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complete Noetherian rings/modules notes, add Hilbert basis theorem, some Noetherian exercises (#2)
* add proof for split exact sequences * port exercise R.2.9 * Noetherian rings: add ex. 3.2 & 3.5 * Hilbert basis theorem, Noetherian module properties * add Noetherian exercises 3.3 & 3.4
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@@ -80,7 +80,7 @@
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\maketitle
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\begin{abstract}
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Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}.
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Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}. For the exercises, I follow the assignments listed at \cite{mit-course}.
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Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
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@@ -178,7 +178,7 @@
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$$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$
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\end{defn}
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\subsection{$\mathbb{Z}$ and $K[X]$, two Principal Ideal Domains}
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\subsection{Z and K[X], two Principal Ideal Domains}
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\begin{lemma}{}
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@@ -696,11 +696,124 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi
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\end{proof}
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\begin{prop}{AM.2.10} \label{2.10}
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Split exact sequence. TODO
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\end{prop}
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\subsection{Sequences}
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\section{Noetherian rings}
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\begin{defn}{R.2.9.a}{Exact Sequence}
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Let a sequence of homomorphisms
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$$L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N$$
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It is \emph{exact} at $M$ if $im(\alpha)=ker(\beta)$.
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ie. $\beta \circ \alpha = 0$ and $\alpha$ maps surjectively to
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$ker(\beta)$.
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\end{defn}
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\begin{defn}{R.2.9.b}{Short Exact Sequence (s.e.s.)} \label{2.9}
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$$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$$
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is exact $\Longleftrightarrow~ L \subset M$ and $N=M / L$.
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Properties:
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\begin{itemize}
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\item $\alpha$ injective
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\item $\beta$ surjective
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\item $\alpha:~ L \Longrightarrow ker \beta$
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\item $\beta$ induces $M/\alpha(L) \longrightarrow N$
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\end{itemize}
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\end{defn}
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\begin{prop}{R.2.10}{Split exact sequence} \label{2.10}
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For the previous s.e.s., 3 equivalent conditions:
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\begin{enumerate}[i.]
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\item $\exists$ isomorphism $M \cong L \oplus N$, with
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\begin{align*}
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\alpha:~ &m \longmapsto (m,0)\\
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\beta:~ &(m, n) \longmapsto n
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\end{align*}
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\item $\exists$ a \emph{section} of $\beta$, that is, a map $s: N \longrightarrow M$ such that $\beta \circ s = id_N$
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\item $\exists$ a \emph{retraction} of $\alpha$, that is, a map $r: M \longrightarrow L$ such that $r \circ \alpha = id_L$
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\end{enumerate}
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If all i, ii, iii are satisfied, it is a split exact sequence.
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\end{prop}
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\begin{proof}
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Intuitively, when a s.e.s. \emph{splits} it means that the middle module $M$ is the direct sum of the other (outer) two modules, ie. $M = L \oplus N$.
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\begin{itemize}
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\item[(i to ii, iii)]
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if $M \cong L \oplus N$ such that $\alpha:~ m \longmapsto (m,0),~~ \beta:~ s(m, n) \longmapsto n$, we can define the maps
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for ii:
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\begin{align*}
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s:~ N &\longrightarrow L \oplus N\\
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s(n) &\longmapsto (0, n)
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\end{align*}
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Then $\beta(s(n))=\beta(0,n)$, so $\beta \circ s = id_N$.
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for iii:
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\begin{align*}
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r:~ L \oplus N &\longrightarrow L\\
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r(m,n) &\longmapsto m
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\end{align*}
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Then $r(\alpha(m))=r(m,0)$, so $r \circ \alpha = id_L$.
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\item[(ii to i)]
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assume $s: N \longrightarrow M$ such that $\beta \circ s = id_M$
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Want to show $M \cong im(\alpha) \oplus im(s)$.
