast}(z) = y^{n}$. We must show that $\Spec(C_{\sep}[X]/\langle X^{n} - z \rangle) \to \Spec(C_{\sep})$ and $j_{\sep}$ are isomorphic as $\bmu_{n, F_{\sep}}$-torsors over $T_{\sep}$. Equivalently, we must exhibit a(n) (iso)morphism of $\mathbb{Z}/n\mathbb{Z}$-graded $C_{\sep}$-algebras $s \colon C_{\sep}[X]/\langle X^{n}-z \rangle \to F_{\sep}[H_{\sep}]$. The condition that $s$ respect the $\mathbb{Z}/n\mathbb{Z}$-grading ensures that the dual morphism of schemes $H \to \Spec(C_{\sep}[X]/\langle X^{n} - z \rangle)$ is $\bmu_{n, F_{\sep}}$-equivariant, hence an isomorphism of $\bmu_{n, F_{\sep}}$-torsors. \\
	By construction, $F_{\sep}[H_{\sep}]$ is the group algebra of $H_{\sep}^{\ast} = P_{\sep}^{\ast} \times_{G_{\sep}^{\ast}} \ZZ/n\ZZ$ over $F_{\sep}$, and $C_{\sep}$ is likewise the group algebra $F_{\sep} \langle T_{\sep}^{\ast} \rangle$. The comorphism $j_{\sep}^{\sharp}$ corresponds to the $\Gamma$-module embedding $j_{\sep}^{\ast} \colon T_{\sep}^{\ast} \hookrightarrow H_{\sep}^{\ast}$. For each $v \in \ZZ/n\ZZ$, put $Q_{v} := ((\pi_{n})_{\sep}^{\ast})^{-1}(v)$. Note that $Q_{v}Q_{v'} \subset Q_{v+v'}$, and $Q_{v} = (y, [1]_{n})^{k_{v}}j^{\ast}(T_{\sep}^{\ast})$, where $k_{v} \in \NN$ is the unique representative for $v$ between $0$ and $n-1$. The $(\ZZ/n\ZZ)$-grading on $H_{\sep}$ arises from the partition
	\[H_{\sep}^{\ast} = \coprod_{v \in \ZZ/n\ZZ} Q_{v}\]
	by setting $R_{v}$ to be the $F_{\sep}$-subspace of $F_{\sep}[H_{\sep}]$ generated by $Q_{v}$. We clearly have $F_{\sep}[H_{\sep}] = \bigoplus_{v \in \ZZ/n\ZZ} R_{v}$, and $R_{v}R_{v'} \subset R_{v+v'}$ follows from $Q_{v}Q_{v'} \subset Q_{v+v'}$. Furthermore, $R_{v}$ is the $C_{\sep}$-submodule of $F_{\sep}[H_{\sep}]$ generated by $(y, [1]_{n})^{k_{v}}$. With this in mind, let $s \colon C_{\sep}[X]/\langle X^{n}-z\rangle \to F_{\sep}[H_{\sep}]$ be the universal morphism of $C_{\sep}$-algebras sending the class of $X$ to $(y, [1]_{n})$. This respects the $(\ZZ/n\ZZ)$-grading on each $C_{\sep}$-algebra, since $(y, [1]_{n})$ belongs to the $[1]_{n}$-graded component of $H_{\sep}$, and $C_{\sep}$ embeds into each algebra as the $[0]_{n}$-graded component.
\end{proof}

Since all vertical arrows of the diagram in Proposition \ref{ExactSeqCharsTors} are isomorphisms, this proves:

\begin{cor}
	The sequence
	\[1 \to G^{\ast}[n] \xrightarrow{~\upsilon_{n}(G)~} \Tors_{\nm}(T, \bmu_{n, F}) \xrightarrow{~\Tors^{\ast}(g)(\bmu_{n, F})~} \Tors_{\nm}(P, \bmu_{n, F})\]
	is exact. \qed
\end{cor}

For any smooth, connected, reductive group $R$ over $F$, define $\tilde{\Lambda}_{n}(R) \colon \Tors_{\nm}(R, \bmu_{n, F}) \to \Inv^{0}_{\hom}(R, K^{M}_{1}/n)$ by $\tilde{\Lambda}_{n}(R) = \Lambda_{n}(R) \circ \lambda_{n}(F[R])^{-1}$. As noted in section \ref{Pushforward}, the last crucial detail in our computation of $\Inv^{1}_{\hom}(G, K^{M}_{1}/n)$ is the following lemma.

