TITLE: How to determine when this is a well-defined homomorphism between cyclic groups
QUESTION [2 upvotes]: I've been having trouble with this concept. I'm not very familiar with cyclic group structure. Question: Let $Z_{36} = \langle x \rangle$. For which integers $a$ does the map $\phi_a$ defined by $\phi_a : \bar{1} \to x^a$ extend to a well-defined homomorphism from $\mathbb{Z}/48\mathbb{Z}$ to $Z_{36}$? I know firstly that $\phi(a \star b) = \phi(a) * \phi(b)$ must hold for all $a,b \in \mathbb{Z}/48\mathbb{Z}$.
Also, is the domain of $\phi_a$ all of $\mathbb{Z}/48\mathbb{Z}$? I'm not sure what's going on. Any help appreciated.

REPLY [0 votes]: First, given that neither 36 nor 48 is prime, these must be the additive cyclic groups (since every element of the group must have an inverse) so it might simplify things to write this out additively as:
$$\phi_a:\mathbb{Z}/48\mathbb{Z}\rightarrow\mathbb{Z}/36\mathbb{Z}$$ defined by
$$\phi_a(1)=ax$$ and the homomorphism quality that $$\phi_a(b+c)=\phi_a(b)+\phi_a(c)$$ as opposed to writing it generally with $*$ and $\star$
Now it seems much more clear that given $\phi_a$ is a homomorphism, we've got:
$$\phi_a(\alpha)=\phi_a(1+...+1)=\phi_a(1)+...+\phi_a(1)=\alpha\phi_a(1)$$
The question now becomes: For which values of $a\in\mathbb{Z}$ does the homomorphism quality hold? Assuming it does hold we see:
$$\phi_a(b+c)=\phi_a(b)+\phi_a(c)=b\phi_a(1)+c\phi_a(1)=b(ax)+c(ax)=(b+c)(ax)$$
which is true for $\forall a \in \mathbb{Z}$. The domain of any function must be the entire set, so yes, the domain is all of $\mathbb{Z}/48\mathbb{Z}$. It's been a while since I've done much group theory, so somebody else should certainly confirm that what I've written is accurate, but as far as I'm concerned, $a$ could be any integer. Cheers!


text:
\newcommand{\milnor}{\mu_0}
\newcommand{\newton}{\nu}

\section{Introduction}
The modern theory of applications of Newton polyhedra to affine B\'ezout problem started from A.\ Kushnirenko's work aimed at answering V.\ I.\ Arnold's question on {\em Milnor numbers} of generic singularities. In \cite{kush-poly-milnor} Kushnirenko gave a beautiful formula for a lower bound of the Milnor number at the origin in terms of volumes of the region bounded by the Newton diagram, and showed that the bound is attained in the case that the singularity is {\em Newton non-degenerate}. In this chapter we reverse the arrow of applications and show that the notion of {\em non-degeneracy at the origin} introduced in \cref{multiplicity-chapter} can be used to derive (and generalize) Kushnirenko's result on Milnor numbers. In particular, based on non-degeneracy at the origin we introduce a non-degeneracy criterion which generalizes Newton non-degeneracy and {\em inner Newton non-degeneracy}\footnote{This terminology is taken from \cite{boubakri-greuel-markwig}.}, the latter introduced by C.\ T.\ C.\ Wall \cite{wall}. We show that in zero characteristic the new criterion is necessary and sufficient for the Milnor number to be the minimum, and the minimum Milnor number can be obtained by Kushnirenko's bound. In positive characteristic this criterion is sufficient, but not necessary.

\section{Milnor number} \label{milnor-defn}

The {\em Milnor number} $\milnor(f)$ of a power series $f$ in $(x_1, \ldots, x_n)$ is the dimension over $\kk$ of the quotient of $\kk[[x_1, \ldots, x_n]]$ by the ideal generated by the partial derivatives of $f$, i.e.\ $\milnor(f) = \multzero{\partialxone{f}}{\partialxn{f}}$. It is a fundamental measure of complexity of the singularity of $V(f)$ at the origin (in the case that $f$ is the Taylor series of a rational function, or in the case that $\kk = \cc$ and $f$ is a analytic at the origin).

\begin{prop} \label{milnor-prop}
Let $f \in \kk[x_1, \ldots, x_n]$ such that $f(\origin) = 0$.
\begin{enumerate}
\item \label{milnor-0}$\milnor(f) = 0$ if and only the origin is a nonsingular point of $V(f)$.
\item \label{milnor-infty} If $V(f)$ has a non-isolated singularity at the origin, then $\milnor(f) = \infty$. The converse holds if $\character(\kk) = 0$.
\end{enumerate}
\end{prop}

\begin{proof}
The first assertion is clear, so we prove the second assertion. If $V(f)$ has a non-isolated singularity at the origin, then the origin is a non-isolated point of $V(\partialxone{f}, \ldots, \partialxn{f})$, so that $\milnor(f) = \infty$ (\cref{int-mult-curve}). Now assume $\milnor(f) = \infty$. Then the origin is a non-isolated point of $V(\partialxone{f}, \ldots, \partialxn{f})$ (\cref{int-mult-curve}) and therefore there is an irreducible curve $C$ containing the origin such that $C \subseteq V(\partialxone{f}, \ldots, \partialxn{f})$ (\cref{closure-curve-lemma}). It suffices to show that $f|_C \equiv 0$ if $\character(\kk) = 0$. Pick a nonsingular point $a \in C$. Then there is an isomorphism $\phi: \kk[[t]] \cong \hatlocal{C}{a}$, and $d(f \circ \phi)/dt = \sum_{j=1}^n (\partialxj{f} ) (d(\phi_j)/dt) \equiv 0 \in \hatlocal{C}{a}$. Since $\character(\kk) = 0$, it follows that $f$ is constant on $C$. Since $\origin \in C$ and $f(\origin) = 0$, it follows that $C \subseteq V(f)$, as required.
\end{proof}

\begin{example}
The converse to assertion \eqref{milnor-infty} of \cref{milnor-prop} may not be true if $p := \character(\kk)$ is positive; consider e.g.\ the case that $n = 1$ and $f(x) = x^p$, or $n = 2$ and $f(x,y) = x^p + y^q$, where $q \geq 2$ is relatively prime to $p$. The latter example in particular shows that Milnor number can be infinite even for {\em isolated} singular points.
\end{example}

In the case that $\kk= \cc$ and the origin is an isolated singular point of $V(f)$, Milnor originally defined $\milnor(f)$ in \cite[Chapter 7]{milnor} in the following way: let $S^{2n-1}_\epsilon$ be the sphere of radius $\epsilon$ centered at the origin of $\cc^n \cong \rr^{2n}$ and $S^{2n-1} := S^{2n-1}_1$ be the unit sphere of $\cc^n$. Given a morphism $g: \cc^n \to \cc^n$ such that the origin is an isolated zero of $g^{-1}(\origin)$, the {\em multiplicity} of $g$ at the origin is the {\em degree of the mapping}\footnote{The degree of a differentiable map $\phi: M \to N$ between oriented differentiable manifolds of the same dimension, where $M$ is compact and $N$ is connected, is the sum of {\em sign} of $df_x$ over all $x \in \phi^{-1}(y)$ for a generic $y \in N$, where $df_x$ is the derivative map from the tangent space of $M$ at $x$ to the tangent space of $N$ at $y$, and the sign of $df_x$ is either $1$ or $-1$ depending on whether $df_x$ preserves or reverses orientation.} $S^{2n-1}_\epsilon \mapsto S^{2n-1}$ given by $z \mapsto g(z)/\norm{g(z)}$ (where $\norm{\cdot}$ is the Euclidean distance). Milnor defined $\milnor(f)$ as the multiplicity at the origin of the map $z \mapsto (\partialxone{f}, \ldots, \partialxn{f})$. Milnor showed that for all sufficiently small $\epsilon$, if $\phi: S^{2n-1}_\epsilon \setminus V(f) \to S^1$ is the map given by $z \mapsto f(z)/\norm{f(z)}$, then each fiber of $\phi$ is a smooth $(2n-2)$-dimensional real manifold with homotopy type of a ``bouquet'' $S^{n-1} \vee \cdots \vee S^{n-1}$ of spheres, and $\milnor(f)$ is precisely the number of spheres in the bouquet. The fact that the multiplicity of a map $g: \cc^n \to \cc^n$ at the origin equals $\multgzero$ was left in \cite[Appendix B]{milnor} as an exercise; a proof can be found in \cite[Chapter I.5]{sing1}.

\section{Generic Milnor number}
Let $\scrA$ be a (possibly infinite) subset of $\znzero$. We write $\scrL_0(\scrA)$ for the set of all power series in $(x_1, \ldots, x_n)$ supported at $\scrA$. For each $j = 1, \ldots, n$, define
\begin{align}
\partialxj{\scrA} := \{\alpha - e_j :\alpha \in \scrA,\ \alpha - e_j \in \znzero,\ p\ \text{does not divide}\ \alpha_j\}
\label{partial-A}
\end{align}
where $e_j$ is the $j$-th standard unit vector in $\zz^n$ and $p := \character(\kk)$. Note that $\partialxj{\scrA}$ is the support of $\partialxj{g}$ for {\em generic}\footnote{``Generic'' refers to elements of a nonempty Zariski open (dense) subset of $\scrL_0(\scrA)$ in the Zariski topology mentioned in \cref{ind-remark}.} $g \in \scrL_0(\scrA)$. Define
\begin{align*}

\milnor(\scrA)
	&:= \min\{\milnor(f): f\in \scrL_0(\scrA)\}

\end{align*}
In \cref{milnor-thm} below we estimate $\milnor(\scrA)$ in terms of the intersection multiplicity $\multGammazero$ at the origin of the {\em Newton diagrams} $\Gamma_j$ of $\partialxj{\scrA}$. Given $f \in \scrL_0(\scrA)$, we say that $f$ is {\em partially $\scrA$-non-degenerate at the origin} if the partial derivatives of $f$ are $(\partialxone{\scrA}, \ldots, \partialxn{\scrA})$-non-degenerate at the origin in the sense of \cref{defn:non-degeneracy-0}; in the case that $\scrA = \supp(f)$, we simply say that $f$ is {\em partially non-degenerate at the origin}. In other words, $f$ is partially non-degenerate at the origin if it satisfies the following property:
\begin{align}
\parbox{0.84\textwidth}{
for each nonempty subset $I$ of $[n]$ and each weighted order $\nu$ centered at the origin on $\kk[x_i: i \in I]$, there is no common zero of $\In_\nu(\partialxj{f}|_{\Ki})$, $j = 1, \ldots, n$, on $\nktorus$.
} \label{partial-non-degeneracy}
\end{align}
Recall that one does {\em not} have to check this condition for {\em all} nonempty subsets of $[n]$ - see \cref{non-degeneracy-0'-thm,non-degeneracy-0'-remark}.

\begin{thm}[{\cite{toricstein}}] \label{milnor-thm}
Let $\Gamma_j := \nd(\partialxj{\scrA})$, $j = 1, \ldots, n$. Assume $\origin \not\in \scrA$. Then
\begin{align}
\milnor(\scrA)  \geq \multGammazero \label{milnor-bound}
\end{align}

Moreover,
\begin{enumerate}
\item \label{milnor-eq}The following are equivalent for all $f \in \scrL_0(\scrA)$:
\begin{enumerate}
\item $\milnor(f) = \multGammazero < \infty$,
\item $f$ is partially $\scrA$-non-degenerate at the origin.
\end{enumerate}
\item\label{milnor-open}  Let $\scrM'_0(\scrA)$ be the set of all $f \in \scrL_0(\scrA)$ which are partially $\scrA$-non-degenerate at the origin. Let $\scrA'$ be any {\em finite} subset of $\scrA$ such that $\partialxj{\scrA'} \supseteq \Gamma_j \cap \partialxj{\scrA}$, $j = 1, \ldots, n$. Then $\scrM'_0(\scrA')$ is a Zariski open subset of $\scrL_0(\scrA')$, and $\scrM'_0(\scrA) = \pi^{-1}(\scrM'_0(\scrA'))$, where $\pi: \scrL_0(\scrA) \to \scrL_0(\scrA')$ is the natural projection.
\item \label{milnor-A-eq} If $\multGammazero = \infty$, then $\scrM'_0(\scrA) = \emptyset$. If $\multGammazero < \infty$, then $\milnor(\scrA) = \multGammazero$ if and only if $\scrM'_0(\scrA)$ is nonempty.

