algebra/inputs/finiteness_conditions.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 $A = (a_{ij}) \Mat(n,n,R)$. 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} %TODO: lernen

\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 $\tilde{A} \implies \exists \tilde{B} \subseteq B$ 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.
    \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}