517 lines
21 KiB
TeX
517 lines
21 KiB
TeX
\subsection{Finitely generated and Noetherian modules}
|
|
|
|
\begin{definition}[Generated submodule]
|
|
Let $R$ be a ring, $M$ an $R$-module, $S \subseteq M$.
|
|
Then the following sets coincide
|
|
\begin{enumerate}
|
|
\item
|
|
\[\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{ finite}, r_s \in R\right\},\]
|
|
\item
|
|
\[\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N,\]
|
|
\item
|
|
The $\subseteq$-smallest submodule of $M$ containing $S$.
|
|
\end{enumerate}
|
|
|
|
This subset of $N \subseteq M$ is called the
|
|
\vocab[Module!Submodule]{submodule of $M $ generated by $S$}.
|
|
If $N= M$ we say that
|
|
\vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}.
|
|
$M$ is finitely generated
|
|
$:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Noetherian $R$-module]
|
|
$M$ is a \vocab{Noetherian} $R$-module
|
|
if the following equivalent conditions hold:
|
|
\begin{enumerate}
|
|
\item
|
|
Every submodule $N \subseteq M$ is finitely generated.
|
|
\item
|
|
Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates.
|
|
\item
|
|
Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a
|
|
$\subseteq$-largest element.
|
|
\end{enumerate}
|
|
\end{definition}
|
|
\begin{proposition}[Hilbert's Basissatz]
|
|
\label{basissatz}
|
|
If $R$ is a Noetherian ring,
|
|
then the polynomial rings $R[X_1,\ldots, X_n]$
|
|
in finitely many variables are Noetherian.
|
|
\end{proposition}
|
|
\subsubsection{Properties of finite generation and Noetherianness}
|
|
|
|
\begin{fact}[Properties of Noetherian modules]
|
|
\begin{enumerate}
|
|
\item
|
|
Every Noetherian module over an arbitrary ring is finitely generated.
|
|
\item
|
|
If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is
|
|
finitely generated.
|
|
\item
|
|
Every submodule of a Noetherian module is Noetherian.
|
|
\end{enumerate}
|
|
\end{fact}
|
|
\begin{proof}
|
|
|
|
\begin{enumerate}
|
|
\item
|
|
By definition, $M$ is a submodule of itself.
|
|
Thus it is finitely generated.
|
|
\item
|
|
Since $M$ is finitely generated,
|
|
there exists a surjective homomorphism $R^n \to M$.
|
|
As $R$ is Noetherian, $R^n$ is Noethrian as well.
|
|
\item
|
|
trivial
|
|
\end{enumerate}
|
|
\end{proof}
|
|
|
|
\begin{fact}
|
|
Let $M, M', M''$ be $R$-modules.
|
|
\begin{enumerate}
|
|
\item
|
|
Suppose $M \xrightarrow{p} M''$ is surjective.
|
|
If $M$ is finitely generated (resp. Noetherian),
|
|
then so is $M''$.
|
|
\item
|
|
Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact.
|
|
If $M'$ and $M ''$ are finitely generated (reps. Noetherian),
|
|
so is $M$.
|
|
\end{enumerate}
|
|
\end{fact}
|
|
\begin{proof}
|
|
\begin{enumerate}
|
|
\item
|
|
Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$.
|
|
Then $p^{-1} M_i''$ yields a strictly ascending sequence.
|
|
If $M$ is generated by $S,
|
|
|S| < \omega$, then $M''$ is generated by $p(S)$.
|
|
\item
|
|
Because of 1.~we can replace $M'$ by $f(M')$
|
|
and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$
|
|
to be exact.
|
|
The fact about finite generation follows from
|
|
Einführung in die Algebra.
|
|
|
|
If $M', M''$ are Noetherian, $N \subseteq M$ a submodule,
|
|
then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$
|
|
are finitely generated.
|
|
Since $0 \to N' \to N \to N'' \to 0$ is exact,
|
|
$N$ is finitely generated.
|
|
\end{enumerate}
|
|
\end{proof}
|
|
\subsection{Ring extensions of finite type}
|
|
|
|
\begin{definition}[$R$-algebra]
|
|
Let $R$ be a ring.
|
|
An $R$-algebra $(A, \alpha)$ is a ring $A$
|
|
with a ring homomorphism $R \xrightarrow{\alpha} A$.
|
|
$\alpha$ will usually be omitted.