\item \label{milnor-existence} If $\character(\kk) = 0$ and $\multGammazero < \infty$, then $\scrM'_0(\scrA)$ is nonempty and $\milnor(\scrA) = \multGammazero$.
\end{enumerate}
\end{thm}

\begin{example} \label{milnor-nonexample}

Assertion \eqref{milnor-existence} of \cref{milnor-thm} may not be true, i.e.\ the bound in \eqref{milnor-bound} may be strict, in the case that $p := \character(\kk) > 0$. E.g.\ let $\scrA := \{(p+1,1), (1, p+1)\} \subset \zzeroo{2}$. Then $\partialxone{\scrA} = \{(p,1), (0, p+1)\}$ and $\partialxii{\scrA}{2} = \{(p+1, 0), (1, p)\}$. If $f_1 = a_{1,1}x_1^p x_2 + a_{1,2}x_2^{p+1}$ and $f_2 = a_{2,1}x_1^{p+1} + a_{2,2}x_1x_2^p$ are generic polynomials supported respectively at $\partialxone{\scrA}$ and $\partial \scrA/x_2$, then $\multpzeronodots{f_1,f_2} = (p+1)^2$. Therefore $\multpzeronodots{\Gamma_1, \Gamma_2} = (p+1)^2$. On the other hand, if $f = ax_1^{p+1}x_2 + bx_1x_2^{p+1}$ is a generic polynomial supported at $\scrA$, then $\partialxone{f} = x_2(ax_1^p + bx_2^p)$ and $\partial f/\partial x_2 = x_1(ax_1^p + bx_2^p)$ so that $\milnor(\scrA) = \multpzeronodots{x_2(ax_1^p + bx_2^p), x_1(ax_1^p + bx_2^p)} = \infty > \multpzeronodots{\Gamma_1, \Gamma_2}$. It is straightforward to check that $\partial f/\partial x_1, \partial f/\partial x_2$ are $\scrA$-degenerate at the origin, i.e.\ $\scrM'_0(\scrA) = \emptyset$.
\end{example}

\begin{proof}[Proof of \cref{milnor-thm}]
Assertions \eqref{milnor-eq}, \eqref{milnor-open} and \eqref{milnor-A-eq} follow from \cref{non-degeneracy-0-thm,finite-0-cor}. Therefore it suffices to show that $\scrM'_0(\scrA)$ is nonempty in the case that $\character(\kk) = 0$. We may assume \woutlog\ that $0 < \milnor(\scrA) < \infty$. Pick any finite subset $\scrA'$ of $\scrA$ satisfying the assumptions of assertion \eqref{milnor-open}. Let $I \subseteq [n]$ and $\nu$ be a weighted order on $\kk[x_i: i \in I]$ centered at the origin. Denote by $\scrL^I_\nu$ the set of all $g \in \scrL_0(\scrA')$ such that
\begin{itemize}
\item $\supp(g) = \scrA'$, and
\item $V(\In_\nu(\partialxone{g}|_{\Ki}), \ldots, \In_\nu(\partialxn{g}|_{\Ki})) \cap \nktorus= \emptyset$
\end{itemize}
It suffices to show that $\scrL^I_\nu$ contains a nonempty Zariski open subset of $\scrL_0(\scrA')$. We may assume without loss of generality that $I = \{1, \ldots, k\}$. Take $g \in \scrL_0(\scrA')$, and express it as
\begin{align*}
g &= g_0(x_1, \ldots, x_k) + \sum_{i = k+1}^n x_ig_i(x_1, \ldots, x_k)  + \cdots
\end{align*}
where the omitted terms have quadratic or higher order in $(x_{k+1}, \ldots, x_n)$. Since $0  < \multGammazero < \infty$, \cref{mult-support} implies that $\partialxj{g}|_{\Ki} \not\equiv 0$ for $l \geq k$ values of $j$; denote them by $1 \leq j_1 < \cdots < j_l \leq n$. If $g_0 \equiv 0$, then each $j_i > k$, and $V(\In_\nu(\partialxone{g}|_{\Ki}), \ldots, \In_\nu(\partialxn{g}|_{\Ki})) = V(\In_\nu(g_{j_1}), \ldots, \In_\nu(g_{j_l})$. Therefore $g \in \scrL^I_\nu$ if $g_{j_1}, \ldots, g_{j_l}$ are {\em BKK non-degenerate}. Since $l \geq k$, \cref{bkk-0} then implies that $\scrL^I_\nu$ contains a nonempty Zariski open subset of $\scrL_0(\scrA')$, as required. So assume $g_0 \not\equiv 0$. Let $h_0 := \In_\nu(g_0)$ and $V'(h_0) :=  V(\partialxone{h_0}, \ldots, \partialxk{h_0}) \cap \nktorus$. Let $\scrA'_0 := \supp(h_0)$. Since $V(\In_\nu(\partialxone{g}|_{\Ki}), \ldots, \In_\nu(\partialxn{g}|_{\Ki})) \subseteq V'(h_0)$, it suffices to show that the set of all polynomials $h \in \scrL_0(\scrA'_0)$ such that $V'(h) = \emptyset$ contains a nonempty Zariski open subset of $\scrL_0(\scrA'_0)$. Let $Z := \{(x, h): x \in \nktorus,\ h \in \scrL_0(\scrA'_0),\ \partialxone{h}(x) = \cdots = \partialxk{h}(x) = 0\} \subset \nktorus \times \scrL_0(\scrA'_0)$, and let $\pi_1: Z \to \nktorus$ and $\pi_2:  Z \to \scrL_0(\scrA'_0)$ be the natural projections. It suffices to show that $\dim(\pi_2(Z)) < \dim \scrL_0(\scrA'_0)$. We prove this by a dimension count. Denote the elements in $\scrA'_0$ by $\alpha_i = (\alpha_{i,1},  \ldots, \alpha_{i,n})$, $i = 1, \ldots, N$, and the coefficients of $x^{\alpha_i}$ (in a polynomial supported at $\scrA'_0$) by $a_i$. Let $x \in \nktorus$. Then $\pi_1^{-1}(x)$ is the subspace of $\scrL_0(\scrA'_0)$ defined by a system of linear equations of the form
\begin{align}
\begin{pmatrix}
\alpha_{1,1}x^{\alpha_1- e_1} & \cdots  & \alpha_{N,1}x^{\alpha_N - e_1} \\
\vdots & & \vdots \\
\alpha_{1,k}x^{\alpha_1- e_k} & \cdots  & \alpha_{N,k}x^{\alpha_N - e_k}
\end{pmatrix}
\begin{pmatrix}
a_1 \\
\vdots\\
a_N
\end{pmatrix}
&= 0 \label{Z_0-eqn-0}
\end{align}
where $e_1, \ldots, e_k$ are the standard unit vectors in $\zz^k$. Since $\bm{\character(\kk) = 0}$ (note: this is the only place the assumption of zero characteristic is used), the rank (as a matrix over $\kk$) of the left-most matrix in \eqref{Z_0-eqn-0} is the same as the rank (as a matrix over $\qq$) of
\begin{align*}
B &:=
\begin{pmatrix}
\alpha_{1,1} & \cdots  & \alpha_{N,1}\\
\vdots & & \vdots \\
\alpha_{1,k} & \cdots  & \alpha_{N,k}
\end{pmatrix}
\end{align*}
Since $\nu$ has positive weights, and the $\alpha_j$ belong to a level set (corresponding to a positive value) of $\nu$, it follows that $\rank(B) = 1 + \dim( \np(h_0))$ and therefore $\dim(\pi_1^{-1}(x)) = N - 1 - \dim( \np(h_0))$. Since this is independent of $x$, it follows that $\dim(Z) = N+n - 1 - \dim(\np(h_0))$. On the other hand, if $x \in V'(h)$ for some $h \in \scrL_0(\scrA'_0)$, then since the support of each of $\partialxj{h}$ is contained in a translation of $\np(h_0)$, it is straightforward to check that $\partialxj{h}(xz^\beta) = 0$ for each $j = 1, \ldots, k$, and each $z \in \nktorus$ and each $\beta \in \zz^n$ which is normal (with respect to the ``dot product'') to $\np(h_0)$. Therefore the dimension of each fiber of $\pi_2$ is at least $n - \dim(\np(h_0))$. It follows that $\dim(\pi_2( Z)) \leq \dim( Z) - n + \dim(\np(h_0)) = N-1$, as required.
\end{proof}

\section{Classical notions of non-degeneracy} \label{classical-section}

\subsection{Newton non-degeneracy}
For a power series $f$ in $(x_1, \ldots, x_n)$, we write $\jjj(f)$ for the ideal generated by the partial derivatives of $f$. We say that $f$ is {\em Newton non-degenerate} iff $\jjj(\In_\nu(f))$ has no zero on $\nktorus$ for each weighted order $\nu$ on $\kk[[x_1, \ldots, x_n]]$ centered at the origin. Newton non-degeneracy is possibly the most studied non-degeneracy property of hypersurface germs: it is a Zariski open condition (in the same sense as partial non-degeneracy at the origin) and, in the characteristic zero case, also nonempty. In this section we discuss its relationship with partial non-degeneracy at the origin. In general Newton non-degeneracy does {\em not} imply ``finite determinacy,'' i.e.\ $f$ can be Newton non-degenerate but still $V(f)$ may have a non-isolated singularity at the origin (take e.g.\ $f:= x_1 \cdots x_n$, $n \geq 2$). However, if the Newton diagram of $f$ is {\em convenient} then Newton non-degeneracy of $f$ implies that the origin is an isolated singularity of $V(f)$ (\cref{Newton-isolated}), which can be resolved by a ``toric modification.'' As a result, the invariants of the singularity can be computed combinatorially in terms of the diagram (see e.g.\ \cite{oka-non-degenerate}). We show in \cref{Newton-to-Milnor} that for isolated singularities, Newton non-degeneracy is a special case of partial non-degeneracy at the origin. However, the following example shows that even in the case of convenient diagrams partial non-degeneracy at the origin does not imply Newton non-degeneracy.

\begin{example}\label{Milnor-not-Newton}
Let $f := x_1 + (x_2 + x_3)^q$, where $q \geq 2$. Then $\nd(f)$ is convenient and $f$ is not Newton non-degenerate (take $\nu$ with weights $(q+1, 1, 1)$ for $(x,y,z)$). However, $f$ is partially non-degenerate at the origin with $\milnor(f) =0$.
\end{example}

We now show that Newton non-degeneracy implies partial non-degeneracy at the origin. The following notation is used in its proof: let $I\subseteq I' \subseteq [n]$, $\nu \in \rnnstar{I}$ and $\nu' \in \rnnstar{I'}$. We say that $\nu$ and $\nu'$ are {\em compatible}

if the weighted order on $\kk[x_i: i \in I]$ induced by $\nu$ and the weighted order on $\kk[x_{i'}: i' \in I']$ induced by $\nu'$ are compatible in the sense of \cref{efficient-section}.

\begin{prop}[{\cite{toricstein}}] \label{Newton-to-Milnor}
Let $f \in \kk[[x_1, \ldots, x_n]]$ and $\Gamma_j := \nd(\partialxj{f})$, $j = 1, \ldots, n$. If $f$ is Newton non-degenerate and $\multGammazero < \infty$, then $f$ is partially non-degenerate at the origin. In particular, if $f$ is Newton non-degenerate and $\nd(f)$ is convenient, then $f$ is partially non-degenerate at the origin.
\end{prop}

\begin{proof}
We start with a direct proof of the second assertion since it is easier to see. Assume $\Gamma := \nd(f)$ is convenient and $f$ is Newton non-degenerate. Pick a nonempty subset $I$ of $[n]$ and a weighted order $\nu$ on $\kk[x_i: i \in I]$ which is centered at the origin. Since $\Gamma$ is convenient, $\Gamma \cap \ri \neq \emptyset$. Therefore we can find a weighted order $\nu'$ on $\kk[x_1, \ldots, x_n]$ such that $\nu'$ is compatible with $\nu$ and $\In_{\nu'}(\Gamma) \subset \ri$. Then $\In_{\nu'}(f)$ depends only on $(x_i: i \in I)$. Since $f$ is Newton non-degenerate, it follows that $\partial(\In_{\nu'}(f))/\partial x_i$, $i \in I$, do not have any common zero in $\nktorus$. But if $i \in I$ is such that $\partial(\In_{\nu'}(f))/\partial x_i$ is not identically zero, then $\partial(\In_{\nu'}(f))/\partial x_i = \partial(\In_{\nu}(f|_{\Ki}))/\partial x_i =  \In_{\nu}(\partialxi{f}|_{\Ki})$. This implies that  $\In_{\nu}(\partialxi{f}|_{\Ki})$, $i \in I$, do not have any common zero on $\Kstari$, as required for partial non-degeneracy of $f$ at the origin. \\

