Last updates: 02 July 2012

Let $f:A\to B$. Let $A\subseteq B$ be commutative rings with $A$ a subring of $B$.

- A
**finitely generated $A$-module**is an $A$-module $M$ such that there exists $n\in {\mathbb{Z}}_{>0}$ and a surjective $A$-module homomorphism such that $M$ is isomorphic to a quotient of ${A}^{\oplus n}\to M$. - A
**finitely generated $A$-algebra**is an $A$-algebra $C$ such that there exists $n\in {\mathbb{Z}}_{>0}$ and a surjective homomorphism of commutative $A$-algebras $A[{x}_{1},\dots {x}_{n}]\to C$. - The ring $B$ is a
**finite**$A-$algebra if $B$ is a finitely generated $A-$module. - The ring $B$ is an $A-$algebra of
**finite type**if $B$ is a finitely generated $A-$algebra. - The map $f$ is
**finite**if $f\left(A\right)$ is a finitely generated $A-$module. - The map $f$ is of
**finite type**if $f\left(A\right)$ is a finitely generated $A-$algebra.

**HW:**
Show that an $A$-module $M$ is finitely generated
if and only if there is a finite subset $S\subseteq M$
such that $M=R\text{-span}\left(S\right)$.

Let $A\subseteq B$ be commutative rings with $A$ a subring of $B$.

- An element $b\in B$ is
**integral**over $A$ if $f\left(b\right)=0$ for some monic polynomial $f\left(x\right)\in A\left[x\right]$. - The ring $B$ is
**integral**over $A$, or $B$ is an**integral extension**of $A$, if every element of $B$ is integral over $A$. - The
**integral closure**of $B$ in $A$ is the largest subring of $B$ which is integral over $A$. - The ring $A$ is
**integrally closed**in $B$ if $A=\{b\in B\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}b\phantom{\rule{.5em}{0ex}}\text{is integral over}\phantom{\rule{.5em}{0ex}}A\}$. - An integral domain $R$ is
**integrally closed**if $R$ is integrally closed in its field of fractions. - An
**algebraic integer**is an element $\alpha \in \u2102$ which is integral over $\mathbb{Z}$.

**HW:**Show that if $A\subseteq B$,
and $N$ is finitely generated as a $B-$module
and $B$ is finitely generated as an $A-$module
then $N$ is finitely generated as an $A-$module.

**HW:**Show that if $A\subseteq B$
and $b\in B$ then $b$ is integral over $A$ if and only if
$$A\left[b\right]=\mathrm{im}({\mathrm{ev}}_{b}:A[x]\to B)$$
is finitely generated as an $A-$module.

**HW:**Show that if $A\subseteq B$
and ${b}_{1},\dots ,{b}_{k}\in B$ are integral over
$A$ then
$A[{b}_{1},\dots ,{b}_{k}]$ is a finitely generated $A$-module.

**HW:**Show that if $A\subseteq B$
then the integral closure of $A$ in $B$ is
$C=\{b\in B\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}b\phantom{\rule{.5em}{0ex}}\text{is integral over}\phantom{\rule{.5em}{0ex}}A\}$.

**HW:**Show that if $A\subseteq B$
then the integral closure $C$ of $A$ in
$B$ is the largest subring of $B$,
$A\subseteq C\subseteq B$,
which is finitely generated as an $A-$module.

**HW: WE NEED A BETTER NOTATION FOR THE INTEGRAL CLOSURE OF A IN B; WHAT DOES
BOURBAKI USE??**

**Example.**
$\mathbb{Z}$ is integrally closed.

Let $A\subseteq B$ be an integral extension.

- Let $\U0001d52d$ be a prime ideal in $A$ and let ${B}_{\U0001d52d}={A}_{\U0001d52d}{\otimes}_{A}B.$ Then ${A}_{\U0001d52d}\subseteq {B}_{\U0001d52d}$ is an integral extension.
- Let $\U0001d51f$ be a prime ideal of $B$ and let $\U0001d51e=\U0001d51f\cap A.$ Then $A/\U0001d51e\subseteq B/\U0001d51f$ is an integral extension.

Let $A$ be a ring and let $M$ be an $A-$module.