\begin{lem} \label{KeyDiagramCommutes}
The diagram
\begin{center}
	\begin{tikzcd}
	{G^{\ast}[n]} \arrow[rr, "\upsilon_{n}(G)"] \arrow[dd, "{\Phi(G, n)}"] &  & {\Tors_{\nm}(T, \bmu_{n, F})} \arrow[rr, "{\Tors^{\ast}(g)(\bmu_{n, F})}"] \arrow[dd, "\tilde{\Lambda}_{n}(T)"] &  & {\Tors_{\nm}(P, \bmu_{n, F})} \arrow[dd, "\tilde{\Lambda}_{n}(P)"] \\
	&  &                                                                        &  &                                                                                &  &                                                                    \\
	{\Inv^{1}_{\hom}(G, K^{M}_{1}/n)} \arrow[rr, "{\Inv(\rho, K^{M}_{1}/n)}"']            &  & {\Inv^{0}_{\hom}(T, K^{M}_{1}/n)} \arrow[rr, "{\Inv(g, K^{M}_{1}/n)}"']                                                        &  & {\Inv^{0}_{\hom}(P, K^{M}_{1}/n)}
	\end{tikzcd}
\end{center}
commutes.
\end{lem}

\begin{proof}
	Unwinding the definitions of $\tilde{\Lambda}_{n}(T)$ and $\tilde{\Lambda}_{n}(P)$, one sees that the commutativity of the right square is a consequence of the functoriality of the pullback map on torsors. To be precise, if $\alpha \colon Y \to X, \beta \colon Z \to Y$ are morphisms of $F$-schemes, then $\Tors^{\ast}(\alpha \circ \beta) = \Tors^{\ast}(\beta) \circ \Tors^{\ast}(\alpha)$. The left square commutes because pullback operation on torsors commutes with changing the group.
\end{proof}

As noted at the end of section \ref{Pushforward}, after a diagram chase, this proves:

\begin{thm} \label{TypeOneGMTModN}
	The map $\Phi(G, n) \colon G^{\ast}[n] \to \Inv^{1}_{\hom}(G, K^{M}_{1}/n)$ is an isomorphism. \qed
\end{thm}

As was the case for type-zero invariants, for any natural number $n$, there is a group morphism $\Inv^{1}(G, \iota_{n}) \colon \Inv^{1}_{\hom}(G, K^{M}_{1}/n) \to \Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)$ given by composition with $\iota_{n}$. For positive integers $n, m$ with $n$ dividing $m$, the maps $\Inv^{1}(G, \iota_{n}), \Inv^{1}(G, \iota_{m})$ are compatible with the map $\Inv^{1}(G, \beta_{n, m}) \colon \Inv^{1}_{\hom}(G, K^{M}_{1}/n) \to \Inv^{1}_{\hom}(G, K^{M}_{1}/m)$ given by composition with $\beta_{n, m}$, and so we obtain a universal induced map
\[\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}(G, \iota_{n}) \colon \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}_{\hom}(G, K^{M}_{1}/n) \to \Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ).\]

\begin{prop} \label{ColimTypeOne}
	The map $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}(G, \iota_{n})$ is an isomorphism.
\end{prop}

\begin{proof}
	Set
	\[u = \displaystyle \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv(\rho, K^{M}_{1}/n), v = \displaystyle \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv(g, K^{M}_{1}/n), \]
	\[u' = \Inv(\rho, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ), v' = \Inv(g, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ).\]
	We have a commutative diagram
	\begin{center}
		\begin{tikzcd} [column sep = 0.2 in]
		{\displaystyle \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}_{\hom}(G, K^{M}_{1}/n)} \arrow[rrr, "u"] \arrow[ddd, "{\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}(G, \iota_{n})}"'] &  &  & {\displaystyle \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}_{\hom}(T, K^{M}_{1}/n)} \arrow[rrr, "v"] \arrow[ddd, "{\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(T, \iota_{n})}"'] &  &  & {\displaystyle \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}_{\hom}(P, K^{M}_{1}/n)} \arrow[ddd, "{\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(P, \iota_{n})}"'] \\
		&  &  &                                                                                &  &  &                                                                                &  &  &                                                                                \\
		&  &  &                                                                                &  &  &                                                                                &  &  &                                                                                \\
		{\Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)} \arrow[rrr, "u'"']                                                                   &  &  & {\Inv^{0}_{\hom}(T, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)} \arrow[rrr, "v'"']                                                                   &  &  & {\Inv^{0}_{\hom}(P, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)}
		\end{tikzcd}
	\end{center}
	whose rows are exact. Since $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(T, \iota_{n})$ and $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{0}(P, \iota_{n})$ are isomorphisms by Proposition \ref{ColimTypeZero}, and $u, u'$ are injective, $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}(G, \iota_{n})$ is an isomorphism.
\end{proof}

\begin{thm} \label{TypeOneGMTQZ}
	The map $\Phi(G) \colon G^{\ast}_{\tors} \to \Inv^{1}_{\hom}(G, K^{M}_{1} \otimes_{\ZZ} \QQ/\ZZ)$ is a group isomorphism.
\end{thm}