Now we prove the first assertion. If $\multGammazero = 0$, then \cref{mult-support} implies that $f$ is partially non-degenerate at the origin. So assume $0 < \multGammazero < \infty$ and $f$ is partially {\em degenerate} at the origin. It suffices to show that $f$ is Newton degenerate. Pick $I \subseteq [n]$ and a {\em primitive} weighted order $\nu$ centered at the origin on $\kk[x_i: i \in I]$ such that $\In_\nu(\partialxi{f}|_{\Ki})$, $i \in I$, have a common zero $(a_1, \ldots, a_n) \in \nktorus$. At first consider the case that $f|_{\Ki} \not\equiv 0$. Then as in the convenient case pick a weighted order $\nu'$ on $\kk[x_1, \ldots, x_n]$ such that $\nu'$ is compatible with $\nu$ and $\In_{\nu'}(f) \in \kk[x_i: i \in I]$. Then for each $j = 1, \ldots, n$, $\partialxj{\In_{\nu'}(f)}$ is either $\In_\nu(\partialxj{f}|_{\Ki})$ or is identically zero. It follows that $(a_1, \ldots, a_n)$ is a common zero of $\jjj(\In_{\nu'}(f))$ on $\nktorus$, so that $f$ is Newton degenerate, as required. Now assume that $f|_{\Ki} \equiv 0$. We may assume that $I = \{1, \ldots, k\}$ for some $k$, $1 \leq k \leq n$, and $\partialxj{f}|_{\Ki} \not\equiv 0$ if and only if $i = k+1, \ldots, k+l$. Then $f$ can be expressed as
\begin{align}
f &= \sum_{j= 1}^l x_{k+j}f_j(x_1, \ldots, x_k) + \sum_{i \geq j \geq 1} x_{k+i}x_{k+j}f_{i,j}(x_1, \ldots, x_n)
\label{fexpression}
\end{align}
Let $h_j := \In_\nu(f_j)$, $j = 1, \ldots, l$, and $B$ be the $k \times l$ matrix with $(i,j)$-th entry $(\partialxi{h_j})(a_1, \ldots, a_k)$. We claim that $\rank(B) < k$. Indeed, let $\nu_i:= \nu(x_i)$, $i = 1, \ldots, k$. For each $j = 1, \ldots, l$, by assumption $(a_1, \ldots, a_k)$ is a common zero of $\In_\nu(\partialxii{f}{k+j}|_{\Ki})= h_j(x_1, \ldots, x_k)$, so that $h_j(a_1t^{\nu_1}, \ldots, a_kt^{\nu_k}) = 0$ for all $t \in \kk$. Note that
\begin{align*}
\frac{d}{dt}(h_j(a_1t^{\nu_1}, \ldots, a_kt^{\nu_k}))
	&= \sum_{i=1}^k \frac{\partial h_j}{\partial x_i}(a_1t^{\nu_1}, \ldots, a_kt^{\nu_k}) \nu_ia_i t^{\nu_i - 1}
	= \sum_{i=1}^k \frac{\partial h_j}{\partial x_i}(a_1, \ldots, a_k) \nu_ia_i t^{\nu(h_j) - 1}
\end{align*}
If $a' := (\nu_1a_1, \ldots, \nu_ka_k) \in \kk^n$, then it follows that $a'B = 0$. Since $\nu$ is primitive, $a' \neq 0$, so that the map $\kk^k \to \kk^l$ given by multiplication by $B$ on the right is {\em not} injective. Therefore $\rank(B) < k$, as claimed. Since $0 < \multGammazero < \infty$, \cref{mult-support} implies that $l \geq k$, so that there is $b = (b_1, \ldots, b_l) \neq 0 \in \kk^l$ such that $B$ times the transpose of $b$ is zero. Let $m$ be the number of nonzero coordinates of $b$. \Woutlog\ we may assume that $b_j \neq 0$ if and only if $j = 1, \ldots, m$. Then we have that
\begin{align}
\sum_{j=1}^m b_j \frac{\partial h_j}{\partial x_i} (a_1, \ldots, a_k) = 0
\end{align}
Fix positive integers $q, q_{k+m+1}, \ldots, q_n$ and let $\nu'$ be the weighted order on $\kk[x_1, \ldots, x_n]$ such that
\begin{align*}
\nu'(x_i)
	&=
		\begin{cases}
		\nu(x_i) &\text{if}\ i = 1, \ldots, k,\\
		q - \nu(f_{i-k}) &\text{if}\ i = k+1, \ldots, k+m,\\
		q_i &\text{if}\ i = k+m+1, \ldots, n.
		\end{cases}
\end{align*}
If $q \gg 1$ and $q_i \gg q$ for $i = k+m+1, \ldots, n$, then identity \eqref{fexpression} implies that $\In_{\nu'}(f) = \sum_{j=1}^m x_{k+j}h_j$. If $b'_1, \ldots, b'_{n-k-m}$ are arbitrary elements in $\ktorus$, it follows that $(a_1, \ldots, a_k, b_1, \ldots, b_m,  b'_1, \ldots, b'_{n-k-m})$ is a zero of $\jjj(\In_{\nu'}(f))$ on $\nktorus$. Therefore $f$ is Newton degenerate, as required.
\end{proof}

\begin{cor}[{Brzostowski and Oleksik \cite[Theorem 3.1]{brzostowski-oleksik}}] \label{Newton-isolated}
Let $\Gamma_j := \nd(\partialxj{f})$, $j = 1, \ldots, n$. Assume $f(\origin) = 0$ and $f$ is Newton non-degenerate. Then the following are equivalent:
\begin{enumerate}
\item $\milnor(f) < \infty$.
\item $\multGammazero < \infty$.
\item $|\{j : \Gamma_j \cap \ri \neq \emptyset\}| \geq |I|$ for each $I \subseteq [n]$.
\end{enumerate}
\end{cor}

\begin{proof}
Combine \cref{mult-support,milnor-thm,Newton-to-Milnor}.
\end{proof}

\subsection{Inner Newton non-degeneracy} \label{inner-section}
A {\em diagram} $\Gamma$ in $\rr^n$ is the Newton diagram of some subset of $\rzeroo{n}$, and a {\em face} $\Delta$ of $\Gamma$ is a compact face of $\Gamma + \rzeroo{n}$. $\Delta$ is called an {\em inner face} if it is not contained in any proper coordinate subspace of $\rr^n$. If $f = \sum_{\alpha \in \znzero} c_\alpha x^\alpha$ is a power series in $(x_1, \ldots, x_n)$ and $\Delta$ is a subset of $\rr^n$, we write $f_\Delta := \sum_{\alpha \in \Delta} c_\alpha x^\alpha$.

We say that $f$ is {\em inner Newton non-degenerate} if there is a {\em convenient} diagram $\Gamma$ such that
\begin{defnlist}
\item no point of $\supp(f)$ ``lies below'' $\Gamma$, i.e.\ $\supp(f) \subseteq \Gamma + \rzeroo{n}$, and

\item for every inner face $\Delta$ of $\Gamma$ and for every nonempty subset $I$ of $[n]$,
\begin{align}
\Delta \cap \ri \neq \emptyset \im V(\jjj(f_\Delta)) \cap \Kstari = \emptyset \label{inner-property}
\end{align}
\end{defnlist}
The difference between Newton non-degeneracy and inner Newton non-degeneracy is most evident in the case of  weighted homogeneous polynomials with isolated singularities, e.g.\ consider the polynomial $f := x_1 + (x_2 + x_3)^q$, where $q \geq 2$, from \cref{Milnor-not-Newton}.  The Newton diagram of $f$ is convenient and has only one inner face, namely the two dimensional face $\Delta$ with vertices $(1, 0,0), (0,q,0), (0,0,q)$. In particular, $f_\Delta = f$ and $\partialxone{f} = 1$, which implies that $f$ is inner non-degenerate. However, as we saw in \cref{Milnor-not-Newton}, $f$ is Newton degenerate. In fact C.\ T.\ C.\ Wall introduced inner Newton non-degeneracy in \cite{wall} in order to find a condition which (in the case of power series with convenient Newton diagrams) is weaker than Newton non-degeneracy, but still wide enough to include all weighted homogeneous polynomials with isolated singularities. We now show that inner Newton non-degeneracy implies partial non-degeneracy at the origin. Given $\Delta \subseteq \rr^n$, we write $I_\Delta$ for the smallest subset of $[n]$ such that $\Delta \subseteq \rii{I_\Delta}$.

\begin{lemma} \label{inner-face-lemma}
Let $\Gamma$ be a convenient diagram, $I \subseteq [n]$ and $\Delta$ be a face of $\Gamma \cap \ri$. Pick an integral element $\nu \in \rnnstar{I}$ centered at the origin such that $\Delta = \In_\nu(\Gamma \cap \ri)$. Then there is an integral $\nu' \in \rnstar$ which is centered at the origin and compatible with $\nu|_{\rii{I_\Delta}}$ such that $\Delta' := \In_{\nu'}(\Gamma)$ is an inner face of $\Gamma$ and $\Delta= \Delta' \cap \rii{I_\Delta}$.
\end{lemma}

\begin{proof}
We proceed by induction on $\epsilon := n - |I_\Delta|$. It is obvious if $\epsilon = 0$. Now assume $\epsilon = 1$. Then \woutlog\ we may assume $I_\Delta = \{1, \ldots, n-1\}$. Pick $\nu \in \rnnstar{I_\Delta}$ such that $\Delta = \In_\nu(\Gamma \cap \rii{I_\Delta})$.  For each pair of positive integers $q,r$, let $\nu_{q,r}$ be the element on $\rnstar$ defined as follows:
$$\langle \nu_{q,r}, (\alpha_1, \ldots, \alpha_n) \rangle
	:= q\langle \nu, (\alpha_1, \ldots, \alpha_{n-1}) \rangle +r\alpha_n$$
If $r \gg q$, then $\In_{\nu_{q,r}}(\Gamma) = \Delta$. On the other hand, since $\nu$ is centered at the origin and $\Gamma$ is convenient, if $q \gg r$, then $\In_{\nu_{q,r}}(\Gamma)$ is a point on the $x_n$-axis. Therefore we can find $q,r$ such that $\In_{\nu_{q,r}}(\Gamma)$ contains $\Delta$ and also a point with positive $n$-th coordinate.  The lemma then holds with $\nu' := \nu_{q,r}$. Now assume $\epsilon \geq 2$. Pick $i' \in [n] \setminus I_\Delta$. The induction hypothesis implies that there is $\tilde \nu \in \rnnstar{ [n] \setminus \{i'\}}$ which is centered at the origin and compatible with $\nu|_{\rii{I_\Delta}}$ such that $\tilde \Delta := \In_{\tilde \nu}(\Gamma  \cap \rii{[n]\setminus \{i'\}})$ is an inner face of $\Gamma  \cap \rii{[n]\setminus \{i'\}}$ and $\Delta = \tilde \Delta \cap \rii{I_\Delta}$. The $\epsilon = 1$ case then implies that there is $\nu' \in \rnstar$ which is centered at the origin and compatible with $\tilde \nu$ such that $\Delta' := \In_{\nu'}(\Gamma)$ is an inner face of $\Gamma$ and $\Delta' \cap \rii{[n]\setminus \{i'\}} = \tilde \Delta$. The lemma then holds with $\nu'$, which completes the proof.
\end{proof}

\begin{prop}[{\cite{toricstein}}]\label{inner-to-Milnor}
If $f \in \kk[[x_1, \ldots, x_n]]$ is inner Newton non-degenerate, then it is partially non-degenerate at the origin.
\end{prop}

\begin{proof}
Assume $f$ is inner Newton non-degenerate with respect to a convenient diagram $\Gamma$. Pick $I \subseteq [n]$ and a weighted order $\nu$ on $\kk[x_i: i \in I]$. Let $\Delta := \In_\nu(\Gamma \cap \ri)$. \Cref{inner-face-lemma} implies that there is $\nu' \in \rnstar$ compatible with $\nu|_{\rii{I_\Delta}}$ such that $\Delta' := \In_{\nu'}(\Gamma)$ is an inner face of $\Gamma$, and $\Delta= \Delta' \cap \rii{I_\Delta}$. Fix $j$, $1 \leq j \leq n$.

