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\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 $S$
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\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 $S$.
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\end{definition}
\begin{definition}[Noetherian $R$-module]
$M$ is a \vocab{Noetherian} $R$-module if the
following equivalent conditions hold:
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\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.
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\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.
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\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.
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\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
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\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 $M$.
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\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.
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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.
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\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.
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\\
\\
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 $\tilde{\alpha} = f \alpha$.
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\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}
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\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 $a_i
\in I$.
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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 $P \mapsto (\alpha(P))(S)$.
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\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.
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\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.
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\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}
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\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 $1$.
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\footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.}
\begin{definition}[Determinant]
Let $A = (a_{ij})
\Mat(n,n,R)$.
We define the determinant by the Leibniz formula
\[
\det(A) \coloneqq \sum_{\pi
\in S_n} \sgn(\pi)
\prod_{i=1}^{n} a_{i, \pi(i)}
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\]
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.
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\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
$P_A(A) = 0$.
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\end{enumerate}
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\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}
%TODO: lernen
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\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:
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\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 $a$.
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\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 $a_i$.
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\end{proposition}
\begin{definition}
\label{intclosure}
Elements that satisfy the conditions from
\ref{propinte} are called
\vocab{integral over} $R$.
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$A / R$ 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 $A$.
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\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.
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\begin{align}
q: R^n & \longrightarrow B \\
(r_1,\ldots,r_n) & \longmapsto \sum_{i=1}^{n} r_i b_i
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\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.
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\end{enumerate}
\end{proof}
\begin{corollary}
\label{cintclosure}
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\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 $R$.
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\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 $R$.
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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 $\tilde{A} \implies \exists
\tilde{B} \subseteq B$ 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.
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\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.
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\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 $a_i \in B$.
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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.
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\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 $B$.
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\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.
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\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
$a^{-1} = \sum_{i=0}^{d-1} \alpha_i
a^{d-1-i} \in A$.
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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 $\exists x \in \tilde{B} : b
\cdot x \cdot
1$.
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\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 $w_{\vec
k}(\alpha) = \sum_{i=1}^{n} k_i
\alpha_i$.
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\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 $\le
\frac{\sqrt{n-1} }{2}$.
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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
$k_i(\alpha_i - \beta_i)$.
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\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
$\ev_a$.
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\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).
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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
$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n}
k_i\alpha_i$.
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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
$w_{\vec{k}}(\alpha) \neq
w_{\vec{k}}(\beta)$.
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Define $b_i \coloneqq a_{i+1} -
a^{k_{i+1}}_1$ for $1 \le i < n$.
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\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.
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We have
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\[
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}}
=
T^{w_{\vec k}(\alpha)} +
\sum_{l = 0}^{w_{\vec k}(\alpha) - 1}
\beta_{\alpha,
l} T^l
\]
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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
$p_\alpha \neq 0$.
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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 $b_i$.
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\end{proof}