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$\forall m \in M$, consider $m - s(\beta(m))$, apply $\beta$ to it:\\
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$\beta(m - s(\beta(m))) = \beta(m) - (\beta \circ s)(\beta(m)) = \beta(m) - \beta(m) = 0$
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Since $ker(\beta) = im(\alpha),~~ \exists! l \in L ~\text{such that}~ \alpha(l) = m - s(\beta(m))$.
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Thus $m = \alpha(l) + s(\beta(m))$.
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\vspace{0.3cm}
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Now, suppose $x \in im(\alpha) \cap im(s)$, then $x = \alpha(l)=s(n)$, apply $\beta$ to it: $\beta(\alpha(l)) = \beta(s(n)) ~\Longrightarrow~ 0=n$.
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If $n=0$, then $s(n)=0$, so the intersection is $\{0\}$.
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\vspace{0.3cm}
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Define
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\begin{align*}
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\phi: L \oplus N &\longrightarrow M\\
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\phi(l,n) &\longmapsto \alpha(l)+s(n)
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\end{align*}
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This isomorphism satisfies the required conditions.
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\item[(iii to i)] similar to the previous one.
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\end{itemize}
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\vspace{0.3cm}
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TL;DR:\\
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$$
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0 \longrightarrow L
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\substack{
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\stackrel{\alpha}{\longrightarrow}\\
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\stackrel{\longleftarrow}{r}
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}
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\substack{M \\[0.5ex] \cong L \oplus N}
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\substack{
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\stackrel{\beta}{\longrightarrow}\\
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\stackrel{\longleftarrow}{s}
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}
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N \longrightarrow 0
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$$
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\begin{align*}
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\alpha:~ &l \longmapsto (l,0)\\
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r:~ &(m,n) \longmapsto m\\
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\alpha &\circ r = id_L\\
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\beta:~ &(l,n) \longmapsto n\\
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s:~ &n \longmapsto (0,n)\\
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\beta &\circ s = id_N
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\end{align*}
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\end{proof}
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\section{Noetherian rings (and modules)}
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\begin{defn}{}{Ascending Chain Condition}
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A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain
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@@ -717,8 +830,8 @@ $\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \
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\item the set $\Sigma$ of ideals of $A$ has the a.c.c.; in other words, every increasing chain of ideals
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$$I_1 \subset I_2 \subset \ldots \subset I_k \subset \ldots$$
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eventually stops, that is $I_k = I_{k+1}=\ldots$ for some $k$.
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\item every nonempty set $S$ of iddeals has a maximal element
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\item every iddeal $I \subset A$ is finitely generated
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\item every nonempty set $S$ of ideals has a maximal element
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\item every ideal $I \subset A$ is finitely generated
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\end{enumerate}
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If these conditions hold, then $A$ is \emph{Noetherian}.
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\end{defn}
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@@ -741,7 +854,7 @@ As in with rings, it is equivalent to say that
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\begin{prop}{R.3.4.P}
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\begin{prop}{R.3.4.P}\label{R.3.4.P}
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Let $0 \longrightarrow L \xrightarrow{\ \alpha \ } M \xrightarrow{\ \beta \ } N \longrightarrow 0$ be a s.e.s. (split exact sequence, \ref{2.10}).
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Then, $M$ is Noetherian $\Longleftrightarrow~ L$ and $N$ are Noetherian.
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@@ -765,7 +878,7 @@ As in with rings, it is equivalent to say that
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$$L \cap M_1 = L \cap M_2 ~\text{and}~ \beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$$
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\end{lemma}
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\begin{proof}
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if $m\in M_2$, then $\beta(m) \in \beta(M_1) = \beta(M_2)$, so that there is an $n \in M_1$ such that $\beta(m)=\beta(m)$.
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if $m\in M_2$, then $\beta(m) \in \beta(M_1) = \beta(M_2)$, so that there is an $n \in M_1$ such that $\beta(m)=\beta(n)$.
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Then $\beta(m-n)=0$, so that
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$$m - n \in M_2 \cap ker(\beta)=M_1 \cap ker(\beta)$$
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@@ -773,6 +886,114 @@ As in with rings, it is equivalent to say that
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Hence $m \in M_1$, thus $M_1 = M_2$.