\begin{proclaim} \label{f-Delta'-claim}
One of the following holds:
\begin{prooflist}
\item either $\partialxj{f_{\Delta'}}$ is identically zero on $\Kii{I_\Delta}$,
\item or $\partialxj{f_{\Delta'}}|_{\Kii{I_\Delta}} = \In_\nu(\partialxj{f}|_{\Ki})$.
\end{prooflist}
\end{proclaim}

\begin{proof}
Assume $\partialxj{f_{\Delta'}}$ is not identically zero on $\Kii{I_\Delta}$. Let $m := \nu'(f) = \nu'(f_{\Delta'})$. Then $\nu'(\partialxj{f_{\Delta'}}) = m - \nu'(x_j)$. There is a positive rational number $\lambda$ such that $\nu'|_{\rii{I_\Delta}} = \lambda \nu|_{\rii{I_\Delta}} $. Then $\nu(\partialxj{f_{\Delta'}}|_{\Kii{I_\Delta}}) = \nu'(\partialxj{f_{\Delta'}}|_{\Kii{I_\Delta}})/\lambda = (m - \nu'(x_j))/\lambda$. Now pick $\alpha \in \supp(\partialxj{f}|_{\Ki}) \setminus \supp(\partialxj{f_{\Delta'}}|_{\Kii{I_\Delta}})$. It suffices to show that $\nu(x^\alpha) > (m - \nu'(x_j))/\lambda$. Let $e_j$ be the unit vector along the (positive) $j$-th axis on $\rr^n$. Pick $\alpha' \in \supp(\partialxj{f_{\Delta'}}|_{\Kii{I_\Delta}})$. Then $\alpha' + e_j \in \Delta'$ and $\alpha + e_j \in \Gamma \setminus \Delta'$. Therefore $\langle \nu', \alpha + e_j \rangle > \langle \nu', \alpha' + e_j \rangle$, so that $\langle \nu', \alpha \rangle > \langle \nu', \alpha' \rangle = m - \nu'(x_j)$. It follows that $\langle \nu, \alpha \rangle  = \langle \nu', \alpha \rangle /\lambda > (m - \nu'(x_j))/\lambda$, as required.
\end{proof}

\Cref{f-Delta'-claim} implies that $V(\In_\nu(\partialxone{f}|_{\Ki}), \ldots, \In_\nu(\partialxn{f}|_{\Ki})) \cap \Kstari$ is contained in the product of $V(\partialxone{f_{\Delta'}}|_{\Kii{I_\Delta}}, \ldots, \partialxn{f_{\Delta'}}|_{\Kii{I_\Delta}}) \cap \Kstarii{I_\Delta}$ with $\Kstarii{I\setminus I_\Delta}$. The inner non-degeneracy of $f$ with respect to $\Gamma$ then implies that $V(\In_\nu(\partialxone{f}|_{\Ki}), \ldots, \In_\nu(\partialxn{f}|_{\Ki})) \cap \Kstari = \emptyset$, as required.
\end{proof}

\begin{cor}[{Wall \cite[Lemma 1.2]{wall}}]
Inner Newton non-degeneracy implies finite determinacy. \qed

\end{cor}

If $p := \character(\kk)$ is nonzero, partial Newton non-degeneracy at the origin is strictly weaker than inner Newton non-degeneracy, e.g.\ $x^p + y^p + x^{p+1} + y^{p+1}$ is partially non-degenerate at the origin, but it is inner Newton {\em degenerate}. We do not know if this is true in zero characteristic (see \cref{non-degenerate-relations}).

\section{Newton number: Kushnirenko's formula for the generic Milnor number}

Let $\Gamma$ be a diagram in $\rr^n$. We write $\bar \Gamma$ for the region bounded by the cone with base $\Gamma$ and apex at the origin, and $\volsubk(\Gamma)$, $0 \leq k \leq n$, for the sum of $k$-dimensional Euclidean volumes of the intersections of $\bar \Gamma$ with the $k$-dimensional coordinate subspaces of $\rr^n$ (in particular, $V_0(\Gamma)$ is defined to be $1$). The {\em Newton number} of $\Gamma$ is
\begin{align*}
\newton(\Gamma)
	&:=
	\begin{cases}
		\sum_{k=0}^n (-1)^{n-k} k! \volsubk(\Gamma)
			&\text{if $\Gamma$ is convenient,} \\
		\sup\{\newton( \Gamma \cup \{me_1, \ldots, me_n\}): m \geq 0\}
			& \text{otherwise,}
	\end{cases}
\end{align*}
where $e_j$ are the unit vectors along the (positive direction of the) axes of $\rr^n$. Let $f \in \kk[[x_1, \ldots, x_n]]$ such that $f(\origin) = 0$. A.\ Kushnirenko proved in \cite{kush-poly-milnor} that $\milnor(f) \geq \newton(\nd(f))$, and $\milnor(f) = \newton(\nd(f))$ if $f$ is Newton non-degenerate and if either $\character(\kk) = 0$ or $\nd(f)$ is convenient. C.\ T.\ C.\ Wall proved in \cite{wall} that if $\character(\kk) = 0$, then the equality $\milnor(f) = \newton(\nd(f))$ continues to hold if $f$ is inner Newton non-degenerate. In this section we present some generalizations of these results.

\subsection{Preliminary results} \label{kushliminary}
Let $\scrA$ be a subset of $\znzero$ not containing the origin. For each $m \geq 0$, let $\scrA_m := \scrA \cup \{(\alpha_1, \ldots, \alpha_n): \sum_{j=1}^n \alpha_j \geq m\}$. For each $j = 1, \ldots, n$, let
\begin{align*}
\scrA'_{m,j} := \{\alpha - e_j :\alpha \in \scrA_m,\ \alpha - e_j \in \znzero\})
\end{align*}
where $e_j$ are the unit vectors along the (positive direction of the) axes of $\rr^n$. Note that $\partialxj{\scrA_m} \subseteq \scrA'_{m,j}$, and the inclusion is proper if $\character(\kk)$ is positive. In any event, \cref{milnor-thm} and the monotonicity of intersection multiplicity (\cref{monotonic-remark-0}) imply that for each $m \geq 1$,
\begin{align}

\milnor(\scrA)
	&\geq \multzero{\partialxone{\scrA}}{\partialxn{\scrA}}
	\geq \multzero{\partialxone{\scrA_m}}{\partialxn{\scrA_m}}
	\geq \multzero{\scrA'_{m,1}}{\scrA'_{m,n}}
\label{kushnirenko-chain}
\end{align}

\begin{lemma} \label{newton'}
Let $\scrA \subseteq \znzero \setminus \{\origin\}$. Define
\begin{align*}
\scrA'_j := \{\alpha - e_j :\alpha \in \scrA,\ \alpha - e_j \in \znzero\}
\end{align*}
where $e_j$ are the $j$-th standard unit vector in $\rr^n$, $j = 1, \ldots, n$. Assume each $\scrA'_j$ is {\em convenient}. Then $\scrA$ is also convenient and $\multzero{\scrA'_1}{\scrA'_n} = \nu(\nd(\scrA))$.
\end{lemma}

\begin{rem}
Kushnirenko's theorem implies that the assumption ``each $\scrA'_j$ is convenient'' in \cref{newton'} is not needed - see \cref{newton''} below.
\end{rem}

\begin{proof}[Proof of \cref{newton'}]
Note that $\multzero{\scrA'_1}{\scrA'_n}  = \multgzero$ for all $g_1, \ldots, g_n \in \kk[[x_1, \ldots, x_n]]$ such that
\begin{prooflist}
\item $\supp(g_j) = \scrA'_j$ for each $j$, and
\item $g_1, \ldots, g_n$ are non-degenerate at the origin.
\end{prooflist}
We will show that there are $g_1, \ldots, g_n$ which satisfy both of the above properties and in addition satisfy $\multgzero = \nu(\nd(\scrA))$. Indeed, choose $g_1, \ldots, g_n$ which satisfy the above properties, and in addition satisfy
\begin{prooflist}[resume]
\item \label{newton':restriction} for each pair of subsets $I, J$ of $[n]$ such that $|I| = |J|$, the restrictions $g_j|_{\Ki}$, $j \in J$ are $(\scrA'_j \cap \ri: j \in J)$-non-degenerate at the origin.
\end{prooflist}
Note that this is possible since each $\scrA'_j$ is convenient and since ``generic'' systems are non-degenerate at the origin (\cref{non-degeneracy-0-thm}). For each $I, J \subseteq [n]$ such that $|I| + |J| = n$, we write
\begin{align*}
\multpnodots{(g_i)_{i \in I}, (x_j)_{j \in J}}{\origin}
	:= \multpnodots{g_{i_1}, \ldots, g_{i_k}, x_{j_1}, \ldots, x_{j_{n-k}}}{\origin}
\end{align*}
where $I = \{i_1, \ldots, i_k\}$ and $J = \{j_1, \ldots, j_{n-k}\}$. Since $\scrA'_j$ are convenient, \cref{mult-support} and property \ref{newton':restriction} imply that $\multpnodots{(g_i)_{i \in I}, (x_j)_{j \in [n]\setminus I}}{\origin}$ is defined for each $I \subseteq [n]$. Consequently \cref{kushnirenko-lemma} implies that
\begin{align*}
\multgzero
	&= \sum_{I \subseteq [n]} (-1)^{n- |I|} \multpnodots{(x_ig_i)_{i \in I}, (x_j)_{j \in [n]\setminus I}}{\origin}
	= \sum_{I \subseteq [n]} (-1)^{n- |I|} \multzero{x_{i_1}g_{i_1}|_{\Ki}} {x_{i_{|I|}}g_{i_{|I|}}|_{\Ki}}

\end{align*}
where $i_1, \ldots, i_{|I|}$ are elements of $I$ for each $I \subseteq [n]$. Fix $I \subseteq [n]$. Since the Newton diagram of the union of the supports of $x_{i_j}g_{i_j}|_{\Ki}$ is $\nd(\scrA) \cap \ri$ (this is where the assumption $\origin \not\in \scrA$ is used!), property \ref{newton':restriction} and \cref{kushniplicity} imply that
\begin{align*}
\multzero{x_{i_1}g_{i_1}|_{\Ki}} {x_{i_{|I|}}g_{i_{|I|}}|_{\Ki}}
	&= \volsub{|I|}(\nd(\scrA) \cap \ri)
\end{align*}
The result now follows from the definition of $\newton(\cdot)$.
\end{proof}

\begin{cor} \label{newton-0}
$\multzero{\scrA'_{m,1}}{\scrA'_{m,n}} = \nu(\nd(\scrA_m))$ for each $m \geq 1$. \qed
\end{cor}

\begin{cor} [{Kushnirenko \cite[Theorem I, part (i)]{kush-poly-milnor}}] \label{milnor-geq-newton}
$\milnor(f) \geq \newton(\nd(f))$ for all $f \in \kk[[x_1, \ldots, x_n]]$. In particular, if $\newton(\nd(f)) = \infty$, then $\milnor(f) = \infty$.
\end{cor}

\begin{proof}
Combine inequation \eqref{kushnirenko-chain} and \cref{newton-0}.
\end{proof}

\subsection{Characteristic zero case} \label{zero-section}
Continue to assume that $\scrA$ is a subset of $\znzero$ not containing the origin.

\begin{thm}\label{milnor=newton-0}
If $\character(\kk) = 0$, then $\milnor(\scrA) = \newton(\nd(\scrA))$.
\end{thm}

\begin{proof}
If $\character(\kk) = 0$, then \cref{milnor-thm} implies that $\milnor(\scrA) = \multzero{\partialxone{\scrA}}{\partialxn{\scrA}}$. At first consider the case that $\multzero{\partialxone{\scrA}}{\partialxn{\scrA}} < \infty$. For $m \gg 1$ then \cref{finite-mult-determinacy} implies that $\multzero{\partialxone{\scrA}}{\partialxn{\scrA}} = \multzero{\partialxone{\scrA_m}}{\partialxn{\scrA_m}}$. Since in zero characteristic $\scrA'_{m,j} = \partialxj{\scrA_m}$, inequation \eqref{kushnirenko-chain} and \cref{newton-0} imply that $\milnor(\scrA) = \newton(\nd(\scrA))$. On the other hand, if $\multzero{\partialxone{\scrA}}{\partialxn{\scrA}} = \infty$, then \cref{finite-mult-determinacy} implies that $\sup_m \multzero{\partialxone{\scrA_m}}{\partialxn{\scrA_m}} = \infty$, so that \cref{newton-0} implies that $\newton(\nd(\scrA)) = \infty$, as required.
\end{proof}

\begin{cor} \label{newton''}
\Cref{newton'} holds even without the assumption that each $\scrA'_j$ is convenient.
\end{cor}

\begin{proof}
In zero characteristic $\milnor(\scrA) = \multzero{\scrA'_1}{\scrA'_n}$ (\cref{milnor-thm}); now use \cref{milnor=newton-0}.
\end{proof}

\begin{cor}[{Cf.\ \cite[Characteristic zero case of Theorem I]{kush-poly-milnor}, \cite[Theorem 1.6]{wall}, \cite[Corollaries 3.10 and 3.11]{brzostowski-oleksik}}]
Let $f \in \kk[[x_1, \ldots, x_n]]$ be such that $f(\origin) = 0$. Assume $\character(\kk) = 0$. Then $\milnor(f) = \newton(\nd(f))$ whenever $f$ is partially non-degenerate at the origin. In particular, if $f$ is either Newton non-degenerate or inner Newton non-degenerate, then $\milnor(f) = \newton(\nd(f))$.
\end{cor}

\begin{proof}
Combine \cref{milnor-thm,Newton-to-Milnor,inner-to-Milnor,milnor=newton-0}.
\end{proof}

