Finiteness conditions

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

Last updates: 14 November 2011

Integral closure

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

• 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.
• An element $c\in B$ is integral over $A$ if $f\left(c\right)=0$ for some monic polynomial $f\left(x\right)\in A\left[x\right]$.
• The ring $B$ is integral over $A$ 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 $A= {b\in B|bis\; integral\; overA}$.
• An integral domain $R$ is integrally closed if $R$ is integrally closed in its field of fractions.
• An algebraic integer is an element $\alpha \in ℂ$ which is integral over $\mathbb{Z}$.

If $A\supseteq B$, and $N$ is finitely generated as a $B-$module by $n_1,...,n_k,$ and $B$ is finitely generated as an $A-$module by $b_1,...,b_l,$ then $N$ is finitely generated as an $A-$module by the products $b_j{n}_i,1\le i\le k,1\le j\le l.$ Hence, if $c_1$ and $c_2$ are integral over $A$, then $A\left[c_1\right]$ and $A\left[c_2\right]$ are finitely generated $A-$modules and $A [c_1,c_2] =A\left[c_1\right]\left[c_2\right]$ is a finitely generated $A-$module. So $c_1+c_2$ and $c_1{c}_2$ are integral over $A$. Thus, the integral closure of $A$ in $B$ is a ring. We have also shown that the integral closure of $A$ in $B$ is $$C= {c\in B|cis\; integral\; overA} .$$ This also shows that $B$ is integral over $A$ (equal to its integral closure) and of finite type over $A$ (finitely generated as an $A-$algebra) if and only if it is a finite $A-$algebra.

If $b\in B$ then $b$ is integral over $B$ if and only if $A\left[b\right]=\mathrm{im}({\mathrm{ev}}_b:A[x]\to B)$ is finitely generated as an $A-$module, i.e. if and only if $b$ is contained in a subring $C$, $A\subseteq C\subseteq B,$ which is finitely generated as an $A-$module. Hence, if $b_1,b_2$ are integral over $A$ then $b_1+b_2\in C_1+C_2$ which is a finitely generated $A-$module and so $b_1+b_2$ is integral. Similarly, $b_1{b}_2$ is an element of the $A-$module generated by the products in $C_1{C}_2$.

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

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

1. 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.
2. Let $𝔟$ be a prime ideal of $B$ and let $𝔞=𝔟\cap A.$ Then $A/𝔞\subseteq B/𝔟$ is an integral extension.

Noetherian rings

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

• 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.
• A composition series of $M$ is a finite chain of submodules $$0=M_0\subseteq M_1\subseteq \cdots \subseteq M_n=Mwith{M}_i/M_{i-1}simple.$$
• The module $M$ is finitely generated if there is a finite subset $S\subseteq M$ such that $M=R\setminus \mathrm{span}\left(S\right)$.
• The ring $R$ is Noetherian if ${}_{R}R$ is Noetherian.
• The ring $R$ is Artinian if ${}_{R}R$ is Artinian.

Let $R$ be a ring, let $M$ be an $R-$module and let $N$ be a submodule of $M$. Then

1. $M$ is Noetherian if and only if $N$ and $M/N$ are Noetherian.
2. $M$ is Artinian if and only if $N$ and $M/N$ are Artinian.
3. $M$ has a composition series if and only if $N$ and $M/N$ have a composition series.
4. If $M$ is finitely generated then $M/N$ is finitely generated.

Examples.

1. 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 composition series, are equivalent to $V$ being finite dimensional.
2. The ring $\mathbb{Z}$ is Noetherian, but not Artinian.

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

1. $M$ has a composition series if and only if $M$ is Noetherian and Artinian.
2. $M$ is Noetherian if and only if every submodule of $M$ is finitely generated.
3. If $R$ is Noetherian and $M$ is finitely generated, then $M$ is Noetherian.

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

1. Any two series $$0\subseteq M_1\subseteq \cdots \subseteq M_r=Mand0\subseteq M_1^{\text{'}}\subseteq \cdots \subseteq M_s^{\text{'}}=M,$$ can be refined to have the same length and the same composition factors.
2. $M$ has a composition series if and only if any series can be refined to a composition series.
3. $M$ has a composition series if and only if $M$ is Noetherian and Artinian.
4. If $M$ has a composition series then any two composition series of $M$ have the same length.

		Proof.
	Suppose $$0=M_0\subseteq M_1\subseteq \cdots \subseteq M_r=Mand0=M_0^{\text{'}}\subseteq M_1^{\text{'}}\subseteq \cdots \subseteq M_s^{\text{'}}=M,$$ 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's 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 $𝔞$ be an ideal of $R\left[x\right]$. Then $$𝔟= {a_k\in R|f\left(x\right)=a_k{x}^k+\cdots +a_0\in 𝔞}$$ is an ideal of $R$. Since $R$ is Noetherian $𝔟$ is finitely generated. Let $b_1$, ..., $b_n$ be generators of $𝔟$ and let $$f_i\left(x\right)=b_i{x}^{k_i}+\cdots +b_i\in 𝔞$$ be polynomials in $𝔞$ corresponding to the generators of $𝔟$. By definition, $⟨f_1,...,f_n⟩\subseteq 𝔞$ and we want to show that equality holds. THERE IS SOMETHING WEIRD IN THIS PROOF. GO OVER IT WITH ARUN. $\square$

(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

1. $A^G$ is a finitely generated $𝔽-$algebra.
2. $A$ is a finitely generated $A^G-$module.

		Proof.
	Let $a_1,...,a_n$ 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 $𝔽 [x_1,...,x_m]$ for some $m$. Thus, by Hilbert's 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 $a_1$, ..., $a_n$ 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. $\square$

Notes and References

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