TITLE: $\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q}$ is separable but not normal, $\mathbb{F}_p(T)/\mathbb{F}_p(T^p)$ is normal but not separable
QUESTION [0 upvotes]: I just begun to learn algebraic extension. In some notes, I see two examples.

Example 1. The polynomial $X^3-2$ has one real root $\sqrt[3]2$ and two non real roots in $\mathbb{C}$. Therefore, the extension $\mathbb{Q}[\sqrt[3]2]/\mathbb{Q}$ (which is separable) is not normal.

How can I show it is separable and not normal?

Example 2. The extension $\mathbb{F}_p(T)/\mathbb{F}_p(T^p)$ (which is normal) is not separable because the minimal polynomial of $T$ is not separable.

How can I show it is normal but not separable?

REPLY [0 votes]: An extension $E/F$ is separable if for every element $\alpha\in E$, the minimal polynomial of $\alpha$ over $F$ has no multiple roots. A test for multiplicities of roots can be carried out by putting in the roots in the formal derivative of the polynomial. More concretely, if $\{\alpha_1,...,\alpha_n\}$ are all the roots of $f(x)$, then $f(x)$ has no multiple roots if and only if $f'(\alpha_i)\neq 0$ for all $i$. In the case of an extension over $\mathbb Q$, or more generally over a field of charactersitic $0$, any extension is separable. This follows from the fact that any irreducible polynomial over $\mathbb Q$ is separable. If $f(x)$ is irreducible and contains a multiple root $\alpha$, then $\alpha$ is also a root of $f'(x)$, hence $f$ divides $f'$ by irreducibility. This is a contradiction since $f'$ is not identically $0$, so it has positive but strictly smaller degree than $f$.
The field extension $\mathbb Q(2^{\frac13})/\mathbb Q$ is not normal because it does not contain all the roots to the polynomial $x^3-2$. As $\mathbb Q(2^{\frac 13})$ is a real field, so it cannot contain the non-real roots.
For the second extension, the minimal polynomial of $T$ is given by $x^p-T^p\in\mathbb F_p(T^p)[x]$. It has multiple roots because in characteristic $p$, $x^p-T^p=(x-T)^p$ in $\mathbb F_p(T)$. (You can prove that this polynomial is irreducible by showing that $(x-T)^k\notin\mathbb F_p(T^p)[x]$ for any $k<p$.) Since this polynomial is non-separable, the field extension is not separable. To see that it is normal, simply notice that $\mathbb F_p(T)=\mathbb F_p(T^p)(T)$, i.e. the field extension is obtained by adjoining the single root $T$. But by the above discussion, $T$ is the only root of the minimal polynomial of $T$. So it is in fact a splitting field, which is normal.


TITLE: Compact perfect sets without regular closed subsets
QUESTION [1 upvotes]: Let $K$ be a non-empty compact perfect set of reals.

Question. Is it possible that $K$ is not totally disconnected and does not contain regular closed subsets?

Intuitively the answer should be negative (removing each time the closed interval subsets).
It is a variant of this question, where I was too late to make the necessary edits to avoid the easy answer.

REPLY [2 votes]: Every subspace contains regular closed subsets: just take the closure of any open set. These are regular closed in the subspace topology.
If $K$ is perfect in the reals, then it either contains an interval or it is totally disconnected. If the former holds, this closed interval is regular closed (in the reals as well) and if the latter, no closed subset of $K$ can be perfect in the reals.


TITLE: Finding the maximal area ellipse in a closed area
QUESTION [1 upvotes]: The task is the following: Find the ellipse with maximal area.
The center of the ellipse is given. Several inequalities are also given, and all of the ellipse's points must satisfy all of the inequalities. So basically I need to find the maximal area ellipse which is in the area enclosed by the inequalities. Here is a picture for illustration: https://ibb.co/LS93208
In my example the center is (Cx,Cy)=(1,1), and the inequalities are the following:

y <= 0.5x + 4
y >= (1/3)x - 3
y >= -0.5x - 4
y <= -0.25x + 4

Any idea how can you solve this? I've tried creating an optimization model where the objective function is the area(width*height) and the three decision variables are width(x2),height(x3),rotation angle(x1) - these 3 determine the ellipse -, but I can't seem to write up the right constraints. I share the model I created, but it is probably wrong. The objective function is x2*x3, you have to maximize it.
https://ibb.co/RSLQ05L
The first 2 equations on the picture is the general ellipse equation, followed by the 4 inequalities(alpha between 0 and 2pi). Then I substituted x(alpha) and y(alpha) into the inequalities(I don't know if it's a right move), and moved everything to left side so the right side is 0. The result is:
https://ibb.co/TbcM2dM
I can't manage to solve this optimization problem in matlab, so I'm on a wrong track probably and you don't solve this task as an optimization problem.