\subsection{The general case} \label{general-section}
We continue to use the notation of \cref{kushliminary}. In particular, $\scrA$ is a subset of $\znzero$ not containing the origin. Let $\scrM'_0(\scrA)$ be as in \cref{milnor-thm} the set of all $f \in \scrL_0(\scrA)$ which are partially $\scrA$-non-degenerate at the origin. \Cref{finite-mult-determinacy,milnor-thm}, inequation \eqref{kushnirenko-chain} and \cref{newton-0} imply the following result.

\begin{prop} \label{milnor=newton}
 $\milnor(\scrA) = \newton(\nd(\scrA))$ if and only if
\begin{enumerate}
\item either $\newton(\nd(\scrA)) = \infty$, or
\item
\begin{enumerate}
\item \label{partial-existence} $\scrM'_0(\scrA)$ is nonempty, and
\item \label{mult-equality}

$\multzero{\partialxone{\scrA_m}}{\partialxn{\scrA_m}} = \newton(\nd(\scrA_m))$ for all $m \gg 1$ not divisible by $p$. \qed
\end{enumerate}
\end{enumerate}
\end{prop}

\begin{rem}
\Cref{mult-support,newton''} imply that the following are equivalent:
\begin{enumerate}
\item $\newton(\nd(\scrA)) = \infty$.
\item $\multzero{\scrA'_1}{\scrA'_n} = \infty$.
\item there is a nonempty subset $I$ of $[n]$ such that $|\{j : \scrA'_j \cap \ri \neq \emptyset\}| < |I|$.
\end{enumerate}
\end{rem}

Let $p := \character(\kk)$. \Cref{milnor-nonexample} shows that if $p$ is positive, then condition \eqref{partial-existence} of \cref{milnor=newton} is nontrivial. The example below shows that condition \eqref{mult-equality} is nontrivial as well.

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[ scale=0.6]
\tikzstyle{dot} = [red, circle, minimum size=3pt, inner sep = 0pt, fill]
\def\p{2}
\pgfmathsetmacro\r{2*\p + 4};
\pgfmathsetmacro\q{2*\p + 2};

\draw [<->] (0,{\q + 0.5}) |- ({\r + 0.5},0);

\node[dot, label=below:{\small $r$}] (r) at (\r,0) {};
\node[dot, label=right:{\small $(p,p)$}] (p) at (\p,\p) {};
\node[dot, label=left:{\small $q$}] (q) at (0,\q) {};

\draw[thick, green] (r) -- (p) -- (q);
\draw[dashed] (0,0) -- (p);
\draw (\p, 1) node [right] {\small $pr/2$};
\draw (1, \p) node {\small $pq/2$};
\end{tikzpicture}
\caption{Subdivision of the area under the Newton diagram from \cref{non-newton-2}}  \label{fig:non-newton-2}
\end{center}
\end{figure}

\begin{example} \label{non-newton-2}
Assume $p$ is positive. Let $f = x_1^r + x_1^{p}x_2^p + x_2^{q}$, where $q,r$ are large positive integers not divisible by $p$. Let $\scrA := \supp(f)$ and $\Gamma := \nd(f)$. It is straightforward to see from \cref{fig:non-newton-2} that $\newton(\Gamma) =  2p(q+r)/2- (r + q) + 1 = (p-1)(q+r) + 1$. On the other hand, $\partial f/\partial x_1 = rx_1^r$ and $\partial f/\partial x_2 = qx_2^{q-1}$, so that $f$ is partially non-degenerate at the origin. It follows that $\scrM'_0(\scrA) \neq \emptyset$ and $\milnor(\scrA) = \milnor(f) = \multpzeronodots{rx_1^{r-1},  qx_2^{q-1}} = (r-1)(q-1)$. It follows that $\milnor(\scrA) > \newton(\nd(\scrA))$ for sufficiently large $q,r$.
\end{example}

We now state a condition which guarantees that condition \eqref{mult-equality} of \cref{milnor=newton} holds. Given a subset $\scrB$ of $\znzero$, let $J_\scrB := \{j \in [n]: \partialxj{\scrB} \neq \emptyset\} = \{j \in [n]:$ there is $(\beta_1, \ldots, \beta_n) \in \scrB$ such that $p$ does not divide $\beta_j\}$. The condition is the following:
\begin{align}
\parbox{.78\textwidth}{
for each $m \gg 1$ not divisible by $p$ and for each face $\Delta$ of $\nd(\scrA_m)$, $J_{\scrA_m \cap \Delta} \neq \emptyset$, and the convex hulls of $\partialxj{(\scrA_m \cap \Delta)}$, $j \in J_{\scrA_m \cap \Delta}$, are dependent.
} \label{milnor=newton-condition-1}
\end{align}

\begin{prop}  \label{milnor=newton-prop-1}
Assume \eqref{milnor=newton-condition-1} holds. Then condition \eqref{mult-equality} of \cref{milnor=newton} also holds. In particular, $\milnor(\scrA) = \newton(\nd(\scrA))$ if either $\newton(\nd(\scrA)) = \infty$ or $\scrM'_0(\scrA)$ is nonempty.
\end{prop}

\begin{proof}
Pick $m \gg 1$ not divisible by $p$. The arguments from the proof of \cref{newton'} show that it suffices to prove that $\multzero{x_1g_1}{x_ng_n} = n! \volsubn(\nd(\scrA_m))$ for power series $g_j$ such that $\supp(g_j) = \partialxj{\scrA_m}$ and $g_1, \ldots, g_n$ are non-degenerate at the origin. Condition \eqref{milnor=newton-condition-1} ensures that the Newton diagram of the union of the supports of $x_jg_j$ is $\nd(\scrA_m)$, and that the condition of assertion \eqref{kushniplicity-equal} of \cref{kushniplicity} is also satisfied. Therefore the result follows from \cref{kushniplicity}.
\end{proof}

Note that condition \eqref{milnor=newton-condition-1} is not necessary for \eqref{mult-equality} of \cref{milnor=newton} to hold - see \cref{newton-problem}.

\begin{cor}[{\cite[Positive characteristic case of Theorem I]{kush-poly-milnor}}] \label{kushnirenko+}
Let $f \in \kk[[x_1, \ldots, x_n]]$ such that $f(\origin) = 0$, $\nd(f)$ is convenient, and $f$ is Newton non-degenerate. Then $\milnor(f) = \newton(\nd(f))$.
\end{cor}

\begin{proof}
Due to \cref{Newton-to-Milnor,milnor=newton-prop-1} it suffices to show that $\scrA:= \supp(f)$ satisfies condition \eqref{milnor=newton-condition-1}. Since $\scrA$ is convenient, $\nd(\scrA_m) = \nd(\scrA)$ for $m \gg 1$. So pick a face $\Delta$ of $\nd(\scrA)$. Define $f_\Delta$ as in \cref{inner-section}. Let $g_j :=\partialxj{f_\Delta}$, $j = 1, \ldots, n$. Since $f$ is Newton non-degenerate, it follows that $g_1, \ldots, g_n$ are BKK non-degenerate. Since $g_1, \ldots, g_n$ have no common zero on $\nktorus$, \cref{bkk-bound-thm,bkk-non-degenerate-thm,positively-mixed} imply that the Newton polytopes of $g_j$ are dependent, as required.
\end{proof}

\section{Open problems}

\subsection{Existence of non-degenerate polynomials}
Let $\scrA$ be a subset of $\znzero$ not containing the origin. The estimate of $\milnor(\scrA)$ from \cref{milnor-thm} is exact if and only if the set $\scrM'_0(\scrA)$ of power series which are partially $\scrA$-non-degenerate at the origin is nonempty. \Cref{milnor-thm} shows that in characteristic zero $\scrM'_0(\scrA)$ is always nonempty, and \cref{milnor-nonexample} shows that in positive characteristic there are $\scrA$ such that $\scrM'_0(\scrA)$ is empty. This motivates the following problem.

\begin{problem} \label{milnor-existence-problem}
In the case that $\character(\kk) > 0$, characterize those $\scrA$ for which $\scrM'_0(\scrA)$ is nonempty. Compute $\milnor(\scrA)$ for those $\scrA$ such that $\scrM'_0(\scrA)$ is empty.
\end{problem}

Let $\scrN'^0_0(\scrA)$ be the set of all power series $f$ supported at $\scrA$ such that $\nd(f) = \nd(A)$, $\nd(\partialxj{f}) = \nd(\partialxj{\scrA})$ for each $j$, and $f$ is Newton non-degenerate. \Cref{Newton-to-Milnor} implies that $\scrN'^0_0(\scrA) \subseteq \scrM'_0(\scrA)$ when $\milnor(\scrA) < \infty$.  The following is therefore a subproblem of \cref{milnor-existence-problem} in that case.

\begin{problem} \label{newton-existence-problem}
In the case that $\character(\kk) > 0$, characterize those $\scrA$ for which $\scrN'^0_0(\scrA)$ is nonempty.
\end{problem}

The proof of \cref{milnor-thm} gives a sufficient condition for existence of Newton non-degenerate polynomials: let $\scrB$ be a finite subset of $\znzero$ and $B$ be the $n \times |\scrB|$ matrix whose columns are the elements of $B$. Let $\rank_\kk(\cdot)$ denote the rank of a matrix over $\kk$. The condition we are interested in is the following:
\begin{align}
\rank_\kk(B) = \dim(\conv(\scrB)) + 1 \label{rank-condition-1}
\end{align}

\begin{lemma} \label{newton-existence-lemma}
Assume \eqref{rank-condition-1} holds. Then the set of polynomials $g$ supported at $\scrB$ such that $\partialxj{g}$, $j = 1, \ldots, n$, have no common zero on $\nktorus$ contains a nonempty Zariski open subset of the space of all polynomials supported at $\scrB$.
\end{lemma}

\begin{proof}
This is the in fact the main content of the proof of \cref{milnor-thm} (the only place where the zero characteristic played a role in that proof was to ensure that \eqref{rank-condition-1} holds).
\end{proof}

\begin{cor} \label{newton-existence-1}
If \eqref{rank-condition-1} holds with $\scrB = \Delta \cap \scrA$ for each face $\Delta$ of $\nd(\scrA)$, then $\scrN'^0_0(\scrA) \neq \emptyset$. \qed
\end{cor}

Assertion \eqref{milnor-eq} of \cref{milnor-thm} implies that for $\scrA$ to admit polynomials which are partially non-degenerate at the origin, it is necessary that
\begin{align}
\multAzero < \infty \label{mult-Gamma-finite}
\end{align}
\Cref{mult-support} implies that \eqref{mult-Gamma-finite} is equivalent to the condition that $|\{j : \scrA_j \cap \ri \neq \emptyset\}| \geq |I|$ for each $I \subseteq [n]$. The following is an immediate corollary of \cref{Newton-to-Milnor,newton-existence-1}.

\begin{cor} \label{milnor-existence-1}
Assume \eqref{mult-Gamma-finite} holds and \eqref{rank-condition-1} holds with $\scrB = \Delta \cap \scrA$ for each face $\Delta$ of $\nd(\scrA)$, then $\scrM'_0(\scrA) \neq \emptyset$. \qed
\end{cor}

\begin{question}
Is the condition from \cref{newton-existence-1} necessary for $\scrN'^0_0(\scrA)$ to be nonempty?
\end{question}

\subsection{Relation among non-degeneracy conditions} \label{non-degenerate-relations}
Given a power series $f$, let us write $(N)$, $(I)$, $(P)$ to denote respectively the conditions that $f$ is Newton non-degenerate, inner Newton non-degenerate and partially non-degenerate at the origin. If $p := \character(\kk)$ is nonzero, then $(P)$ does not imply $(I)$, e.g. $x^p + y^p + x^{p+1} + y^{p+1}$ is partially non-degenerate at the origin, but it is inner Newton degenerate. This observation together with the discussion from \cref{classical-section} implies the relations depicted in \cref{fig:relations}, where ``$(N_{\milnor < \infty})$'' denotes the condition that $f$ is Newton non-degenerate and $\milnor(f) < \infty$.