\begin{proof}
	Let $n, m$ be positive integers with $n$ dividing $m$, and let $\tau_{n, m} \colon \bmu_{n, F} \to \bmu_{m, F}$ be the canonical embedding. We claim that the diagram
	\begin{center}
		\begin{tikzcd}
		{G^{\ast}[n]} \arrow[ddd, "{\Phi(G, n)}"'] \arrow[rrr, "{\sigma_{n, m}}"]     &  &  & {G^{\ast}[m]} \arrow[ddd, "{\Phi(G, m)}"] \\
		&  &  &                                           \\
		&  &  &                                           \\
		{\Inv^{1}_{\hom}(G, K^{M}_{1}/n)} \arrow[rrr, "{\Inv^{1}(G, \beta_{n, m})}"'] &  &  & {\Inv^{1}_{\hom}(G, K^{M}_{1}/m)}
		\end{tikzcd}
	\end{center}
	commutes, where $\sigma_{n, m}$ is the group morphism given by composition with $\tau_{n, m}$. Indeed, it is sufficient to show that $\Sigma_{m}(L) \circ \beta_{n, m}(L) = \Tors_{\ast}(\tau_{n, m})(L) \circ \Sigma_{n}(L)$ for any field extension $L/F$. Fixing $[y] \in L$, put $U = \Spec(L[X]/\langle X^{n} -y \rangle), V = \Spec(L[X]/\langle X^{m} - y^{m/n} \rangle)$. The morphism of $L$-schemes $U \times \bmu_{m, L} \to V$ defined functorially by
	\[U(R) \times \bmu_{m, L}(R) \to V(R), (u, z) \mapsto uz\]
	for any $L$-algebra $R$ is constant on $\bmu_{n, L}^{\tau_{n, m}}$-orbits, and so descends to a morphism of $L$-schemes $(U \times \bmu_{m, L})/(\bmu_{n, L}^{\tau_{n, m}}) \to V$, which one may check is $\bmu_{m, L}$-equivariant. This establishes that $\Tors_{\ast}(\tau_{n, m})(L)(U) = V$. \\
	The universally induced map $\operatornamewithlimits{colim}\limits_{n \in \NN} \Phi(G, n) \colon G^{\ast}_{\tors} \to \operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}(G, K^{M}_{1}/n)$ is an isomorphism, as $\Phi(G, n)$ is an isomorphism for each $n$. Since $\Phi(G)$ is just the composition of $\operatornamewithlimits{colim}\limits_{n \in \NN} \Phi(G, n)$ with the $\operatornamewithlimits{colim}\limits_{n \in \NN} \Inv^{1}(G, \iota_{n})$, it is an isomorphism by Proposition \ref{ColimTypeOne}.
\end{proof}

\bibliographystyle{amsalpha}
\bibliography{References}

config: arxiv


TITLE: Different graph results for Integral of $ \frac{1}{x^2-8x+1}$
QUESTION [0 upvotes]: I had to do the integral of $ \frac{1}{x^2-8x+1} $.
The approach I followed is in the following MIT video, they didn't finish writing the final integral but I did. However, after checking the answer from Wolfram and Integral-Calculator.com I get results that look a little bit different in a graph. Is my result wrong? How can the graphs be so different even between the answer of Wolfram and Integral-Calculator.com?
I graphed them all in Desmos as well: desmos.com/calculator/fftyt801sf
My result using the approach from MIT
$$ -\frac{\sqrt{15}}{15}\cdot\ln\left(\left|\frac{x-4+\sqrt{15}}{\sqrt{\left(x-4\right)^{2}-15}}\right|\right) + C $$

Wolfram

Integral Calculator.com

REPLY [0 votes]: Since
$$
x^2-8x+1=(x-4)^2-15=(x-4-\sqrt{15})(x-4+\sqrt{15})
$$
we have
\begin{eqnarray}
\frac{1}{x^2-8x+1}&=&\frac{1}{(x-4-\sqrt{15})(x-4+\sqrt{15})}\cr
&=&\frac{1}{2\sqrt{15}}\left(\frac{1}{x-4-\sqrt{15}}-\frac{1}{x-4+\sqrt{15}}\right).
\end{eqnarray}
It follows that
\begin{eqnarray}
\int \frac{1}{x^2-8x+1}\,dx&=&
\frac{1}{2\sqrt{15}}\int \left(\frac{1}{x-4-\sqrt{15}}-\frac{1}{x-4+\sqrt{15}}\right)\, dx\cr
&=&\frac{1}{2\sqrt{15}}\left(\ln|x-4-\sqrt{15}|-ln|x-4+\sqrt{15}|\right)+C\cr
&=&\frac{1}{2\sqrt{15}}\ln\left|\frac{x-4-\sqrt{15}}{x-4+\sqrt{15}}\right|+C
\end{eqnarray}