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\end{proof}
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\begin{cor}{R.3.5}{Properties of Noetherian modules.}\label{R.3.5}
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\begin{enumerate}[i.]
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\item if $\forall i \in [r],~~M_i$ are Noetherian modules, then
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$\bigoplus_{i=1}^r M_i$ is Noetherian.
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\item if $A$ a Noetherian ring, then an $A$-module $M$ is Noetherian iff it is finite over $A$.
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\item if $A$ a Noetherian ring, $M$ a finite module, then any submodule $N \subset M$ is again finite.
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\item if $A$ a Noetherian ring, and $\psi: A \longrightarrow B$ a ring homomorphism such that $B$ is a finite $A$-module, then $B$ is a Noetherian ring.
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\end{enumerate}
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\end{cor}
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\begin{proof}
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\begin{enumerate}[i.]
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\item a direct sum $M_1 \oplus M_2$ is a particular case of an exact sequence.
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Then, Proposition \ref{R.3.4.P} proves this statement when $r=2$. The case $r>2$ follows by induction.
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\item if $M$ finite, then $\exists~$ surjective homomorphism
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$$A^r \longrightarrow M \longrightarrow 0$$
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for some $r$, so that $M$ is a quotient
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$$M \cong A^r / N$$
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for some submodule $N \subset A^r$.
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$A^r$ is a Noetherian module by i., so $M$ is Noetherian due Proposition \ref{R.3.4.P}.
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Conversely, $M$ Noetherian implies $M$ finite.
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item as in previous implications:\\
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$M$ finite and $A$ Noetherian $\Longrightarrow$ $M$ is Noetherian,\\
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$\Longrightarrow$ since $N \subseteq M$, then $N$ is Noetherian too\\
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$\Longrightarrow$ which implies that $N$ is a finite $A$-module.
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\item $B$ is Noetherian as an $A$-module; but ideals of $B$ are submodules of $B$ as an $A$-submodule, so that $B$ is a Noetherian ring.
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\end{enumerate}
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\end{proof}
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\vspace{0.5cm}
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\begin{thm}{R.3.6}{Hilbert basis theorem} \label{hilbert-basis}
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if $A$ a Noetherian ring, then so is the polynomial ring $A[x]$.
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\end{thm}
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\begin{proof}
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Prove that any ideal $I \subset A[x]$ is fingen.
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Define auxiliary sets $J_n \subset A$ by
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$$J_n = \{ a \in A ~|~ \exists f \in I ~\text{s.th.}~ f = a x^n + b_{n-1}x^{n-1} + \ldots b_0 \}$$
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ie. $J_n$ is the set of leading coefficients of $I$ of degree $n$.
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$J_n$ is an ideal, since $I$ is an ideal.
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$J_n \subset J_{n+1}$, since for $f \in I$ also $x f \in I$.
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Therefore $J_1 \subset J_2 \subset \ldots \subset J_k \subset \ldots$ is an increasing chain of ideals.
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Using the assumption that $A$ is Noetherian, deduce that $J_n = J_{n+1}$ for some $n$.
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For each $m \leq n, ~~ J_m \subset A$ is fingen, ie.
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$$J_m = (a_{m,1}, \ldots a_{m, r_m})$$
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By definition of $J_m$, for each $a_{m,j}$ with $1 \leq j \leq r_m$,\\
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$\exissts$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
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$$\Longrightarrow~~ \{ f_{m,j} \}_{m<n; 1 \leq j \leq r_m}$$
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the set of elements of $I$.
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Claim: this finite set ($\{ f_{m,j} \}$) generates $I$.
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$\forall f \in I$, if $\deg f =m$, then its leading coefficient is $a \in J_m$,
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hence if $m \geq n$, then $a \in J_m=J_n$, so that
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$$a = \sum b_i a_{n,i} ~~ \text{with}~ b_i \in A$$
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and
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$$f - \sum b_i X^{m-n} \cdot f_{n, i}$$
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has degree $<m$.