- The module $M$ is
**finitely generated**if there exists $n\in {\mathbb{Z}}_{>0}$ and ${m}_{1},\dots ,{m}_{n}\in M$ such that $$M=\text{span}\{{m}_{1},\dots ,{m}_{n}\},$$ where $\text{span}\{{m}_{1},\dots ,{m}_{n}\}$ is the $A$-submodule of $M$ generated by $\{{m}_{1},\dots ,{m}_{n}\}$. - The module $M$ is
**simple**if it has no submodules except $M$ and $0$. - A
**finite composition series**of $M$ is a finite chain of submodules $$0={M}_{0}\subseteq {M}_{1}\subseteq \cdots \subseteq {M}_{n}=M\phantom{\rule{2em}{0ex}}\text{with}\phantom{\rule{.5em}{0ex}}{M}_{i}/{M}_{i-1}\phantom{\rule{.5em}{0ex}}\text{simple}.$$ - The module $M$ is
**Noetherian**if every ascending chain of submodules is eventually constant. - The module $M$ is
**Artinian**if every descending chain of submodules is eventually constant. - The ring $A$ is
**Noetherian**if $A$ is Noetherian as an $A$-module. - The ring $A$ is
**Artinian**if $A$ is Artinian as an $A$-module.Let $A$ be a ring, let $M$ be an $R-$module and let $N$ be a submodule of $M$. Then

- If $M$ is finitely generated then $M/N$ is finitely generated.
- $M$ has a finite composition series if and only if $N$ and $M/N$ have finite composition series.
- $M$ is Noetherian if and only if $N$ and $M/N$ are Noetherian.
- $M$ is Artinian if and only if $N$ and $M/N$ are Artinian.

**Examples.**- Let $\mathbb{F}$ be a field. An $\mathbb{F}-$module is a vector space $V$ over $\mathbb{F}$. Any of the conditions (i) $V$ is Noetherian, (ii) $V$ is Artinian, or (iii) $V$ has a finite composition series, are equivalent to $V$ being finite dimensional. An infinite dimensional vector space is neither Noetherian or Artinian.
- The ring $\mathbb{Z}$ is Noetherian, but not Artinian.

Let $A$ be a ring and let $M$ be an $A-$module.

- $M$ has a finite composition series if and only if $M$ is Noetherian and Artinian.
- $M$ is Noetherian if and only if every submodule of $M$ is finitely generated.
- If $R$ is Noetherian and $M$ is finitely generated then $M$ is Noetherian.

*Proof.*(a) Follows from Theorem 2.4 below.

(b) ⇐: Assume that every submodule of $M$ is finitely generated. Let ${N}_{1}\subseteq {N}_{2}\subseteq \cdots $ be an ascending chain. Then $\bigcup {N}_{i}$ is a finitely generated submodule of $M.$ Let ${x}_{1},...,{x}_{k}$ be generators and let ${l}_{1},...,{l}_{k}$ be such that ${x}_{i}\in {N}_{{l}_{i}}.$ Then ${x}_{1},...,{x}_{k}\in {N}_{r}$ where $r=\mathrm{max}\{{l}_{1},...,{l}_{k}\}.$ So $\bigcup {N}_{i}={N}_{r}$ and ${N}_{r}={N}_{r+1}={N}_{l}$ for all $l>r.$ So $M$ in noetherian.

(b) ⇒: Assume that $M$ is noetherian and let $N$ be a submodule of $M.$ Then $$\{P\subseteq N\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}P\phantom{\rule{.1em}{0ex}}is\; finitely\; generated\}$$ has a maximal element ${P}_{\mathrm{max}}.$ If ${P}_{\mathrm{max}}\ne N$ let $x\in N\setminus {P}_{\mathrm{max}}.$ Then $P\subseteq \u27e8{P}_{\mathrm{max}},x\u27e9\subseteq N$ and $\u27e8{P}_{\mathrm{max}},x\u27e9$ is finitely generated, which is a contradiction to the maximality of ${P}_{\mathrm{max}}.$ So ${P}_{\mathrm{max}}=N.$ So every submodule of $M$ is finitely generated.$\square $

(Jordan-Hölder theorem.) Let $M$ be an $A-$module.

- Any two series $$0\subseteq {M}_{1}\subseteq \cdots \subseteq {M}_{r}=M\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}0\subseteq {M}_{1}^{\text{'}}\subseteq \cdots \subseteq {M}_{s}^{\text{'}}=M,$$ can be refined to have the same length and the same composition factors.
- $M$ has a composition series if and only if any series can be refined to a composition series.
- $M$ has a composition series if and only if $M$ is Noetherian and Artinian.
- If $M$ has a composition series then any two composition series of $M$ have the same length.

