AeCodimension/main.tex

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\begin{document}

\title{Some thoughts on finite extensions of rings}

\author{Others}
\address{
  Other authors\dots
}
%\email{brandhorst@math.uni-sb.de}
%
\author{Matthias Zach}
\address{
  Matthias Zach:
  RPTU Kaiserslautern-Landau,
  67663 Kaiserslautern,
  Germany 
}

%\email{zach@mathematik.uni-kl.de}
\begin{abstract}
  Dah.
\end{abstract}
\maketitle

We are considering complex analytic maps $f \colon (\CC^n, 0) \to (\CC^p, 0)$ which 
are \textit{finite}. In particular, we may assume $n \geq p$. By definition, $f$ is 
finite if and only if its pullback $\varphi \colon \OO_p \to \OO_n$ is a finite 
extension of rings, i.e. $\OO_n = \CC\{x_1,\dots,x_n\}$ is a finite 
$\OO_p = \CC\{y_1,\dots,y_p\}$-module via $\varphi$. 

\begin{definition}
  Let $\varphi \colon R \to S$ be a morphism of commutative rings with unit. An element 
  $x \in S$ is called \textit{integral} over $R$ if there exists a \textit{monic}
  polynomial 
  \[
    P = X^r + c_{r-1} \cdot X^{r-1} + \dots + c_1 \cdot X + c_0 \in R[X]
  \]
  such that $P(x) = 0$ in $S$. 
\end{definition}

\begin{proposition}
  \label{prp:Finiteness}
  Let $\varphi \colon R \to S$ be a morphism of commutative, Noetherian rings with unit. 
  The set of elements in $S$ which are integral over $R$ is a subalgebra of $S$. 
  Moreover, $S$ is finite over $R$ if and only if $S$ is generated 
  as an $R$-algebra by finitely many integral elements. 
\end{proposition}
\noindent
For a proof see \cite[Theorem 4.2 and Corollary 4.5]{Eisenbud95}.

\medskip

For complex analytic mappings $f$ as above, we are working with rings of 
convergent power series in several variables. In practice, however, we will 
mostly encounter maps which admit a \textit{polynomial representative}
\[
  f \colon \CC^n \to \CC^p, \quad (x_1,\dots,x_n) \mapsto (y_1,\dots, y_p) = 
  (f_1(x), \dots, f_p(x)), \quad f_j(x) \in \CC[x_1,\dots,x_n].
\]
It seems to be no significant restriction for practical applications to 
assume that also this polynomial representative is actually finite. 
For finite extensions of polynomial rings we do have algorithmic methods 
to compute with the structure of $S$ as an $R$-module.

\begin{proposition}
  Let $\varphi \colon \CC[y] \to \CC[x]$ be a mapping of polynomial 
  rings given by $y_j \mapsto f_j(x_1,\dots,x_n)$. Let 
  \[
    J = \langle y_j - f_j(x) : j = 1,\dots, p\rangle \subset \CC[y, x]
  \]
  and $LJ$ the \textit{leading ideal} of $J$ with respect to a monomial 
  ordering which is an elimination ordering on the $x$-variables. 
  Then $\varphi$ is finite if and only if $\langle x_1,\dots, x_n\rangle^N \subset LJ$ 
  for some $N \in \NN$. In this case, the set of monomials 
  $\{ x^\alpha : \alpha \in \NN^n, \, x^\alpha \notin LI\}$
  are a generating set for $\CC[x]$ as a $\CC[y]$-module.
\end{proposition}
\noindent
TODO: Find a reference for the proof.
\medskip

Based on this result we can implement seamless transitions for elements $h \in S$ 
between their representation as polynomials $h \in \CC[x]$ 
and module elements $v \in R \cdot e_1 + \dots + R \cdot e_m$ with respect to a 
chosen system of generators over $R$. 

\medskip
Working with power series in computer algebra is an almost impossible undertaking 
beyond simple arithmetic. Recall that in general it must be deemed impossible to even decide 
equality of two given power series. Thus, the naive hope at this point would be 
that \textit{localizing} at the origins of the domain and codomain of $f$, 
respectively, would bring us sufficiently close to working with power 
series by restricting ourselves to those which are in fact \textit{rational functions}. 

\begin{proposition}
  \label{prp:NonFiniteness}
  Let $\varphi \colon R \to S$ be a finite extension of commutative Noetherian rings with 
  unit. Let $P \subset R$ be a prime ideal and $Q \subset S$ prime with 
  $\varphi^{-1}(Q) = P$ (i.e. $Q$ is a \textit{prime over} $P$). 
  Then the naturally induced morphism on the localizations 
  \[
    \phi \colon R_P \to S_Q
  \]
  is finite if and only if $Q$ is the only prime in $S$ with that property.
\end{proposition}

\begin{proof}
  According to Proposition 
  \ref{prp:Finiteness}, we have finiteness of $\phi$ if and only if an arbitrary element 
  $\frac{a}{b} \in S_Q$ is integral over $R_P$. Now by assumption both $a$ and $b$ are 
  integral over $R$ and hence over $R_P$ and since $\frac{a}{b} = a \cdot \frac{1}{b}$ 
  it suffices to consider elements of the form $\frac{1}{b}$ for $b \notin Q$. 

