Finiteness conditions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 02 July 2012

Finitely generated modules

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 $n \in ℤ_{> 0}$ 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 $n \in ℤ_{> 0}$ 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 $f (A)$ is a finitely generated $A -$ module.
The map $f$ is of finite type if $f (A)$ 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 $M = R -span (S)$ .

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

An element $b \in B$ is integral over $A$ if $f (b) = 0$ for some monic polynomial $f (x) \in A [x]$ .
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 $A = {b \in B | b is integral over A}$ .
An integral domain $R$ is integrally closed if $R$ is integrally closed in its field of fractions.
An algebraic integer is an element $α \in ℂ$ which is integral over $ℤ$ .

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 $A [b] = im ({ev}_{b} : A [x] \to B)$ 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 $C = {b \in B | b is integral over A}$ .

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. $ℤ$ is integrally closed.

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

Let $𝔭$ be a prime ideal in $A$ and let $B_{𝔭} = A_{𝔭} \otimes_{A} B .$ Then $A_{𝔭} \subseteq B_{𝔭}$ is an integral extension.
Let $𝔟$ be a prime ideal of $B$ and let $𝔞 = 𝔟 \cap A .$ Then $A / 𝔞 \subseteq B / 𝔟$ is an integral extension.

Structure of the lattice of submodules

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

The module $M$ is finitely generated if there exists $n \in ℤ_{> 0}$ and $m_{1}, \dots, m_{n} \in M$ such that $M = span {m_{1}, \dots, m_{n}},$ where $span {m_{1}, \dots, m_{n}}$ 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 $0 = M_{0} \subseteq M_{1} \subseteq \dots \subseteq M_{n} = M with M_{i} / M_{i - 1} simple .$
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 $𝔽$ be a field. An $𝔽 -$ module is a vector space $V$ over $𝔽$ . 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 $ℤ$ 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 $N_{1} \subseteq N_{2} \subseteq \dots$ be an ascending chain. Then $⋃ N_{i}$ 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 $r = \max {l_{1}, ..., l_{k}} .$ So $⋃ N_{i} = N_{r}$ 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 ${P \subseteq N \| P is finitely generated}$ has a maximal element $P_{\max} .$ If $P_{\max} \neq N$ let $x \in N ∖ P_{\max} .$ Then $P \subseteq ⟨ P_{\max}, x ⟩ \subseteq N$ and $⟨ P_{\max}, x ⟩$ is finitely generated, which is a contradiction to the maximality of $P_{\max} .$ So $P_{\max} = N .$ So every submodule of $M$ is finitely generated. $□$

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

Any two series $0 \subseteq M_{1} \subseteq \dots \subseteq M_{r} = M and 0 \subseteq M_{1}^{'} \subseteq \dots \subseteq M_{s}^{'} = M,$ 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 $0 = M_{0} \subseteq M_{1} \subseteq \dots \subseteq M_{r} = M and 0 = M_{0}^{'} \subseteq M_{1}^{'} \subseteq \dots \subseteq M_{s}^{'} = M,$ are chains of submodules of $M$ . Change $M_{i} \subseteq M_{i + 1}$ to $M_{i} = (M_{0}^{'} + M_{i}) \cap M_{i + 1} \subseteq (M_{1}^{'} + M_{i}) \cap M_{i + 1} \subseteq \dots \subseteq (M_{s}^{'} + M_{i}) \cap M_{i + 1} = M_{i + 1}$ and change $M_{j}^{'} \subseteq M_{j + 1}^{'}$ to $M_{j} = (M_{0} + M_{j}^{'}) \cap M_{j + 1}^{'} \subseteq (M_{1} + M_{j}^{'}) \cap M_{j + 1}^{'} \subseteq \dots \subseteq (M_{r} + M_{j}^{'}) \cap M_{j + 1}^{'} = M_{j + 1}^{'} .$ Claim: $\frac{(M_{j}^{'} + M_{i - 1}) \cap M_{i}}{(M_{j - 1}^{'} + M_{i - 1}) \cap M_{i}} ≅ \frac{(M_{i} + M_{j - 1}^{'}) \cap M_{j}^{'}}{(M_{i - 1} + M_{j - 1}^{'}) \cap M_{j}^{'}} .$ This claim will be established by Lemma 2.6. $□$

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

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

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

		Proof.
	To show: Every ideal of $R [x]$ is finitely generated. Let $𝔞$ be an ideal of $R [x]$ . Let $𝔟 = {a_{k} \in R \| f (x) = a_{k} x^{k} + \dots + a_{0} \in 𝔞}$ an ideal of $R$ . Since $R$ is Noetherian, $𝔟$ is finitely generated. Let $b_{1}, \dots, b_{n}$ be generators of $𝔟$ . Let $\begin{array}{lcl} f_{1} (x) & = & b_{1} x^{k_{1}} + \dots + b_{10} \in 𝔞 \\ f_{2} (x) & = & b_{2} x^{k_{2}} + \dots + b_{20} \in 𝔞, \\ ⋮ \\ f_{n} (x) & = & b_{n} x^{k_{n}} + \dots + b_{n 0} \in 𝔞, \end{array}$ be polynomials in $𝔞$ corresponding to the generators of $𝔟$ . Then $⟨ f_{1},, \dots, f_{n} ⟩ \subseteq 𝔞$ Let $f \in 𝔞$ , $f = a x^{k} + \dots + a_{0}$ . If $k \geq \max (k_{1}, \dots, k_{n})$ then $a = \sum_{i = 1}^{n} r_{i} b_{i} for some r_{i} \in R, and f - \sum_{i = 1}^{n} r_{i} f_{i} x^{k - r_{i}} \in 𝔞 and has lower degree.$ Thus, if $f \in 𝔞$ then $f = g + h with g \in 𝔞 and \deg (h) < \max (k_{i}) .$ If $M = R -span {1, x, x^{2}, \dots, x^{k - 1}}, then 𝔞 = (𝔞 \cap M) + ⟨ f_{1}, \dots, f_{n} ⟩ .$ Since $M$ is finitely generated, $M$ is Noetherian. So $𝔞 \cap M$ is finitely generated as an $R$ -module. Let $g_{1}, \dots, g_{m}$ generate $𝔞 \cap M$ . Then $g_{1}, \dots, g_{m}, f_{1}, \dots, f_{n}$ generate $𝔞$ . So $R [x]$ is Noetherian. $□$

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

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

		Proof.
	Let $a_{1}, \dots, a_{n}$ be generators of $A$ as an $𝔽 -$ algebra. Let $B = ⟨ c_{i}_{j} \in A^{G} \| c_{i}_{j} are coefficients of \prod_{g \in G} (x - g a_{i}) ⟩ .$ Then $B$ is a finitely generated $𝔽 -$ algebra. So $B$ is a quotient of $𝔽 [x_{1}, \dots, x_{m}]$ for some $m$ . Thus, by the Hilbert basis theorem, $B$ is Noetherian. If $a \in A$ then $a$ satisfies the polynomial $\prod_{g \in G} (x - g a) \in A^{G} [x]$ 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 $𝔽 -$ algebra. $□$

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.

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