*Proof.*Suppose $$0={M}_{0}\subseteq {M}_{1}\subseteq \cdots \subseteq {M}_{r}=M\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}0={M}_{0}^{\text{'}}\subseteq {M}_{1}^{\text{'}}\subseteq \cdots \subseteq {M}_{s}^{\text{'}}=M,$$ are chains of submodules of $M$. Change ${M}_{i}\subseteq {M}_{i+1}$ to $${M}_{i}=({M}_{0}^{\text{'}}+{M}_{i})\cap {M}_{i+1}\subseteq ({M}_{1}^{\text{'}}+{M}_{i})\cap {M}_{i+1}\subseteq \cdots \subseteq ({M}_{s}^{\text{'}}+{M}_{i})\cap {M}_{i+1}={M}_{i+1}$$ and change ${M}_{j}^{\text{'}}\subseteq {M}_{j+1}^{\text{'}}$ to $${M}_{j}=({M}_{0}+{M}_{j}^{\text{'}})\cap {M}_{j+1}^{\text{'}}\subseteq ({M}_{1}+{M}_{j}^{\text{'}})\cap {M}_{j+1}^{\text{'}}\subseteq \cdots \subseteq ({M}_{r}+{M}_{j}^{\text{'}})\cap {M}_{j+1}^{\text{'}}={M}_{j+1}^{\text{'}}.$$ Claim: $$\frac{({M}_{j}^{\text{'}}+{M}_{i-1})\cap {M}_{i}}{({M}_{j-1}^{\text{'}}+{M}_{i-1})\cap {M}_{i}}\cong \frac{({M}_{i}+{M}_{j-1}^{\text{'}})\cap {M}_{j}^{\text{'}}}{({M}_{i-1}+{M}_{j-1}^{\text{'}})\cap {M}_{j}^{\text{'}}}.$$ This claim will be established by Lemma 2.6. $\square $

(Modular Law) If $A$, $B$, $C$ are submodules of $M$, and $B\subseteq C$, then $$C+(A\cap B)=(C+A)\cap B.$$

*Proof.*If $c+a\in C+(A\cap B)$ then $c+a\in (C+A)\cap B$.

If $b=c+a\in (C+A)\cap B$ then $b=c+a=c+(b-c)\in C+(A\cap B)$.$\square $

(Zassenhaus Isomorphism) If $V\subseteq U$ and $V\text{'}\subseteq U\text{'}$ are submodules of $M$ then $$\frac{(U+V\text{'})\cap U\text{'}}{(V+V\text{'})\cap U\text{'}}\cong \frac{U\cap U\text{'}}{(U\cap V\text{'})+(U\text{'}\cap V)}\cong \frac{(U\text{'}+V)\cap U}{(V\text{'}\cap V)\cap U}.$$

(Hilbert basis theorem) Let $R$ be a commutative Noetherian ring. Then $R\left[x\right]$ is a commutative Noetherian ring.

*Proof.*To show: Every ideal of $R\left[x\right]$ is finitely generated.

Let $\U0001d51e$ be an ideal of $R\left[x\right]$. Let $$\U0001d51f=\{{a}_{k}\in R\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}f\left(x\right)={a}_{k}{x}^{k}+\cdots +{a}_{0}\in \U0001d51e\}$$ an ideal of $R$. Since $R$ is Noetherian, $\U0001d51f$ is finitely generated. Let ${b}_{1},\dots ,{b}_{n}$ be generators of $\U0001d51f$. Let $$\begin{array}{lcl}{f}_{1}\left(x\right)& =& {b}_{1}{x}^{{k}_{1}}+\cdots +{b}_{10}\in \U0001d51e\\ {f}_{2}\left(x\right)& =& {b}_{2}{x}^{{k}_{2}}+\cdots +{b}_{20}\in \U0001d51e,\\ \vdots \\ {f}_{n}\left(x\right)& =& {b}_{n}{x}^{{k}_{n}}+\cdots +{b}_{n0}\in \U0001d51e,\end{array}$$ be polynomials in $\U0001d51e$ corresponding to the generators of $\U0001d51f$. Then $$\u27e8{f}_{1},,\dots ,{f}_{n}\u27e9\subseteq \U0001d51e$$ Let $f\in \U0001d51e$, $f=a{x}^{k}+\cdots +{a}_{0}$.