\begin{center}
\begin{figure}[h]

\tikzcdset{arrow style=tikz, diagrams={>={Computer Modern Rightarrow[width=8pt,length=3pt]}}}

\begin{subfigure}[b]{0.3\textwidth}
\begin{tikzcd}[column sep=tiny]
& (P)
\arrow[ld, double, "/" anchor=center, shift left = 0.5ex] \arrow [ld, double, shift right=0.5ex, <-]
\arrow[rd, double,  shift right = 0.5ex, swap, "?"] \arrow [rd, double, shift left= 0.5ex, <-]
& \\
(N_{\milnor < \infty})
\arrow[double, rr, "?" , shift left=.5ex] \arrow[double, rr, "/" anchor=center, <-, shift right=0.5ex]
&
&(I)
\end{tikzcd}
\caption{zero characteristic}
\label{fig:relations-0}
\end{subfigure}

\begin{subfigure}[b]{0.3\textwidth}
\begin{tikzcd}[column sep=tiny]
& (P)
\arrow[ld, double, "/" anchor=center, shift left = 0.5ex] \arrow [ld, double, shift right=0.5ex, <-]
\arrow[rd, double,  shift right = 0.5ex, "/" anchor=center] \arrow [rd, double, shift left= 0.5ex, <-]
& \\
(N_{\milnor < \infty})
\arrow[double, rr, "?" , shift left=.5ex] \arrow[double, rr, "/" anchor=center, <-, shift right=0.5ex]
&
&(I)
\end{tikzcd}
\caption{positive characteristic}
\label{fig:relations-+}
\end{subfigure}
\caption{Relation among non-degeneracy conditions}  \label{fig:relations}
\end{figure}
\end{center}

\begin{problem}
Determine if the question-marked implications from \cref{fig:relations} are valid.
\end{problem}

We now show that in zero characteristic the implication $(P) \im (I)$ does hold in dimension $\leq 3$.

\begin{prop} \label{Milnor-to-inner}
Let $f \in \kk[x_1, \ldots, x_n]]$ be partially non-degenerate at the origin. If $n \leq 3$ and $\character(\kk)$ is zero, then $f$ is also inner Newton non-degenerate.
\end{prop}

\begin{proof}
Since all the $\partialxj{f}$ can not be identically zero on any axis, $\nd(f)$ satisfies the following property:
\begin{align}
\parbox{0.6\textwidth}{
the distance from any axis to $\nd(f)$ can not be greater than $1$.
}\label{ndistance}
\end{align}
This leads to the possibilities of \cref{fig-partial-2} in the case that $n = 2$. If $\nd(f)$ is convenient, take $\Gamma = \nd(f)$, otherwise take $\Gamma$ to be the union of $\nd(f)$ and edges (with appropriate slopes) from the end points of $\nd(f)$ to some points on the axes (e.g.\ the ``dashed edges'' in \cref{fig-partial-2}). It is straightforward to check that $f$ is inner Newton non-degenerate with respect to $\Gamma$.

\begin{figure}[h]
\begin{center}

\begin{tikzpicture}[scale=0.3]
\def\shiftone{12}

\def\opazero{0.5}
\def\tx{2}
\def\ty{3}
\def\gridx{7.5}

\def\gridy{5.5}
\def\colorzero{green}
\def\colorone{red}

\draw [gray,  line width=0pt] (-0.5,-0.5) grid (\gridx,\gridy);
\draw [<->] (0, \gridy) |- (\gridx, 0);

\draw[ultra thick, \colorone]  (7,0) -- (4,1) -- (2,2) -- (1,3) -- (0,5);
\fill[\colorzero, opacity=\opazero ]  (7,0) -- (4,1) -- (2,2) -- (1,3) -- (0,5) -- (0,\gridy) -- (\gridx,\gridy) -- (\gridx,0) -- cycle;

\begin{scope}[shift={(\shiftone,0)}]
	\draw [gray,  line width=0pt] (-0.5,-0.5) grid (\gridx,\gridy);
	\draw [<->] (0, \gridy) |- (\gridx, 0);
	\draw[ultra thick, \colorone]  (7,0) -- (4,1) -- (2,2) -- (1,3);
	\fill[\colorzero, opacity=\opazero ]  (7,0) -- (4,1) -- (2,2) -- (1,3) -- (1,\gridy) -- (\gridx,\gridy) -- (\gridx,0) -- cycle;
	\draw[ thick, dashed] (1,3) -- (0,4);

\end{scope}

\begin{scope}[shift={(2*\shiftone,0)}]
	\draw [gray,  line width=0pt] (-0.5,-0.5) grid (\gridx,\gridy);
	\draw [<->] (0, \gridy) |- (\gridx, 0);
	\draw[ultra thick, \colorone]  (4,1) -- (2,2) -- (1,3) -- (0,5);
	\fill[\colorzero, opacity=\opazero ] (4,1) -- (2,2) -- (1,3) -- (0,5)-- (0,\gridy) -- (\gridx,\gridy) -- (\gridx,1) -- cycle;
	\draw[ thick, dashed] (4,1) -- (6,0);

\end{scope}

\begin{scope}[shift={(3*\shiftone,0)}]
	\draw [gray,  line width=0pt] (-0.5,-0.5) grid (\gridx,\gridy);
	\draw [<->] (0, \gridy) |- (\gridx, 0);
	\draw[ultra thick, \colorone]  (4,1) -- (2,2) -- (1,3);
	\fill[\colorzero, opacity=\opazero ]   (4,1) -- (2,2) -- (1,3) -- (1,\gridy) -- (\gridx,\gridy) -- (\gridx,1) -- cycle;
	\draw[ thick, dashed] (1,3) -- (0,4);
	\draw[ thick, dashed] (4,1) -- (6,0);

\end{scope}

\end{tikzpicture}

\end{center}
\caption{Four possibilities for $\nd(f)$ in dimension two} \label{fig-partial-2}
\end{figure}

Now assume that $n = 3$. If $\nd(f)$ does not touch (at least) one of the three coordinate hyperplanes, then $f$ would be divisible by some $x_i$. The partial non-degeneracy of $f$ at the origin would then imply that $f = x_ig$ such that $g(0) \neq 0$, i.e.\ $\nd(f) = \{e_i\}$, where $e_i$ is the $i$-th standard unit vector in $\rr^3$. In that case $f$ would be inner Newton non-degenerate with any convenient diagram with a vertex at $e_i$. Therefore we may assume that $\nd(f)$ touches every coordinate hyperplane. Let $\Gamma$ be the Newton diagram of $f + x_1^N + x_2^N + x_3^N$ for some $N \gg 1$. We claim that $f$ is inner Newton non-degenerate with respect to $\Gamma$. Indeed, let $\Delta$ be an inner face of $\Gamma$ and $I \subseteq \{1, 2, 3\}$ be such that $\Delta \cap \ri \neq \emptyset$. We will check that \eqref{inner-property} holds for $\Delta$ and $I$. Let $\Delta' := \Delta \cap \ri$. If $\Delta' \cap \nd(f) = \emptyset$, then $\Delta' = \{Ne_i\}$ for some $i$. Assume \woutlog\ that $i = 1$. Since $\Delta \cap \nd(f) \neq \emptyset$ (since e.g.\ $\Delta$ is an inner face), it follows due to property \eqref{ndistance} that there is an element on $\Delta$ of the form $(k,1,0)$ or $(k,0,1)$ for some $k \geq 0$. Then $x_1^k \in \jjj(f_\Delta)$, and therefore $V(\jjj(f_\Delta))$ does not contain any point on $x_1$-axis other than the origin, as required. It remains to consider the case that $\Delta'' := \Delta' \cap \nd(f) \neq \emptyset$. If $|I| = 1$ or $|I| = 3$, it is straightforward to check that violation of \eqref{inner-property} implies violation of partial non-degeneracy of $f$. So we may assume \woutlog\ $I = \{1, 2\}$. Since $\nd(f)$ is not contained in the $(x_1,x_2)$-plane, it follows that $\dim(\Delta'') \leq 1$. If $\dim(\Delta') = 0$, then $f_\Delta = cx_1^{\alpha_1}x_2^{\alpha_2} + x_3g$ for some polynomial $g$, $(\alpha_1, \alpha_2) \in \znzero \setminus \{\origin\}$ and $c \neq 0$. It is then clear that either $\partialxone{f_\Delta}$ or $\partialxii{f_\Delta}{2}$ does not vanish at any point on $\Kstari$. So assume $\dim(\Delta'') = 1$. It then follows that $\Delta'' = \Delta'$, and if $N$ is sufficiently large, then $\Delta$ is in fact a (two dimensional) face of $\nd(f)$ containing $\Delta'$, see \cref{fig:3inner-Delta}.

\begin{center}
\begin{figure}[h]
\def\scale{0.6}
\def\viewx{75}
\def\viewy{15}
\def\colorzero{blue}
\def\colorzerob{black}
\def\colorone{orange}
\def\coloroneb{red}
\def\coloreta{green}
\begin{subfigure}[b]{0.3\textwidth}
\begin{tikzpicture}[scale=\scale]
\pgfplotsset{every axis title/.append style={at={(0,-0.2)}}, view={\viewx}{\viewy}, axis lines=middle, enlargelimits={false}}

\begin{axis}[
	xticklabels = \empty, yticklabels = \empty, zticklabels = \empty,
	xtick = \empty, ytick = \empty, ztick = \empty
]
	\addplot3[fill=\colorzero,opacity=\opazero] coordinates{(1,6,0) (6,1,0) (4,0,3) (1,1,5) (0,3,4) (1,6,0)};
	\addplot3 [ultra thick, \colorzerob] coordinates{(1,6,0) (6,1,0)};
	\addplot3 [ultra thick, \coloreta, ->] coordinates{(3,4,0) (4,5,0)};

\end{axis}
\node[anchor = west] at (2.5,2.5) {\picfontsize $\Delta$};
\node[anchor = west] at (2.75,0) {\picfontsize $\Delta'$};
\node[anchor = west] at (5.5,0.5) {\picfontsize $\nu$};
\end{tikzpicture}
\caption{$\Delta$}  \label{fig:3inner-Delta}
\end{subfigure}
\begin{subfigure}[b]{0.3\textwidth}
\begin{tikzpicture}[scale=\scale]
\pgfplotsset{every axis title/.append style={at={(0,-0.2)}}, view={\viewx}{\viewy}, axis lines=middle, enlargelimits={false}}

\begin{axis}[
	xticklabels = \empty, yticklabels = \empty, zticklabels = \empty,
	xtick = \empty, ytick = \empty, ztick = \empty
]
	\addplot3[fill=white,opacity=0] coordinates{(1,6,0) (6,1,0) (4,0,3) (1,1,5) (0,3,4) (1,6,0)};
	\addplot3[fill=\colorzero,opacity=\opazero] coordinates{(2,4,0) (4,2,0) (4,0,2) (1,1,4) (0,3,3) (2,4,0)};
	\addplot3 [ultra thick, \coloreta, ->] coordinates{(3,3,0) (4,4,0)};

\end{axis}
\node[anchor = west] at (1,2.5) {\picfontsize $\partialxii{\Delta}{3}$};
\node[anchor = west] at (4.5,0.5) {\picfontsize $\nu$};
\end{tikzpicture}
\caption{Case 1: $\partialxii{\Delta}{3}$\\ touches $(x_1,x_2)$-plane}  \label{fig:3inner-Delta3-1}
\end{subfigure}
\begin{subfigure}[b]{0.3\textwidth}
\begin{tikzpicture}[scale=\scale]
\pgfplotsset{every axis title/.append style={at={(0,-0.2)}}, view={\viewx}{\viewy}, axis lines=middle, enlargelimits={false}}

\begin{axis}[
	xticklabels = \empty, yticklabels = \empty, zticklabels = \empty,
	xtick = \empty, ytick = \empty, ztick = \empty
]
	\addplot3[fill=white,opacity=0] coordinates{(1,6,0) (6,1,0) (4,0,3) (1,1,5) (0,3,4) (1,6,0)};
	\addplot3[fill=\colorzero,opacity=\opazero] coordinates{(4,0,2) (1,1,4) (0,3,3) (1,3,2) (4,0,2)};
	\addplot3[fill=\colorone,opacity=\opazero] coordinates{(4,0,2) (1,3,2) (2,5,0) (4,3,0) (4,0,2)};
	\addplot3 [ultra thick, \coloroneb] coordinates{(4,0,2) (1,3,2)};
	\addplot3 [ultra thick, \coloroneb] coordinates{(2,5,0) (4,3,0)};
	\addplot3 [ultra thick, \coloreta, ->] coordinates{(3,4,0) (4,5,0)};