|
|
In general $\alpha$ is not assumed to be injective.
|
|
\\
|
|
\\
|
|
An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq A$.\\
|
|
A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is
|
|
a ring homomorphism with $\tilde{\alpha} = f \alpha$.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Generated (sub)algebra, algebra of finite type]
|
|
Let $(A, \alpha)$ be an $R$-algebra.
|
|
\begin{align*}
|
|
\alpha: R[X_1,\ldots,X_m] & \longrightarrow A[X_1,\ldots,X_m]\\
|
|
P = \sum_{\beta \in \N^m} p_\beta X^{\beta}
|
|
& \longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta}
|
|
\end{align*}
|
|
is a ring homomorphism.
|
|
We will sometimes write $P(a_1,\ldots,a_m)$ instead of
|
|
$(\alpha(P))(a_1,\ldots,a_m)$.
|
|
|
|
Fix $a_1,\ldots,a_m \in A^m$.
|
|
Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$.
|
|
The image of this ring homomorphism is the $R$-subalgebra of $A$
|
|
\vocab[Algebra!generated subalgebra]{generated by the $a_i$}.
|
|
$A$ is \vocab[Algebra!of finite type]{of finite type}
|
|
if it can be generated by finitely many $a_i \in I$.
|
|
|
|
For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the
|
|
intersection of all subalgebras containing $S$ \\
|
|
$=$ the union of subalgebras generated by finite $S' \subseteq S$\\
|
|
$= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$.
|
|
\end{definition}
|
|
\subsection{Finite ring extensions} % LECTURE 2
|
|
\begin{definition}[Finite ring extension]
|
|
Let $R$ be a ring and $A$ an $R$-algebra.
|
|
$A$ is a module over itself and the ringhomomorphism $R \to A$
|
|
allows us to derive an $R$-module structure on $A$.
|
|
$A$ \vocab[Algebra!finite over]{is finite over} $R$ /
|
|
the $R$-algebra $A$ is finite / $A / R$ is finite
|
|
if $A$ is finitely generated as an $R$-module.
|
|
\end{definition}
|
|
\begin{fact}[Basic properties of finiteness]
|
|
\begin{enumerate}[A]
|
|
\item
|
|
Every ring is finite over itself.
|
|
\item
|
|
A field extension is finite as a ring extension
|
|
iff it is finite as a field extension.
|
|
\item
|
|
$A$ finite $\implies$ $A$ of finite type.
|
|
\item
|
|
$A / R$ and $B / A$ finite $\implies$ $B / R$ finite.
|
|
\end{enumerate}
|
|
\end{fact}
|
|
\begin{proof}
|
|
\begin{enumerate}[A]
|
|
\item
|
|
$1$ generates $R$ as a module
|
|
\item
|
|
trivial
|
|
\item
|
|
Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module.
|
|
Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra.
|
|
\item
|
|
Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module
|
|
and $B$ by $b_1,\ldots,b_n$ as an $A$-module.
|
|
For every $b$ there exist $\alpha_j \in A$
|
|
such that $b = \sum_{j=1}^{n} \alpha_j b_j$.
|
|
We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$
|
|
for some $\rho_{ij} \in R$
|
|
thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$
|
|
and the $a_ib_j$ generate $B$ as an $R$-module.
|
|
\end{enumerate}
|
|
|
|
\end{proof}
|
|
|
|
\subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements?
|
|
This generalizes some facts about matrices to matrices with elements from
|
|
commutative rings with $1$.
|
|
\footnote{Most of this even works in commutative rings without $ 1$,
|
|
since $1$ simply can be adjoined.}
|
|
\begin{definition}[Determinant]
|
|
Let $A = (a_{ij}) \in \Mat(n,n,R)$.
|
|
We define the determinant by the Leibniz formula
|
|
\[
|
|
\det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)}.
|
|
\]
|
|
|
|
Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$,
|
|
where $M_{ij}$ is the determinant of the matrix resulting from $A$
|
|
after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column.
|
|
\end{definition}
|
|
\begin{fact}
|
|
\begin{enumerate}
|
|
\item
|
|
$\det(AB) = \det(A)\det(B)$.
|
|
\item
|
|
Development along a row or column works.
|
|
\item
|
|
Cramer's rule:
|
|
$A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$.
|
|
$A$ is invertible iff $\det(A)$ is a unit.