REPLY [0 votes]: Just in case anyone interested, my optimization model was correct and could be solved in matlab(fseminf function), also can be generalized to higher dimension


TITLE: Factorising a determinant
QUESTION [0 upvotes]: I have read many questions on this website earlier regarding factorization of a determinant but i have one small doubt that has been troubling me a lot. For example if you had to factorize this determinant
\begin{vmatrix}
x&y&z\\
x^2&y^2&z^2\\
x^3&y^3&z^3\\
\end{vmatrix}
I understood that if x=y the determinant evaluates to 0 and therefore by factor theorem $x-y$ is a factor. However when we put y=x then we also get the determinant as 0 so shouldn't $y-x$ also be a factor thereby making $(x-y)^2$ as a factor of the determinant?

REPLY [2 votes]: Here's a way to factorise your determinant. By your approach, you should be able to see why $x,y,z,x-y,y-z,x-z$ must all be its factors. So the det must have
$$xyz(x-y)(y-z)(x-z)$$
as a factor. So we have
$$\det A=f(x,y,z)xyz(x-y)(y-z)(x-z)$$
Now if $f$ viewed as a polynomial in $x$ has $\deg_x f>0$, then $\det A$ viewed as a polynomial in $x$ would have degree greater than or equal to $4$, which is impossible. Same for $y,z$, so $f$ must be constant. To determine its value, again we view $\det A$ as a polynomial in $x$, what should be its leading coefficient?
As a caveat: this particular method works only for particular determinants, because it can only find the linear factors of the determinant, but in general the factors of a determinant aren't all necessarily linear, of course.


TITLE: For a surjective bounded linear operator, does $U^{*}U = I_X$ imply $UU^{*} = I_Y$?
QUESTION [0 upvotes]: Let X, Y be Hilbert spaces and $U\in\mathcal{B}(X,Y)$ be a surjective bounded linear operator between the two spaces. Given that the adjoint $U^{*}\in\mathcal{B}(Y,X)$ is such that $U^{*} U = I_X$ does it follow that $U U^{*} = I_Y$?

REPLY [0 votes]: Since $U^*U=1$, $U$ is injective. If $U$ is also surjective, then $U$ is invertible by the Open Mapping Theorem. Hence $U^*=U^{-1}$, and $UU^*=1$.
Note that this is general for Banach spaces $X$,$Y$.


TITLE: Why the degree of $p/|p|$ is zero on $\partial W^\prime?$
QUESTION [0 upvotes]: Let $W$ be a smooth compact region in $\mathbb{C}$ whose boundary contains no zeros of the polynomial $p$. $p$ has only   finitely many roots, $z_0, \dots, z_n$ in $W$. Around each $z_i$, circumscribe a small closed disk $D_i$, making the disks all disjoint from each other and from the boundary. Then $p/|p|$ is defined on
  $$W^\prime = W - \bigcup_{i=0}^n D_i.$$
  Thus,
  $$\partial W^\prime = \partial W - \bigcup_{i=0}^n \partial D_i.$$

Then why the degree of $p/|p|$ is zero on $\partial W^\prime?$
One thought...
$$\frac{p(z)}{|p(z)|}\quad \text{and}\quad \frac{z^m}{|z^m|}=\left(\frac{z}{r}\right)^m$$
are homotopic maps of $S \to S^1$. Thus $p/|p|$ must have the same degree as $(z/r)^m$ - namely, $m$. So I find trouble concluding degree is $0$ from the boundary condition.
A second thought...
Should I use the fact that $\deg\Big(\frac{p(z)}{|p(z)|}\Big) = I\Big(\frac{p(z)}{|p(z)|},\{y\}\Big), y \in \partial W^\prime$, the I am not sure how to show $\Big(\frac{p(z)}{|p(z)|}\Big)^{-1}(\{y\}) = 0.$ In fact what I got is not necessarily equal to 0, consider the preimage of a point on the unit circle, whose preimage may be multiple points.

REPLY [2 votes]: The map $p/|p|$ is a smooth map $W'\to S^1$, hence by the Boundary Theorem its degree on the boundary is $0$. This argument works in any dimension.


text:
\subsection{Error estimates}
\label{sec:est}
\noindent Regarding the equations \eqref{eq:bie_mu} and \eqref{eq:bie_u}, they both contain terms $(z_j-z_i)^{-1}$. For $M^{(1)}$ and $M^{(2)}$ the limits when $i=j$ are well-defined, and for the term $\Re\left\{ \frac{z_j^\prime}{z_j-z_i}\right\}$ singularity subtraction is used, see \eqref{eq:numm_udisc}. However, when droplets get close to each other, i.e. when $\|z_i-z_j\|\ll 1$ for points $z_i$ and $z_j$ on different interfaces, the integrals are said to become near-singular. Analytically this is not an issue, but numerically large errors are introduced as the integrand gets harder to resolve.