If $k\ge \mathrm{max}({k}_{1},\dots ,{k}_{n})$ then $$a=\sum _{i=1}^{n}{r}_{i}{b}_{i}\phantom{\rule{0.5em}{0ex}}\text{for some}\phantom{\rule{0.5em}{0ex}}{r}_{i}\in R,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f-\sum _{i=1}^{n}{r}_{i}{f}_{i}{x}^{k-{r}_{i}}\phantom{\rule{2em}{0ex}}\in \U0001d51e\phantom{\rule{1em}{0ex}}\text{and has lower degree.}$$ Thus, if $f\in \U0001d51e$ then $$f=g+h\phantom{\rule{1em}{0ex}}\text{with}\phantom{\rule{0.5em}{0ex}}g\in \U0001d51e\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}\mathrm{deg}\left(h\right)<\mathrm{max}\left({k}_{i}\right).$$ If $$M=R\text{-span}\{1,x,{x}^{2},\dots ,{x}^{k-1}\},\phantom{\rule{1em}{0ex}}\text{then}\phantom{\rule{1em}{0ex}}\U0001d51e=(\U0001d51e\cap M)+\u27e8{f}_{1},\dots ,{f}_{n}\u27e9.$$ Since $M$ is finitely generated, $M$ is Noetherian. So $\U0001d51e\cap M$ is finitely generated as an $R$-module. Let ${g}_{1},\dots ,{g}_{m}$ generate $\U0001d51e\cap M$. Then ${g}_{1},\dots ,{g}_{m},{f}_{1},\dots ,{f}_{n}$ generate $\U0001d51e$. So $R\left[x\right]$ is Noetherian.$\square $

(Finite generation of invariants.) Let $\mathbb{F}$ be a field and let $A$ be a finitely generated $\mathbb{F}-$algebra. Let $G$ be a finite group acting on $A$ by automorphisms. Then

- ${A}^{G}$ is a finitely generated $\mathbb{F}-$algebra.
- $A$ is a finitely generated ${A}^{G}-$module.

*Proof.*Let ${a}_{1},\dots ,{a}_{n}$ be generators of $A$ as an $\mathbb{F}-$algebra. Let $$B=\u27e8c_{i}{}_{j}\in {A}^{G}\phantom{\rule{0.5em}{0ex}}\left|\phantom{\rule{0.5em}{0ex}}c_{i}{}_{j}\phantom{\rule{.5em}{0ex}}\text{are coefficients of}\phantom{\rule{.5em}{0ex}}\prod _{g\in G}\right(x-g{a}_{i})\u27e9.$$ Then $B$ is a finitely generated $\mathbb{F}-$algebra. So $B$ is a quotient of $\mathbb{F}[{x}_{1},\dots ,{x}_{m}]$ for some $m$. Thus, by the Hilbert basis theorem, $B$ is Noetherian.

If $a\in A$ then $a$ satisfies the polynomial $$\prod _{g\in G}(x-ga)\in {A}^{G}\left[x\right]$$ and so $A$ is an integral extension of ${A}^{G}$. Thus, since ${a}_{1},\dots ,{a}_{n}$ are generators of $A$, we have that $A$ is a finitely generated $B-$module. So ${A}^{G}$ is a finitely generated $B-$module. So ${A}^{G}$ is a finitely generated $\mathbb{F}-$algebra.$\square $

## Notes and References

Composition series and the Jordan-Hölder theorem are treated in [Bou, Alg. Ch. I § 4.7] and in [AM, Proposition 6.7]. Chain conditions, Noetherian rings and Artinian rings are covered in [AM] Chapters 6, 7 and 8. The Hilbert basis theorem is [AM, Theorem 7.5] and [Bou, Comm. Alg. Ch. III § 2 No. 10] and the Finite generation of invariants theorem is [AM, Ch. 7 Ex. 5] and [Bou, Comm. Alg. Ch. V § 1 No. 9, Theorem 2]. An alternative, efficient treatment is found in [Ben], where the Hilbert basis theorem is [Ben, Theorem 1.2.4] and the Finite generation of invariants theorem is [Ben, Theorem 1.3.1].

The basics of noetherian and artinian modules and rings are treated in [Bou, Alg. Ch. 8 § 1].

## References

[AM] M. Atiyah and I.G. Macdonald,

*Introduction to commutative algebra*, Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont. 1969 ix+128 pp. MR0242802.[Ben] D.J. Benson,

*Polynomial invariants of finite groups*, London Mathematical Society Lecture Note Series 190, Cambridge University Press, Cambridge, 1993. x+118 pp. ISBN: 0-521-45886-2 MR1249931.[Mac] I.G. Macdonald,

*Algebraic geometry. Introduction to Schemes*, W. A. Benjamin, Inc., New York-Amsterdam 1968 vii+113 pp. MR0238845.