\end{axis}
\node[anchor = west] at (1,3.25) {\picfontsize $\partialxii{\Delta}{3}$};
\node[anchor = east] at (1.1,2) {\picfontsize $\Lambda''$};
\node at (2.75,1.75) {\picfontsize $\Lambda$};
\node[anchor = east] at (3.6,0.3) {\picfontsize $\Lambda'$};
\node[anchor = west] at (5.5,0.5) {\picfontsize $\nu$};
\end{tikzpicture}
\caption{Case 2: $\partialxii{\Delta}{3}$ does not touch $(x_1,x_2)$-plane}  \label{fig:3inner-Delta3-2}
\end{subfigure}
\caption{Scenarios when $\dim(\Delta'') = 1$} \label{fig:3inner}
\end{figure}
\end{center}

If $\partialxii{\Delta}{3}$ touches the $(x_1,x_2)$-plane, then applying the partial non-degeneracy condition \eqref{partial-non-degeneracy} with $I = \{1,2\}$ and $\nu$ equal to the ``inner normal'' to $\Delta'$ in $\ri$ shows that $V(\jjj(f_\Delta)) \cap \Kstari = \emptyset$. So assume that $\partialxii{\Delta}{3}$ does not touch $(x_1,x_2)$-plane. In this case we claim that $V(\partialxone{f_{\Delta'}},\partialxii{f_{\Delta'}}{2}) \cap \Kstari = \emptyset$. Indeed, assume to the contrary that there is a common zero $a = (a_1,a_2,0)$ of $\partialxone{f_{\Delta'}},\partialxii{f_{\Delta'}}{2}$ on $\Kstari$. The partial non-degeneracy implies that the intersection of $\partialxii{\Gamma}{3}$ with the $(x_1,x_2)$-plane is nonempty, and if $\Lambda'$ is the face of the intersection determined by $\nu$, then $(\partialxii{f}{3})_{\Lambda'}$ does {\em not} vanish at $a$. It is straightforward to check that there is a face $\Lambda$ of $\partialxii{\Gamma}{3}$ which contains $\Lambda'$ and also intersects $\partialxii{\Delta}{3}$. Let $\Lambda'' := \Lambda \cap \partialxii{\Delta}{3}$. If $(\partialxii{f}{3})_{\Lambda''}$ does not vanish after substituting $x_1 = a_1$ and $x_2 = a_2$, then one can find a zero of $(\partialxii{f}{3})_{\Lambda}$ of the form $(a_1,a_2,b)$ for some $b \in \kk^*$, and this would contradict partial non-degeneracy condition \eqref{partial-non-degeneracy} with $I = \{1,2,3\}$ and $\nu$ being an inner normal to $\Lambda$ in $\rr^3$. On the other hand, if $(\partialxii{f}{3})_{\Lambda''}$ vanishes after substituting $x_1 = a_1$ and $x_2 = a_2$, then $\Lambda''$ must be an edge parallel to $\Delta'$ and the partial non-degeneracy condition \eqref{partial-non-degeneracy} is violated with $I = \{1,2,3\}$ and $\nu$ being an inner normal to $\Lambda''$ in $\rr^3$. This proves that $V(\partialxone{f_{\Delta'}},\partialxii{f_{\Delta'}}{2}) \cap \Kstari = \emptyset$, as claimed. But then it is clear that $V(\jjj(f_\Delta)) \cap \Kstari$ is also empty, as required.
\end{proof}

\subsection{Conditions for validity of Kushnirenko's formula in positive characteristic} \label{newton-problem}
Consider the case that $p := \character(\kk) > 0$. The characterization of the conditions under which $\milnor(\scrA) = \newton(\nd(\scrA))$ established in \cref{milnor=newton}, in particular condition \eqref{mult-equality} of \cref{milnor=newton}, is {\em not} explicit. It would be interesting to find simple criteria under which this condition holds. The criterion from \cref{milnor=newton-prop-1} is sufficient, but not necessary, e.g.\ if $f = x + y^p$, then \eqref{milnor=newton-condition-1} fails (and $f$ is not Newton non-degenerate as well), but $\milnor(f) = \newton(\nd(f)) = 0$. Note that $f$ is {\em inner Newton non-degenerate}. Y.\ Boubakri, G.-M.\ Greuel and T.\ Markwig claim in \cite[Theorem 3.5]{boubakri-greuel-markwig} that $\milnor(f) = \newton(\nd(f))$ if $f$ is inner Newton non-degenerate, i.e.\ Wall's result \cite[Theorem 1.6]{wall} extends to positive characteristics.

\begin{problem}
Determine necessary and sufficient conditions for $\milnor(f) = \newton(\nd(f))$ to hold when $\character(\kk) > 0$. In particular, is inner Newton non-degeneracy necessary for this equality?
\end{problem}

\subsection{Monotonicity of Newton number}
\Cref{milnor-thm,milnor=newton-0} and the monotonicity of intersection multiplicity (\cref{monotonic-remark-0}) implies that the Newton number $\newton(\Gamma)$ of a diagram $\Gamma$ grows monotonically with $\Gamma$. Even though the arguments in this implication involve nontrivial algebraic geometry, the monotonicity of Newton numbers is a purely ``elementary'' convex geometric statement. It is a problem of V.\ I.\ Arnold \cite[Problem 1982-16]{arnold-problems} to find an elementary proof of this statement. S.\ K.\ Lando wrote in the commentary of that problem that he gave an elementary proof for $n = 2$. For $n = 3$ an elementary proof was given by S.\ Brzostowski, T.\ Krasi{\'n}ski and J.\ Walewska \cite{brzostowski-krasinski-walewska}. Since the monotonicity of the formula \eqref{mult-formula} for intersection multiplicity is obvious (see \cref{monotonic-remark-0}), one possible strategy would be to find an elementary proof of the following identity (which is an immediate consequence of \cref{milnor-thm,milnor=newton-0}):
\begin{align*}
\newton(\Gamma)
	&=  \sum_{I \in \tAone}
		\multzerostar{\Gamma^{I}_1, \Gamma^{I}_{j_2}}{\Gamma^{I}_{j_{|I|}}}
		\times
		\multzero{\pi_{[n]\setminus I}(\Gamma_{j'_1})}{\pi_{[n]\setminus I}(\Gamma_{j'_{n-|I|}})}
\end{align*}
where the right hand side is as in \cref{milnor=newton-0} with $\Gamma_j := \nd(\{\alpha - e_j:\alpha \in \Gamma + \rzeroo{n},\ \alpha - e_j \in \znzero\})$, $j = 1, \ldots, n$.

config: arxiv


TITLE: Why are Killing horizons related to event horizons?
QUESTION [2 upvotes]: According to Carroll (6.3 Killing horizons):

In general, if a Killing vector field $\chi^\mu$is null along some hypersurface $\Sigma$, we say that $\Sigma$ is a Killing horizon of $\chi^\mu$.

where a Killing vector field is a vector field such that the metric is conserved i.e. $\chi^\mu=(\partial_{\sigma_*})^\mu\implies \partial_{\sigma_*}g_{\mu\nu}=0$.
He then goes on to say, under the necessary assumptions, that 'Every event horizon of $\Sigma$ in a stationary, asymptotically flat spacetime is a killing horizon for some Killing vector field $\chi^\mu$.'
This last fact is not clear to me and I generally don't understand what Killing horizons have to do with event horizons. This makes it hard for me to understand why Killing horizons are even useful.
This question is related but doesn't answer why these two types of horizons are related.

REPLY [1 votes]: The following is just my quick and dirty understanding after skimming Hawking and Ellis pp. 323-340, and looking at this talk, and this SE question.
We want to talk about the notion that gravitational collapse is complete, and a black hole has reached an equilibrium.
To a distant observer, this is defined by whether the spacetime is stationary, i.e., has a timelike Killing vector "near infinity."
To an observer who intends to poke and prod the black hole from nearby, we want some other definition. However, this is hard, because if we poke the black hole with a stick, we lose our stick. We would like to say that somehow this observer doesn't see the horizon "changing over time," but the horizon is a null surface, so this doesn't quite make sense. You can't visit the horizon once and then visit it again later to see if it's changed. The way we get around this is by the definition of a Killing horizon. A black hole in equilibrium is one whose event horizon is also a Killing horizon. (Not all Killing horizons are event horizons. A light cone in Minkowski space is a Killing horizon.)
The Hawking rigidity theorem says that if the spacetime is a vacuum spacetime and is analytic and regular, then if a distant observer says it's in equilibrium (stationarity), then a nearby observer does as well (horizon is Killing). This is a non-obvious theorem, because due to time dilation, the distant observer may have a hard time judging whether infalling matter has really fallen through the horizon "now."
Another way of stating the result of the theorem is that we actually have two independent Killing vectors at the horizon: the null one and another one that constitutes the extension of the timelike one that exists at infinity.
It's clearly necessary that we assume a vacuum spacetime, or at least have some kind of energy condition. For if not, then we could write down a metric for something like a Schwarzschild metric that oscillates in size, and there would be some matter field that would allow this, which we could find from the Einstein field equations.
There seems to be a lot of work going into strengthening the theorem by weakening the assumptions about analyticity and so on. An early version by Carter, before Hawking's, required axial symmetry. Hawking's proof looks long and high-tech to me, and it seems to involve some case splitting between spinning and static black holes, the former having an ergosphere which seems to complicate things. People seem to suspect that analyticity isn't essential (or maybe can be replaced by something more physically motivated?). Analyticity makes the proofs easier because apparently a theorem by Nomizu says that if there's analyticity, we can successfully extend the timelike Killing vector from infinity to the entire manifold (although it will not necessarily be timelike everywhere, e.g., inside the horizon of a Schwarzschild black hole).


TITLE: Clarification of multipole expansion for a point charge
QUESTION [2 upvotes]: In Griffith's electrodynamic: 3.4.2 He pointed out that the monopole term is the exact potential for a single point charge.
However I was under the impression that different configuration of a charge distribution can act as a point charge from superposition thus allowing other multipoles to exist?
If not how do I prove that a single point charge only has monopole?