To demonstrate how these errors grow, Stokes equations are solved in a fixed domain with Dirichlet boundary conditions. This problem contains the same near-singular behaviour as \eqref{eq:bie_u}. Given any analytical solution to Stokes equations (for example, that generated by a point source forcing as shown in Figure~\ref{fig:est_domu}), the Dirichlet velocity data can be evaluated at the boundary and used as boundary conditions for the numerical method. The boundary integral formulation and exact problem setting is described in \ref{sec:app_est}, the domain and solution $u(z)$ are shown in Figure~\ref{fig:est_domu}.
\begin{figure}[h!]
	\subfigure[Domain $\Omega$, source points and solution $u(z)$.]{
	\centering
	\includegraphics[width=0.49\textwidth]{graphics/bigplot_u_NEW180607.png}
		\label{fig:est_domu}}
	\hfill
	\subfigure[Error in first quadrant of domain.]{
	\includegraphics[width=0.49\textwidth]{graphics/smallplot_error_nps25_NEW180607.png}\label{fig:est_err}}
	\caption{Example of analytical solution $u(z)$ and error to \eqref{eq:est_u} given sources ($x_1$,$f_1$), ($x_2,f_2$) and ($x_3$,$f_3$) (black markers) in a domain $\Omega$. Solution is computed with a composite 16-point Gauss-Legendre quadrature with 25 panels.}
\end{figure}

The relative error of $u(z)$ compared to the exact solution, using $25$ Gauss-Legendre panels, is shown in Figure~\ref{fig:est_err}. Here only the first quadrant is shown, as the error behaves identical in the other three. It is clear that at an evaluation point $z_0$ close to the boundary, errors are large and other treatments of the integrals are needed for these cases. In this paper a special quadrature is used, which is explained in \S\ref{sec:specq}. How close to the panel the errors become large depends on the refinement of the discretisation.  In order to know when special treatment is needed, the quadrature errors of such near-singular integrals can be estimated. This was originally done by \citeauthor{AfKlinteberg2017}, where estimates for the quadrature errors for Laplace's and Helmholtz equations were derived in \cite{AfKlinteberg2017,klintebergadapt}. Using the same approach based on contour integration and calculus of residues for integrals of the type appearing in \eqref{eq:bie_u}, the errors when computing $u(z)$ can be estimated also for Stokes equations. Details of this can be found in \ref{sec:app_est}. The estimates follow the error levels remarkably well, as is shown in Figure~\ref{fig:est_est}. Both errors and estimates for $25$ and $50$ panels are shown, and it is clear that refining the interface discretisation makes the region of large errors narrower, but will not eliminate it.
\begin{figure}[h!]
	\centering
	\subfigure[$25$ panels]{
	\includegraphics[width=0.49\textwidth,trim={1cm 2cm 0 2cm},clip]{graphics/smallplot_est_nps25_NEW180607.png}}
	\;
	\hspace{-1cm}
	\subfigure[$50$ panels]{
	\includegraphics[width=0.49\textwidth,trim={1cm 2cm 0 2cm},clip]{graphics/smallplot_est_nps50_NEW180607.png}}
	\caption{Computed error estimates in black for error levels $10^{-p}$, where $p=14,12,\hdots,2,0$. Measured errors in colour.}
	\label{fig:est_est}
\end{figure}

config: arxiv


TITLE: Prove or confute the follow proposition
QUESTION [2 upvotes]: I have to prove or confute the follow proposition:
Let $F$ a field and $ f: $$\mathbb Z \rightarrow F$ an homomorphism of rings such that $f(1_\mathbb Z ) =1_F$.
If $f$ isn't injective $\Rightarrow$ $F$ is finite

This is my work :
$f$ is an homomorphism of rings $\Rightarrow Ker f $ is an ideal of $\mathbb Z \Rightarrow $ $Ker f = n \mathbb Z $
$n \in \mathbb N$ is uniquely determined and is the characteristic of the field $F$
Such that $f$ is not injective $\Rightarrow Ker f \neq  (0)$ then $n \neq 0$
Now I can't conclude that $F$ is infinite neither that is finite.
If I take $\mathbb Z_2(x)$ I know that is a field infinite with characteristic 2, so I found an infinite field with finite characteristic.
Is it correct? I forgot something important to say?
How can I prove that $\mathbb Z_2 (x) $ is an infinite field?