|
|
\item
|
|
Caley-Hamilton:
|
|
If $P_A = \det(T \cdot \mathbf{1}_n - A)$%
|
|
\footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$},
|
|
then $P_A(A) = 0$.
|
|
\end{enumerate}
|
|
\end{fact}
|
|
\begin{proof}
|
|
All rules hold for the image of a matrix under a ring homomorphism if they hold
|
|
for the original matrix.
|
|
The converse holds in the case of injective ring homomorphisms.
|
|
Caley-Hamilton was shown for algebraically closed fields in LA2 using the
|
|
Jordan normal form.
|
|
Fields can be embedded into their algebraic closure,
|
|
thus Caley-Hamilton holds for fields.
|
|
Every domain can be embedded in its field of quotients $\implies$
|
|
Caley-Hamilton holds for domains.
|
|
|
|
In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$
|
|
where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain)
|
|
under the morphism $S \to A$ of evaluation defined by
|
|
$X_{i,j} \mapsto a_{i,j}$.
|
|
Thus Caley-Hamilton holds in general.
|
|
\end{proof}
|
|
|
|
\subsection{Integral elements and integral ring extensions} %LECTURE 2
|
|
\begin{proposition}[on integral elements]
|
|
\label{propinte}
|
|
Let $A$ be an $R$-algebra, $a \in A$.
|
|
Then the following are equivalent:
|
|
\begin{enumerate}[A]
|
|
\item
|
|
$\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$.
|
|
\item
|
|
There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$.
|
|
\end{enumerate}
|
|
If $a_1, \ldots, a_k \in A$ satisfy these conditions,
|
|
there is a subalgebra of $A$ finite over $R$ and containing all $a_i$.
|
|
\end{proposition}
|
|
\begin{definition}
|
|
\label{intclosure}
|
|
Elements that satisfy the conditions from \ref{propinte} are called
|
|
\vocab{integral over} $R$.
|
|
$A / R$ is \vocab[Algebra!integral]{integral},
|
|
if all $a \in A$ are integral over $R$.
|
|
The set of elements of $A$ integral over $R$ is called the
|
|
\vocab{integral closure} of $R$ in $A$.
|
|
\end{definition}
|
|
\begin{proof}
|
|
\hskip 10pt
|
|
\begin{enumerate}
|
|
{\color{gray} \item[B $\implies$ A]
|
|
Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$
|
|
and finite over $R$.
|
|
Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module.
|
|
\begin{align*}
|
|
q: R^n & \longrightarrow B\\
|
|
(r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i
|
|
\end{align*}
|
|
is surjective.
|
|
Thus there are $\rho_{i} = \left(r_{i,j}\right)_{j=1}^n \in R^n$
|
|
such that $a b_i = q(\rho_i)$.
|
|
Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns.
|
|
Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$.
|
|
By induction it follows that
|
|
$q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$.
|
|
Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$
|
|
and using Caley-Hamilton,
|
|
we obtain $P(a) \cdot q(v) = 0$. $P$ is monic.
|
|
Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$.
|
|
Thus $P(a) = 0$ and $a$ satisfies A.}
|
|
\item[B $\implies$ A]
|
|
if $R$ is Noetherian.\footnote{This suffices in the exam.}
|
|
Let $a \in A$ satisfy B.
|
|
Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$.
|
|
Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$
|
|
with $0 \le i < n$.
|
|
As a finitely generated module over the Noetherian ring $R$,
|
|
$B$ is a Noetherian $R$-module.
|
|
Thus the ascending sequence $M_n$ stabilizes at some step $d$
|
|
and $a^d \in M_d$.
|
|
Thus there are $(r_i)_{i=0}^{d-1} \in R^d$
|
|
such that $a^d = \sum_{i=0}^{d-1} r_ia^i$.
|
|
\item[A $\implies$ B]
|
|
Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A,
|
|
i.e.~$a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$
|
|
with $r_{i,j} \in R$.
|
|
Let $B \subseteq A$ be the sub-$R$-module generated by
|
|
$a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$
|
|
with $0 \le \alpha_i < d_i$.
|
|
$B$ is closed under $a_1 \cdot $ since
|
|
\[
|
|
a_1a^{\alpha} =
|
|
\begin{cases}
|
|
a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1, \\
|
|
\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1.
|
|
\end{cases}
|
|
\]
|
|
By symmetry, this hold for all $a_i$.