REPLY [2 votes]: The multipole coefficients associated with a $1/|r|$ distribution $\rho$ depends on the choice of origin. For example, if you have a point charge and you choose the origin to be at that point charge, then it will have a pure monopole character. However, if you choose the origin to be elsewhere, it will have nonzero expansion coefficients other than the monopole. This is an artifact of your choice of coordinate system.
To make this rigorous, let $\mathbf{I}=\{I_0^0,I_1^{-1},I_1^{0},I_1^1,...\}$ and $\mathbf{R}=\{R_0^0,R_1^{-1},R_1^{0},R_1^1,...\}$ be the set of irregular and regular solid harmonics. Then the potential $V(\mathbf{r})$ due to $\rho$ admits the exterior and interior multipole expansions
$$V=\sum_{J=0}^\infty \mathbf{I}_J\rangle\left\langle\mathbf{R}_J,\rho\right\rangle\qquad\text{when }|\mathbf{r}|>r_\text{max}
\\
V=\sum_{J=0}^\infty \mathbf{R}_J\rangle\left\langle\mathbf{I}_J,\rho\right\rangle\qquad\text{when }|\mathbf{r}|<r_\text{min}
$$
or in matrix notation,
$$V=\mathbf{I}\mathbf{R}^\dagger\rho\qquad\text{when }|\mathbf{r}|>r_\text{max}
\\
V=\mathbf{R}\mathbf{I}^\dagger\rho\qquad\text{when }|\mathbf{r}|<r_\text{min}.$$
In the case where $\rho$ is purely real, we can use the real solid harmonics $\mathbf{I}'$ and $\mathbf{R}'$, which are related to the standard solid harmonics by a unitary block diagonal matrix $\mathbf{U}$ via $\mathbf{I}'=\mathbf{I}\mathbf{U}$ from which we obtain the analogous real expansions
$$V=\mathbf{I}\mathbf{R}^\dagger\rho=\mathbf{I}\mathbf{U}\mathbf{U}^\dagger\mathbf{R}^\dagger\rho=[\mathbf{I}'][\mathbf{R}']^\dagger\rho=[\mathbf{I}'][\mathbf{R}']^\mathsf{T}\rho\qquad\text{when }|\mathbf{r}|>r_\text{max}
\\
V=\mathbf{R}\mathbf{I}^\dagger\rho=\mathbf{R}\mathbf{U}\mathbf{U}^\dagger\mathbf{I}^\dagger\rho=[\mathbf{R}'][\mathbf{I}']^\dagger\rho=[\mathbf{R}'][\mathbf{I}']^\mathsf{T}\rho\qquad\text{when }|\mathbf{r}|<r_\text{min}$$
which has the advantage that the list of multipole moments $[\mathbf{I}']^\mathsf{T}\rho$ or $[\mathbf{R}']^\mathsf{T}\rho$ are purely real.
So, why does a point charge not located at the origin have moments other than a monopole? It's for the same reason why a washing machine with a raccoon inside of it will shake around when it's on a wash cycle: it's not balanced, as the charges (or mass) are not located at the center of the relevant coordinate system.
As an explicit proof of why a point charge not located at the origin can't have a pure monopole moment, suppose otherwise. Then a test charge will be uniformly accelerated towards the center of the coordinate system, instead of towards the point charge. This is a contradiction. Therefore, there must be higher moments involved.
Alternatively,  a detailed justification can also be obtained by applying the addition theorem for spherical harmonics, but hopefully the proof given in the previous paragraph is sufficiently illuminating to show why higher moments will appear when a point charge is not located at the chosen origin.
Here's a numerical example to compute the moments of a single point charge located at spherical coordinate $(R,\pi/2,0)$ in Mathematica (it also computes the potential $V$ at an arbitrary point and compares it to the potential obtained from direct application of $V=1/|\mathbf{r}-\mathbf{r}_0|$):
SolidHarmonicI[l_, m_, r_, \[Theta]_, \[Phi]_] :=
  Sqrt[(4 \[Pi])/(2 l + 1)]
    SphericalHarmonicY[l, m, \[Theta], \[Phi]]/r^(l + 1);
SolidHarmonicR[l_, m_, r_, \[Theta]_, \[Phi]_] :=
  Sqrt[(4 \[Pi])/(2 l + 1)] r^
   l SphericalHarmonicY[l, m, \[Theta], \[Phi]];
SphToCart = CoordinateTransform["Spherical" -> "Cartesian", #] &;
r = {R, \[Pi]/2, 0};(*Spherical coordinates of point charge*)

Q[L_, m_] := ((-1)^m SolidHarmonicR[L, -m, ##] & @@
    r) q;(*Exterior multipole moment or order (L,m)*)
MatrixForm[
 Table[Q[L, m], {L, 0, 5}, {m, -L, L}]]
rule = {R -> 1.2, q -> 2,
   rtest -> 5.2, \[Theta]test -> 1.2, \[Phi]test -> 2.3};
Chop[Sum[SolidHarmonicI[L, m, rtest, \[Theta]test, \[Phi]test] Q[L,
     m], {L, 0, 5}, {m, -L, L}] /. rule]
q/Norm[SphToCart@r -
    SphToCart@{rtest, \[Theta]test, \[Phi]test}] /. rule

0.332219
0.332273
$$\left(
\begin{array}{c}
 \{q\} \\
 \left\{\frac{q R}{\sqrt{2}},0,-\frac{q R}{\sqrt{2}}\right\} \\
 \left\{\frac{1}{2} \sqrt{\frac{3}{2}} q R^2,0,-\frac{q R^2}{2},0,\frac{1}{2} \sqrt{\frac{3}{2}} q R^2\right\}
   \\
 \left\{\frac{1}{4} \sqrt{5} q R^3,0,-\frac{1}{4} \sqrt{3} q R^3,0,\frac{1}{4} \sqrt{3} q R^3,0,-\frac{1}{4}
   \sqrt{5} q R^3\right\} \\
 \left\{\frac{1}{8} \sqrt{\frac{35}{2}} q R^4,0,-\frac{1}{4} \sqrt{\frac{5}{2}} q R^4,0,\frac{3 q
   R^4}{8},0,-\frac{1}{4} \sqrt{\frac{5}{2}} q R^4,0,\frac{1}{8} \sqrt{\frac{35}{2}} q R^4\right\} \\
\end{array}
\right)$$

Note that there are nonzero moments of all orders whenever $R\neq 0$. However, the potential at the test location is correct up to parts per thousand accuracy when the sum runs up to $L=4$.

how do I prove that a single point charge only has monopole?

Set $R=0$ in the above triangle of numbers. Everything vanishes except the monopole term.


TITLE: Analytic isomorphism of affine nonsingular curves
QUESTION [1 upvotes]: I'm getting confused on an exercise in Appendix B of Hartshorne's Algebraic Geometry:

If two affine nonsingular curves are isomorphic as complex manifolds, then they're isomorphic as algebraic varieties.

I don't know how the affine and nonsingular condition can be used here. Any proof or reference is appreciated.

REPLY [0 votes]: The field of rational functions are isomorphic as well as the ring of regular functions.
The non-singular requirement is to make sure the ring of regular functions is the same on both side.
An isomorphism of affine variety is the same as an isomorphism of their coordinate ring.
For example $x $ is in $\Bbb{C}[x]$ but not in $\Bbb{C}[x^2,x^3]$ ($\cong \Bbb{C}[y,z]/(y^3-z^2)$ the coordinate ring of $V(y^3-z^2)$) whereas in both case the function field is $\Bbb{C}(x)$ and the underlying complex manifold is $\Bbb{C}$.
Thus there is no affine variety isomorphism $\Bbb{A^1_C} \to V(y^3-z^2)$. This is because $V(y^3-z^2)$ is singular at $(0,0)$.

From the isomorphic field of rational functions you can construct a common compact Riemann surface $C$ (isomorphic to a non-singular projective curve) containing both $A,B$,  that $A,B$ are non-singular implies that their coordinate ring is the whole ring of meromorphic functions on $C$ whose poles are located on $C-A,C-B$.


TITLE: Mass sliding down a rotating sphere
QUESTION [0 upvotes]: I've been studying rotating frames of reference, and I faced a problem regarding a particular type of sum.
Imagine there is a point mass, sliding off a sphere, and eventually looses contact. I'm supposed to find the angle with the horizontal at which this happens. This problem is relatively simple and can be solved using a combination of free body diagrams of the forces and conservation of energy.
The angle comes out to be $\sin^{-1}\frac{2}{3}$. At this angle with the horizontal, the body loses contact.
Now imagine, the sphere is rotating about a central axis, with constant angular velocity $\omega$. I'm supposed to find the centrifugal force acting on the point mass. In order to do that, I need to also find the angle at which the body loses contact. Moreover, I need to find the force with respect to a frame attached to a sphere.
I try to draw a free-body diagram of the point mass as before, and I get the following equation :
$\frac{mv^2}{R} = mg\sin\theta - N - m\omega^{2}R\cos^2\theta$
Here I've assumed $\theta$ to be angle with the horizontal (latitude), and the last term is the component of the centrifugal force acting in the direction of the normal reaction.
I need to find out the centrifugal force i.e. $m\omega^2a = m\omega^2R\cos\theta$.
Hence I need to find $cos\theta$ here.
I try to use the law of conservation of energy as before,
$mgR-mgR\sin\theta = \frac{1}{2}mv^2$
Plugging this back into our force equation, I have :
$2mg(1-\sin\theta) = mg\sin\theta - N - m\omega^{2}R\cos^2\theta$.
Normally, as in the non-rotating case, I'd put $N=0$ at the point where the mass loses contact, and then find out the angle, as the normal force would disappear there. However, I'm stuck here, because of the extra term.
I've seen in many solutions, it is written "from a non-rotating frame, we can say $\omega$ is $0$", and then ignore the final term, and then calculate the angle to be exactly the same as the non-rotating sphere case i.e. $\sin^{-1}\frac{2}{3}$.
Then they plug this back to the expression for centrifugal force i.e. $m\omega^2R\cos\theta$.
However, I'm still unable to figure out, why are they suddenly considering $\omega=0$, even in the case of the rotating sphere. It can not be because of the frame being attached to the rotating sphere, as that would make $\omega=0$, but even the centrifugal force would disappear in that case. I'm not being able to understand this, or I'm missing something extremely obvious.
Is this because, from the outside, i.e. non-rotating inertial frame perspective, the centrifugal force does not exist, as it only exists in a non-inertial frame. So, from the outside, the angle of losing contact doesn't matter if the sphere is rotating or not. Is that why we put $\omega = 0$ ? I don't think that analogy is correct. What is the correct reasoning of plugging $\omega=0$ ?
Why is the angle of losing contact, same for the rotating sphere and the non-rotating sphere.
Here is a link to the question, however, on this website, it is done in a rather convoluted way.
Any help in understanding this would be highly appreciated.

REPLY [1 votes]: A pseudo-force (m$v^2$/R) can be used (along with other radial forces) to determine the radial acceleration of an object in a coordinate system which is rotating with the object. In this case, the object is moving along a vertical arc about the center of the sphere. The coordinate system rotating with it is rotating about a horizontal axis (not with the sphere).  In that system R is constant (as long as the object stays on the surface), and one can say that N +  m$v^2$/R – mg sin(θ) = 0.


TITLE: What does the mysterious constant marked by C on a slide rule indicate?
QUESTION [76 upvotes]: Years ago, before everyone (or anyone) had electronic calculators, I had a pocket slide rule which I used in secondary school until the first TI-30 cane out.
Recently I dug it out. Here's a photo of one end of it.

As you can see, there's a number $C$ marked at about $1.128$ (times some power of $10$; with a slide rule you supply that yourself) on the C and D scales. Reading across to the A scale, its square is about $1.27$. By the C1 scale (which reads reciprocals of the C scale) its reciprocal is about $0.886$ (times some power of $10$).
The only two special numbers marked are $C$ and $\pi$.
I'm not sure whether it's some frequently used constant that's used (eg) in some branch of engineering, or a number which is useful for some trick for using the slide rule.
Unlike $\pi$, which is marked on most of the scales, this mysterious $C$ only appears on the C and D scales, which are the main ones used for multiplication and division.
If you need me to, I can give more explanation of the various scales on the rule and how calculations are done. That might give some clues as to what $C$ is for.
I'm sure the instructions explained what $C$ was, but I last saw those in the 1970s.
Has anyone any idea what $C$ is and why it would be useful on a slide rule?

REPLY [80 votes]: I found the answer by googling "slide rule markings"! It took me straight to the Glossary of the International Slide Rule Museum, which gives C its own entry:

C - Gauge mark found on the C and D scales denoting $\sqrt{4/\pi} = 1.128$ for calculating the area of a circle and the volume of a cylinder. Place the C mark on the C scale over the diameter of a circle on the D scale. The area of the circle is found above the index on the A scale. If this is the base of a cylinder, without moving the slide, move the cursor to the height of the cylinder on the B scale. The volume is read on the A scale. This gauge mark was rendered obsolete with the advent of multi-lined cursors.

And there is, of course, so much more at that site.


TITLE: Does linearity require a field, ring, or group?
QUESTION [0 upvotes]: Given any set $S$, a function $f : S^n \to S^m$ is linear iff for all $a, b \in S, x, y \in S^n, f(ax+by) = af(x) + b(y)$.  This presupposes, of course, that addition and scalar multiplication in $S$ are defined.  What requirements do we impose on these operations? Must they constitute a field? A ring?  Since different definitions of addition and multiplication may be valid and result in fields, are there functions which are linear in any field definition, or are we always talking implicitly about a specific, fixed field?
Thinking about it further, the question as posed may be trivial, because we've only defined linearity, but haven't given it any significance.  Does the definition below tighten things:
Given a set $F$ of functions from $S^n \to S^m$, $L$ is the subset of $F$ such that:

$f, g \in F \implies f \circ g \in F$
There is a one-to-one correspondence between elements of $L$ and matrices, a matrix defined as $n \cdot m$ values from $S$.

What restrictions on addition and multiplication are necessary for that definition to work? To allow standard matrix multiplication to implement the functions? To allow composition to be implemented as matrix multiplication?

REPLY [3 votes]: The most general context over which linear algebra is usually done is in that of modules over a ring. Let $R$ be a ring (commutative, with unity), and $M$ be a set endowed with operations of scalar multiplication $R\times M\to M$ and addition $+: M\times M\to M$. If these operations are nice and compatible with all the usual properties of associativity, commutativity, distributivity, existence of identity, etc. then $M$ is said to be a module over $R$.
If $M$ and $M'$ are modules over a ring $R$ then a map $f: M\to M'$ is said to be linear if $f(ax+by)=af(x)+bf(y)$ for all $x,y\in M$ and $a,b\in R$.