REPLY [0 votes]: The field $\Bbb Z_2(x)$ is an infinite field since $1,x,x^2,x^3\ldots\in\Bbb Z_2(x)$. And this proves that the statement is false.


TITLE: Inequality for trace of product of matrices
QUESTION [3 upvotes]: Assume that $A \in \mathbb{R}^{n \times n}$ is a symmetric matrix and $B \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix. Is the following statement true
$$
\lambda_{\mathrm{min}} \operatorname{tr} A \le \operatorname{tr} (AB) \le \lambda_{\mathrm{max}} \operatorname{tr} A \, ?
$$
Here, $\lambda_{\mathrm{min}}$ denotes the smallest eigenvalue of $B$ and $\lambda_{\mathrm{max}}$ denotes the largest eigenvalue of $B$.

REPLY [7 votes]: The statement is not true, since $A=-I_2$ and $B=\text{diag}(1,2)$ is an obvious counterexample. But if we assume that $A$ is positive definite and $B$ is symmetric, then the statement is true. (In fact, $A\ge O$ is also necessary for the statement to be true.) We begin with the following observation.

If $A,B$ are symmetric, positive definite matrices, then $\text{tr}(AB)\ge 0$.

Let $\sqrt{A}$ denote the square root of $A$. Then,
$$\begin{align*}
\text{tr}(AB)&=\text{tr}(\sqrt{A}\cdot\sqrt{A}^TB)\\&=\text{tr}(\sqrt{A}^TB\cdot\sqrt{A})\\
&=\sum_{i=1}^n e_i^T \sqrt{A}^TB\sqrt{A}e_i\ge 0
\end{align*}$$
since $B$ is positive definite. $\blacksquare$
Knowing this, note that if $A\ge O$ and $B$ is symmetric, then$$
B-\lambda_{\text{min}}I\ge O,\quad \ \lambda_{\text{max}}I-B\ge 0.
$$ Thus,$$
\text{tr}(A(B-\lambda_{\text{min}}I))=\text{tr}(AB)-\lambda_{\text{min}}\text{tr}(A)\ge 0,
$$ and
$$
\text{tr}(A(\lambda_{\text{max}}I-B))=\lambda_{\text{max}}\text{tr}(A)-\text{tr}(AB)\ge 0.
$$


TITLE: Punctured spectrums of local rings
QUESTION [8 upvotes]: Let $A$ be a local ring with the unique maximal ideal $\mathfrak{m}$. The punctured spectrum of $A$ is the open subset $\text{Spec}(A)\setminus \{\mathfrak{m}\}$.  I have seen many papers (for instance Horrocks' papers) studying vector bundles over algebraic varieties (in particular, projective spaces) by putting them over a punctured spectrum of a local ring.
However, I am wondering why punctured spectrum is better than varieties in this satiation. I feel that geometric pictures of varieties are clearer than that of a punctured spectrum. More essential, are the categories of coherent sheaves over a variety and its punctured spectrum equivalent? How much information about coherent sheave (in particular, vector bundles) can be recovered from punctured spectrums? For second question, I am thinking examples that reflect some relations between those two gadgets.

REPLY [6 votes]: Here is a general principle. If a problem is formulated over a general (say noetherian) base, by localization arguments one can often reduce to the case of a local base. Arguing by induction on the dimension of the base, we can exploit the fact that the punctured spectrum, while often not affine, is of smaller dimension!  Then we can try to "extend" the solution over the punctured spectrum to a solution over the local spectrum (by taking into account local cohomology or other tools).  I think this brilliant idea is due to Grothendieck (truly exploiting the concept of a scheme).
For example, observe that the punctured spectrum of a 2-dimensional regular local ring is a (typically non-affine!) Dedekind scheme, over which flatness is a hands-on concept. Much deeper, the proof of Zariski's Main Theorem in EGA IV$_3$ goes via dimension induction on the base using the above trick with a punctured spectrum having smaller dimension.
For the study of coherent sheaves, the categories on the spectrum and punctured spectrum are not at all equivalent, though the situation for vector bundles is more accessible; the problem is governed by local cohomology and depth considerations (so the situation is more tractable over a regular base).  This is studied in depth (no pun intended) in the middle of SGA2.


TITLE: Mutually singular measures.
QUESTION [2 upvotes]: We say measures $\mu,  \nu$ are mutually singular on $E$ if $E$ can be written as disjoint union, $E= E_1\cup E_2$ with $\mu(E_1) = \nu(E_2) = 0$.