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Similarly, if $m \leq n$, then $a \in J_m$, so that
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$$a = \sum b_i a_{m, i} ~~\text{with}~ b_i \in A$$
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and
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$$f - \sum b_i f_{n, i}$$
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has degree $<m$.
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\vspace{0.3cm}
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By induction on $m$, $f$ can be written as a linear combination of finitely many elements.
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Thus, any ideal of $A[x]$ is finitely generated.
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\end{proof}
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\begin{cor}{R.3.6.C}
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if $A$ a Noetherian ring, and $\psi: A \longrightarrow B$ a ring homomorphism such that $B$ is a fingen extension ring of $\psi(A)$, then $B$ is Noetherian.
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In particular, any fingen algebra over $\mathbb{Z}$ or over a field $K$ is Noetherian.
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\end{cor}
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\begin{proof}
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the assumption is that $B$ is a quotient of a polynomial ring,
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$$B \cong A[x_1, \ldots, x_n] / I$$
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for some ideal $I$.
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By the Hilbert basis theorem \ref{hilbert-basis} and induction,\\
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$A$ being Noetherian implies that $A[x_1, \ldots, x_n]$ is Noetherian.
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And by Corollary \ref{R.3.5}(iv),\\
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$A[x_1, \ldots, x_n]$ being Noetherian implies that $A[x_1, \ldots, x_n]/I$ is Noetherian.
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\end{proof}
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\newpage
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\section{Exercises}
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@@ -813,7 +1034,7 @@ The exercises that start with \textbf{R} are the ones from the book \cite{reid},
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Let
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$$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
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The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$.
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The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
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\begin{enumerate}[i.]
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\item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
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@@ -1001,6 +1222,195 @@ The exercises that start with \textbf{R} are the ones from the book \cite{reid},
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\subsection{Exercises Chapter 2}
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\begin{ex}{R.2.9}
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$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
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\end{ex}
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\begin{proof}
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Denote the generators of $L$ and $N$ respectively as
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\begin{align*}
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\{l_1, \ldots, l_k \} &\subseteq L\\
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\{n_1, \ldots, n_p \} &\subseteq N
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\end{align*}
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By s.e.s. definition,
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\begin{itemize}
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\item[-] $\alpha$ is injective (one-to-one), so
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$$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
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\item[-] $\beta$ is surjective (onto), so
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$$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
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\end{itemize}
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We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
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Let $m \in M$, then $\beta(m) \in N$, and
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$$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$
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Take $m' \in M$, with $m' = \sum a_j y_j$, then
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$$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$
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Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus
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$$(m-m') \in ker(\beta)$$
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By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$.
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Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$.
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Since $\{l_i\}_k$ generate $L$,
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$$l = \sum^k b_i l_i$$
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thus
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$$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$
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Rearrange,
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$$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$
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So, $L$ provides $k$ generators for the kernel part of $M$, $N$ provides $p$ "lifts" for the quotient part of $M$; thus $M$ is generated by $k+p$ elements.\\
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Thus $M$ is finitely generated over $A$.
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\end{proof}
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\subsection{Exercises Chapter 3}
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\begin{ex}{R.3.2}
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$K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
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Prove that $A$ is Noetherian and Artinian.
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\end{ex}
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\begin{proof}
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$dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
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\begin{enumerate}
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\item Noetherian:\\
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let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
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Since each $I_i$ is a subspace, we have
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$$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
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where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Noetherian.
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\item Artinian:\\
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Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
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then
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$$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
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where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Artinian.