|
|
By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$,
|
|
$B$ is invariant under $a^{\alpha}\cdot $.
|
|
Since these generate $B$ as an $R$-module,
|
|
$B$ is multiplicatively closed.
|
|
Thus A holds.
|
|
Furthermore we have shown the final assertion of the proposition.
|
|
\end{enumerate}
|
|
\end{proof}
|
|
\begin{corollary}
|
|
\label{cintclosure}
|
|
\begin{enumerate}
|
|
\item[Q]
|
|
Every finite $R$-algebra $A$ is integral.
|
|
\item[R]
|
|
The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$.
|
|
\item[S]
|
|
If $A$ is an $R$-algebra,
|
|
$B$ an $A$-algebra and $b \in B$ integral over $R$,
|
|
then it is integral over $A$.
|
|
\item[T]
|
|
If $A$ is an integral $R$-algebra and $B$ any $A$-algebra,
|
|
$b \in B$ integral over $A$,
|
|
then $b$ is integral over $R$.
|
|
\end{enumerate}
|
|
\end{corollary}
|
|
\begin{proof}
|
|
\begin{enumerate}
|
|
\item[Q]
|
|
Put $ B = A $ in B.
|
|
\item[R]
|
|
For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$,
|
|
hence integral over $R$.
|
|
From B it follows,
|
|
that the integral closure is closed under ring operations.
|
|
\item[S]
|
|
trivial
|
|
\item[T]
|
|
Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$.
|
|
Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$,
|
|
such that all $a_i \in \tilde{A}$.
|
|
$b$ is integral over $\tilde{A}$
|
|
Hence $\exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and $b \in \tilde{B}$.
|
|
Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite,
|
|
$\tilde{B} / R$ is finite and $b$ satisfies B.
|
|
\end{enumerate}
|
|
\end{proof}
|
|
|
|
\subsection{Finiteness, finite generation and integrality}
|
|
% some more remarks on finiteness, finite generation and integrality
|
|
|
|
\begin{fact}[Finite type and integral $\implies$ finite]
|
|
\label{ftaiimplf}
|
|
If $A$ is an integral $R$-algebra of finite type,
|
|
then it is a finite $R$-algebra.
|
|
\end{fact}
|
|
\begin{proof}
|
|
Let $A $ be generated by $\left( a_i \right)_{i=1}^{n}$ as an $R$-algebra.
|
|
By the proposition on integral elements (\ref{propinte}),
|
|
there is a finite $R$-algebra $B \subseteq A$ such that all $a_i \in B$.
|
|
We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra.
|
|
\end{proof}
|
|
\begin{fact}[Finite type in tower]
|
|
If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra
|
|
of finite type, then $B$ is an $R$-algebra of finite type.
|
|
\end{fact}
|
|
\begin{proof}
|
|
If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$,
|
|
then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$.
|
|
\end{proof}
|
|
{
|
|
\color{red}
|
|
\begin{fact}[About integrality and fields]
|
|
\label{fintaf}
|
|
Let $B$ be a domain integral over its subring $A$.
|
|
Then $B$ is a field iff $A$ is a field.
|
|
\end{fact}
|
|
}
|
|
\begin{proof}
|
|
Let $B$ be a field and $a \in A \setminus \{0\} $.
|
|
Then $a^{-1} \in B$ is integral over $A$,
|
|
hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$
|
|
for some $\alpha_i \in A$.
|
|
Multiplication by $a^{d-1}$ yields
|
|
$a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
|
|
|
|
On the other hand, let $B$ be integral over the field $A$.
|
|
Let $b \in B \setminus \{0\}$.
|
|
As $B$ is integral over $A$,
|
|
there is a sub-$A$-algebra $\tilde{B} \subseteq B$,
|
|
$b \in \tilde{B}$ finitely generated as an $A$-module,
|
|
i.e.~a finite-dimensional $A$-vector space.
|
|
Since $B$ is a domain,
|
|
$\tilde{B} \xrightarrow{b\cdot } \tilde{B}$
|
|
is injective,
|
|
hence surjective,
|
|
thus $\exists x \in \tilde{B} : b \cdot x \cdot 1$.
|
|
\end{proof}
|
|
\subsection{Noether normalization theorem}
|
|
\begin{lemma}
|
|
\label{nntechlemma}
|
|
Let $S \subseteq \N^n$ be finite.