Prove that the following condition implies $\nu$ and $\mu$ are mutually singular on $E$:  Given any $\epsilon >0$, there are disjoint measurable $E_1$ and $E_2$ with $E=E_1\cup E_2$ and $\mu(E_1)<\epsilon$, $\nu(E_2)<\epsilon$.

I think it might need to use some $\limsup$ of sets, but I don't know how to construct $E_1$ and $E_2$ with measure $0$ respectuvely.

REPLY [3 votes]: Yes, $\limsup$ is a good idea. Let's use $A$ and $B$ for the sets rather than proliferating indices on $E$.
Choose a sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ of positive numbers. By assumption, for every $n$ there are disjoint measurable $A_n, B_n$ with $E = A_n \cup B_n$ and $\mu(A_n) < \varepsilon_n$, $\nu(B_n) < \varepsilon_n$. Now let
$$A = \limsup_{n \to \infty} A_n\quad \text{and}\quad B = \limsup_{n \to \infty} B_n\,.$$
Then $A,B$ are measurable with $A \cup B = E$ (if $x \notin A$, there is an $N_x$ such that $x \notin A_n$ for all $n \geqslant N_x$, and hence $x \in B$), but in general we have $A \cap B \neq \varnothing$. That doesn't matter however. If $\nu(B) = 0$, then also $\nu(E\setminus A) =  0$ since $E\setminus A \subset B$, so we can take $E_1 = A$ and $E_2 = E \setminus A$.
It remains to see that we can choose $(\varepsilon_n)$ in such a way that $\mu(A) = 0 = \nu(B)$. Sufficient for that is
$$\sum_{n = 0}^{\infty} \varepsilon_n < +\infty\,.$$


TITLE: Find a basis for a set of vectors
QUESTION [0 upvotes]: I'm trying to complete the following exercise:
Let $V$ be the set of vectors $(x_1,x_2,x_3,x_4)\in\mathbb{R}^4$ such that $$2x_1-3x_2-x_3+x_4=0$$ $$x_1-x_2+2x_3-x_4=0$$ Show that $V$ is a subspace of $\mathbb{R}^4$ and find a basis for $V$.
I already showed that $V$ is a subspace of $\mathbb{R}^4$, but I'm having trouble finding a basis for $V$. Any help is welcome

REPLY [1 votes]: The sum of the two equations gives
$$x_3=4x_2-3x_1$$
the second one yields to
$$x_4=x_1-x_2+2x_3$$
$$=x_1-x_2+2(4x_2-3x_1)$$
$$=-5x_1+7x_2$$
thus
$$(x_1,x_2,x_3,x_4)\in V\iff$$
$$(x_1,x_2,x_3,x_4)=$$
$$(x_1,x_2,-3x_1+4x_2,-5x_1+7x_2)=$$
$$x_1(1,0,-3,-5)+x_2(0,1,4,7)=$$
$$x_1\vec{u}+x_2\vec{v}$$
$\vec{u} $ and $ \vec{v} $ are clearly independent, so $ (\vec{u},\vec{v}) $ is a basis of $ V$ and dim$(V)=2$.


TITLE: Why is the set of rationals $Q$, as a subspace of $R$ with the usual (Euclidean) topology, is not locally compact
QUESTION [0 upvotes]: I know this question is very popular but i couldnt find direct proof by using the definition of locally compactness. Every proof is with sequences. How can we proof this by using just definition of locally compactness?

REPLY [1 votes]: Let $A\subseteq\Bbb R$, and suppose that $A$ is locally compact. Then each $x\in A$ has a compact nbhd $K_x$. The set $K_x$ has the same relative topology in $\Bbb R$ that it has in $A$, so it is a compact subset of $\Bbb R$ and therefore a closed set in $\Bbb R$. Since $K_x$ is a nbhd of $x$ in $A$, there is some open interval $(a,b)$ in $\Bbb R$ such that $x\in(a,b)\cap A\subseteq K_x$, and $K_x$ is closed in $\Bbb R$, so $\operatorname{cl}_{\Bbb R}\big((a,b)\cap A\big)\subseteq K_x$.
Now suppose further that $A$ is dense in $\Bbb R$. Then $\operatorname{cl}_{\Bbb R}\big((a,b)\cap A\big)=[a,b]$, so $x\in(a,b)\subseteq K_x\subseteq A$, and $A$ is therefore open in $\Bbb R$.
We have now shown that any locally compact, dense subset of $\Bbb R$ is a dense open subset of $\Bbb R$, and its complement is therefore a nowhere dense closed subset of $\Bbb R$. Since the irrationals are neither closed nor nowhere dense in $\Bbb R$, $\Bbb Q$ cannot be locally compact.