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\end{enumerate}
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\end{proof}
|
||||
|
||||
\begin{ex}{R.3.5}
|
||||
Let $0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ an exact sequence. Let $M_1, M_2 \subseteq M$ be submodules of $M$.
|
||||
|
||||
Prove if the following holds or not:
|
||||
$$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
|
||||
\end{ex}
|
||||
\begin{proof}
|
||||
Counterexample showing that it does not hold:
|
||||
|
||||
Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
|
||||
|
||||
Set, for $l \in L,~ (m_1, m_2) \in M$,
|
||||
\begin{align*}
|
||||
\alpha:~ &l \longmapsto (l, 0)\\
|
||||
\beta:~ &(m_1, m_2) \longmapsto m_2
|
||||
\end{align*}
|
||||
|
||||
So we have
|
||||
$$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
|
||||
|
||||
Then,
|
||||
\begin{align*}
|
||||
M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
|
||||
M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
|
||||
\end{align*}
|
||||
|
||||
(Geometric interpretation: $M_1,~ M_2$ are the \emph{diagonal line} and \emph{y-axis} respectively; and $\alpha,~\beta$ capture information about the \emph{vertical} components (x-axis, y-axis respectively), but not about the \emph{diagonal} way a submodule is embedded in $M$).
|
||||
|
||||
Then,
|
||||
\begin{align*}
|
||||
\beta(M_1) &= \{ x ~|~ x \in K\} = K \\
|
||||
\beta(M_2) &= \{ x ~|~ x \in K\} = K
|
||||
\end{align*}
|
||||
thus, $\beta(M_1) = \beta(M_2)$.
|
||||
|
||||
For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
|
||||
for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
|
||||
thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
|
||||
|
||||
|
||||
So we've seen that
|
||||
\begin{align*}
|
||||
\beta(M_1) = \beta(M_2)\\
|
||||
\alpha^{-1}(M_1)=\alpha^{-1}(M_2)
|
||||
\end{align*}
|
||||
while having $M_1 \neq M_2$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
\begin{ex}{R.3.3}
|
||||
Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
|
||||
Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian.
|
||||
\end{ex}
|
||||
\begin{proof}
|
||||
\begin{enumerate}[i.]
|
||||
\item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
|
||||
$\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
|
||||
\item Take the canoncial homomorphism
|
||||
$$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
|
||||
by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
|
||||
|
||||
$\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$.
|
||||
|
||||
Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
|
||||
|
||||
Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
|
||||
|
||||
We know that any submodule of a Noetherian module is Noetherian, thus, since
|
||||
\begin{itemize}
|
||||
\item $A/I_i$ is Noetherian by hypothesis of the exercise
|
||||
\item $A \cong im(\phi)$
|
||||
\item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
|
||||
\end{itemize}
|
||||
then, $A$ is Noetherian.
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
|
||||
\begin{ex}{R.3.4}
|
||||
Prove that if A is a Noetherian ring and M a finite A-module, then there
|
||||
exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$.
|
||||
That is, M has a presentation as an A-module in terms of finitely many generators and relations.
|
||||
\end{ex}
|
||||
\begin{proof}
|
||||
since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$.
|
||||
|
||||
Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$:
|
||||
\begin{align*}
|
||||
\beta: A^p &\longrightarrow M\\
|
||||
(a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i
|
||||
\end{align*}
|
||||
|
||||
Let $K=ker(\beta)$. By the 1st Isomorphism Theorem,
|
||||
$$M \cong A^p / K$$
|
||||
|
||||
Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module.
|
||||
|
||||
Every submodule of a Noetherian module is fingen.
|
||||
|
||||
$\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen.
|
||||
|
||||
Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$.
|
||||
|
||||
Define $\psi: A^q \longrightarrow K$.
|
||||
|
||||
Compose it with the inclusion map $i: K \longrightarrow A^p$,
|
||||
$$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$
|
||||
|
||||
So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where
|
||||
\begin{itemize}
|
||||
\item $\beta$ is surjective
|
||||
\item $im(\alpha)=ker(\beta)$
|
||||
\end{itemize}
|
||||
so that it is a exact sequence, thus, $M$ has a finite presentation.
|
||||
\end{proof}
|
||||
|
||||
\bibliographystyle{unsrt}
|
||||
\bibliography{commutative-algebra-notes.bib}
|
||||
|
||||
|
||||
Reference in New Issue
Block a user