|
|
Then there exists $\vec k \in \N^n$ such that $k_1 =1$
|
|
and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$
|
|
for $\alpha \neq \beta \in S$,
|
|
where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Intuitive:
|
|
For $\alpha \neq \beta$ the equation
|
|
$w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$
|
|
($\kappa \in \R^{n-1}$) defines a codimension $1$
|
|
affine hyperplane in $\R^{n-1}$.
|
|
It is possible to choose $\kappa$ such that all $\kappa_i$ are
|
|
$> \frac{1}{2}$
|
|
and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$
|
|
from the union of these hyperplanes.
|
|
By choosing the closest $\kappa'$ with integral coordinates,
|
|
each coordinate will be disturbed by at most $\frac{1}{2}$,
|
|
thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$.
|
|
|
|
More formally:\footnote{The intuitive version suffices in the exam.}
|
|
Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\}$.
|
|
We can choose $k$ such that $k_i > (i-1) M k_{i-1}$.
|
|
Suppose $\alpha \neq \beta$.
|
|
Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$.
|
|
Then the contributions of $\alpha_j$
|
|
(resp.~$\beta_j$)
|
|
with $1 \le j < i$ to $w_{\vec k}(\alpha)$
|
|
(resp. $w_{\vec k}(\beta)$)
|
|
cannot undo the difference $k_i(\alpha_i - \beta_i)$.
|
|
\end{proof}
|
|
|
|
\begin{theorem}[Noether normalization]
|
|
\label{noenort}
|
|
Let $K$ be a field and $A$ a $K$-algebra of finite type.
|
|
Then there are $a = (a_i)_{i=1}^{n} \in A$ which
|
|
are algebraically independent over $K$,
|
|
i.e.~the ring homomorphism
|
|
\begin{align*}
|
|
\ev_a: K[X_1,\ldots,X_n] & \longrightarrow A\\
|
|
P & \longmapsto P(a_1,\ldots,a_n)
|
|
\end{align*}
|
|
is injective.
|
|
$n$ and the $a_i$ can be chosen such that $A$ is finite over the image of
|
|
$\ev_a$.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
Let $(a_i)_{i=1}^n$ be a minimal number of elements
|
|
such that $A$ is integral over its $K$-subalgebra
|
|
generated by $a_1, \ldots, a_n$.
|
|
(Such $a_i$ exist, since $A$ is of finite type).
|
|
Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
|
|
If suffices to show that the $a_i$ are algebraically independent.
|
|
Since $A$ is of finite type over $K$ and thus over $\tilde{A}$,
|
|
by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite),
|
|
$A$ is finite over $\tilde{A}$.
|
|
Thus we only need to show that the $a_i$ are algebraically independent
|
|
over $K$.
|
|
Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\}$
|
|
such that $P(a_1,\ldots,a_n) = 0$.
|
|
Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and
|
|
$S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$.
|
|
For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define
|
|
$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
|
|
|
|
By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$
|
|
such that $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have
|
|
$w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
|
|
|
|
Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$.
|
|
\begin{claim}
|
|
$A$ is integral over the subalgebra $B$ generated by the $b_i$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
By the transitivity of integrality,
|
|
it is sufficient to show that the $a_i$ are integral over $B$.
|
|
For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$.
|
|
Thus it suffices to show this for $a_1$.
|
|
Define
|
|
$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
|
|
b_{n-1} + T^{k_n}) \in B[T]$.
|
|
We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$.
|
|
Hence it suffices to show that the leading coefficient of $Q$ is a unit.
|
|
|
|
We have
|
|
\[
|
|
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} =
|
|
T^{w_{\vec k}(\alpha)} +
|
|
\sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
|
|
\]
|
|
with suitable $\beta_{\alpha, l} \in B$.
|
|
|
|
By the choice of $\vec k$, we have
|
|
\[
|
|
Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)}
|
|
+ \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j
|
|
\]
|
|
with $q_j \in B$ and
|
|
$\alpha$ such that $w_{\vec k}(\alpha)$ is maximal subject
|
|
to the condition $p_\alpha \neq 0$.
|
|
Thus the leading coefficient of $Q$ is a unit.
|
|
\end{subproof}
|
|
|
|
This contradicts the minimality of $n$, as $B$ can be generated by $< n$
|
|
elements $b_i$.
|
|
\end{proof}
|