  Suppose we found a monic polynomial 
  \[
    F = X^r + c_{r-1} \cdot X^{r-1} + \dots + c_1 \cdot X^1 + c_0 \in R_P[X]
  \]
  with $F\left(\frac{1}{b}\right) = 0$ in $S_Q$. Then we can clear denominators to 
  obtain 
  \[
    F' = c'_r \cdot X^r + c'_{r-1} \cdot X^{r-1} + \dots + c'_1 \cdot X^1 + c'_0 \in R[X]
  \]
  with $F'\left(\frac{1}{b}\right) = 0$ and $c'_r \notin P$.
  But then, by definition of localization at $Q$, 
  \[
    u \cdot \left( c'_r + c'_{r-1} \cdot b + \dots + c'_1 \cdot b^{r-1} + c'_0 \cdot b^r
    \right) = 0
  \]
  in $S$ for some $u \notin Q$. This expression can be split as 
  \begin{equation}
    \label{eqn:Splitting}
    -u \cdot c'_r = u \cdot b \cdot \left(c'_{r-1} + c'_{r-2} \cdot b + \dots + c'_1 \cdot b^{r-2} + c'_0 \cdot b^{r-1} \right).
  \end{equation}

  Now suppose $Q'\neq Q$ was another prime with $\varphi^{-1}(Q') = P = \varphi^{-1}(Q)$. 
  Choosing $b \in Q' \setminus Q$ leads to a contradiction in (\ref{eqn:Splitting}):
  Clearly, the right hand side, containing $b$ as a factor, is an element in $Q'$. 
  Comparing with the left hand side, we find that 
  $c'_r \notin P = \varphi^{-1}(Q) = \varphi^{-1}(Q')$ and, hence, $\varphi(c'_r) \in S$ is 
  neither an element of $Q$, nor of $Q'$. This leaves us with $u \in Q'$. 

  To arrive at a contradiction, let 
  \[
    \mathrm{ord}_{Q'} \colon S \to \NN_0, \quad a \mapsto \mathrm{max} 
    \left\{k \in \NN_0 : a \in Q'^{(k)}\right\}
  \]
  where $Q'^{(k)}$ denotes the $k$-th \textit{symbolic power}, i.e. 
  the preimage of $\left(Q'_{Q'}\right)^k$ in $S$ under the localization 
  map $S \to S_{Q'}$. The map $\mathrm{ord}_{Q'}$ is a \textit{valuation} 
  and it is now easy to see that the valuation of the left hand side of 
  (\ref{eqn:Splitting}) is strictly less than the valuation of the 
  right hand side. 

  TODO: Find an argument to prove the converse.
% On the contrary, if $Q$ was the only prime over $P$, we can argue as follows. 
% Let $b \in S \setminus Q$ be arbitrary and 
% \[
%   G = X^r + a_{r-1} \cdot X^{r-1} + a_{r-2} \cdot X^{r-2} + \dots + a_1 \cdot X + a_0 \in 
%   R[X]
% \]
% a monic polynomial with $G(b) = 0$ in $S$. We may choose $G$ to be of minimal 
% degree with this property. Then 
% \[
%   G^\vee = 1 + a_{r-1} \cdot X + a_{r-2} \cdot X^2 + \dots + a_1 \cdot X^{r-1} + a_0 \cdot X^r
% \]
% satisfies $G^\vee\left(\frac{1}{b}\right) = 0$ in $S_Q$. Provided that we are allowed 
% to divide by $a_0$ in $R_P$, we would thus get a monic polynomial 
% in $R_P[X]$ annihilating $\frac{1}{b}$.
%
% Now suppose $a_0 \in P$ so that $\varphi(a_0) \in Q$. Expanding $G(b) = 0$ 
% then yields 
% is a unit in $
% Now suppose $\varphi(a_0) \in Q$, so
% $a_0 \in P = \varphi^{1}(Q)$. 

\end{proof}

\begin{example}
  Consider the map 
  \[
    f \colon \CC \to \CC, \quad x \mapsto y = x - x^3.
  \]
  Clearly, $\varphi \colon \CC[y] \to \CC[x]$ is finite and, in fact, 
  $\CC[x]$ is a free $\CC[y]$-module with generating set $1, x, x^2$. 

  For the analytification of this map $f \colon (\CC, 0) \to (\CC, 0)$ 
  the implicit function theorem yields that there exists a local inverse 
  \[
    y = x - x^3, \quad x = y + \sum_{k = 2}^\infty c_k \cdot y^k \in \CC\{y\}
  \]
  as a convergent power series in $y$. Therefore $\varphi^h \colon \CC\{y\} \to \CC\{x\}$ 
  is in fact an \textit{isomorphism} and $\CC\{x\}$ is generated as a $\CC\{y\}$-module 
  by the unit element $1 \in \CC\{x\}$. 

  Thus we find a discrepancy of $3 - 1 = 2$ in the ranks of the two modules over the 
  respective rings. One could hope that the localization 
  \[
    \phi \colon \CC[y]_{\langle y\rangle} \to \CC[x]_{\langle x \rangle}
  \]
  provides us with a reasonable approximation to $\varphi^h$, but 
  Proposition \ref{prp:NonFiniteness} tells us that the elements 
  \[
    M_p = \left\langle\left(\frac{1}{1 - x}\right)^k : k \leq p\right\rangle_{\CC[y]_{\langle y\rangle}}
  \]
  form an infinite, ever increasing sequence of 
  $\CC[y]_{\langle y \rangle}$-submodules of $\CC[x]_{\langle x \rangle}$ and 
  that, hence, $\CC[x]_{\langle x \rangle}$ does not have a finite 
  $\CC[y]_{\langle y\rangle}$-module structure. 
\end{example}
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