How can I complete the proof of Noetherianity of I. S. Cohen?

Question

Theorem (I. S. Cohen).

If $R$ is an unital commutative ring, and for each ideal prime $\mathfrak{p}\in Spec(R)$ we know $\mathfrak{p}$ is finitely-generated as $R$-mod then $R$ is Noetherian.

Proof.

We use the Zorn's Lemma.

Let $\mathcal{C}$ $:=$ ${\lbrace{\mathfrak{a}\in\mathcal{I}deals(R)\mid \mathfrak{a}\neq R \wedge \mathfrak{a} \text{ is not finitely-generated as }R\text{-mod}}\rbrace}$.

Now we reason by reductio ad absurdum.

Let $\mathcal{C}\neq\emptyset$.

Now for any ascendent (in the inclusion sense) chain of ideals $C\in\mathcal{C}$ where $C={\lbrace C_i \rbrace}_{i \in I}$ for some set of indexes $I$, we have that $\bigcup C$ is a proper ideal of $R$, and $\bigcup C \in \mathcal{C}$, because it is not, the finite set of generators of $\bigcup C$ was into some ideal of the chain, we say $C_{i_0}$, and this ideal of the chain was finitely-generated as $R$-mod. Absurd. Then $\bigcup C$ is not finitely-generated as $R$-mod, then $\bigcup C \in \mathcal{C}$, and it is maximal for its chain. Then $\mathcal{C}$ is an inductive set. Let $\mathfrak{m}$ a maximal ideal in $\mathcal{C}$.

The following step is to use the fact that any ideal of $\mathcal{C}$ is not prime. As $\mathfrak{m}$ is in $\mathcal{C}$, it is not in $Spec(R)$, and if the proper ideal $\mathfrak{a}$ is such that $\mathfrak{m} \subsetneq \mathfrak{a}$ then $\mathfrak{a}$ is finitely-generated as $R$-mod.

Let $x,y\in R\setminus \mathfrak{m}$ and $xy\in \mathfrak{m}$. They exist because the mentioned ideal is neccesarily non-prime. Then we can see that the ideals $\mathfrak{m}+{\langle x\rangle}_R$ is finitely-generated because contains strictily $\mathfrak{m}$. Let ${\langle\lbrace u_1,\ldots,u_n,x\rbrace\rangle}_R = \mathfrak{m}+{\langle x \rangle}_R$, where ${\lbrace u_1,\ldots,u_n\rbrace}\subseteq \mathfrak{m}$. In the other hand $\mathfrak{m} \subsetneq \left( \mathfrak{m} : x\right)$ because $y \notin \mathfrak{m}$ and $y \in \left( \mathfrak{m} : x\right)$. Let ${\langle\lbrace v_1,\ldots,v_m,y\rbrace\rangle}_R = \left( \mathfrak{m} : x\right)$.

Now it is easy note that

$\langle \lbrace {\phantom{\rangle\lbrace}}u_1 v_1, u_1 v_2, \ldots ,u_1 v_m,u_1 y,$

$u_2 v_1, u_2 v_2, \ldots ,u_2 v_m,u_2 y,$

$\ldots \ldots,$

$u_n v_1, u_n v_2, \ldots ,u_n v_m,u_n y,$

$ x v_1, x v_2, \ldots , x v_m, x y {\phantom{\langle\lbrace} \rbrace \rangle}_R = \left( \mathfrak{m} + {\langle x \rangle}_R\right)\left( \mathfrak{m} : x\right) \subseteq \mathfrak{m}$

We can augment the set of generators,

$G := \lbrace {\phantom{\rangle\lbrace}} u_1 , u_2 , \ldots , u_{n-1} , u_n,$

$u_1 v_1, u_1 v_2, \ldots ,u_1 v_m,u_1 y,$

$u_2 v_1, u_2 v_2, \ldots ,u_2 v_m,u_2 y,$

$\ldots \ldots,$

$u_n v_1, u_n v_2, \ldots ,u_n v_m,u_n y,$

$ x v_1, x v_2, \ldots , x v_m, x y {\phantom{\lbrace} \rbrace}$

The only problem I have is that $\left( \mathfrak{m} + {\langle x \rangle}_R\right)\left( \mathfrak{m} : x\right) \subseteq {\langle G \rangle}_R \subseteq \mathfrak{m}$, but the reverse $\mathfrak{m} \subseteq {\langle G \rangle}_R$(?) is neccesary to complete the proof (I believe). If we would prove in some way that $\mathfrak{m}$ is finitely-generated then we have obtained the contradicition that we expected, and then $\mathcal{C}=\emptyset$ and we can derive that $R$ is finitely-generated as $R$-mod.

I know something in front of me but I do not see. Some explanation? Thanks.

Answer 1

The Theorem you mention is Theorem 3.4 in Matsumura's Commutative Ring Theory. I provide a sketch of the given (very elegant) proof by Matsumura in the form of a hint:

Let $\Gamma$ be the set of ideals that are not finitely generated. If $\Gamma \neq \emptyset$ then we have a maximal element $I$ of $\Gamma$ and elements $x,y$ of the ring such that $x \not\in I, y \not\in I, xy \in I$. Now notice that the ideals $I+Ay$ and $(I:y)$ are finitely generated. Write $I+Ay=(u_1,\dots,u_m,y), (I:y) = (v_1,\dots,v_n)$. Can you now complete the proof? (i can provide further details upon request)

Edit:

More specifically, let $I+Ay = (\xi_1,\dots,\xi_m)$. Since $\xi_i \in I+Ay$ we can write $\xi_i = u_i + c_i y$ with $u_i \in I$. Then we see that $I+Ay=(u_1,\dots,u_m,y)$. Also $(I:y) = (v_1,\dots,v_n)$. Now let $z \in I$. Then $z = a_1 u_1 + \cdots + a_m u_m + a y$. Hence $a \in (I:y)$ and so $a = d_1 v_1 + \cdots + d_n v_n$, which gives $z=a_1 u_1 + \cdots + a_m u_m + d_1 v_1 y + \cdots + d_n v_n y$ and so $I=(u_1,\dots,u_m,yv_1,\dots,yv_n)$.

Answer 2

Theorem (I. S. Cohen).

If $R$ is an unital commutative ring, and for each ideal prime $\mathfrak{p}\in Spec(R)$ we know $\mathfrak{p}$ is finitely-generated as $R$-mod then $R$ is Noetherian.

Proof.

We use the Zorn's Lemma.

First we build the set of all hipotetical ideals of $R$ that they are not finitely-generated as $R$-mod, that we will call $\mathcal{C}$.

$$\mathcal{C}:={\lbrace{\mathfrak{a}\in\mathcal{I}deals(R)\mid \mathfrak{a}\neq R \wedge \mathfrak{a} \text{ is not finitely-generated as }R\text{-mod}}\rbrace}$$

Then, the intersection of $\mathcal{C}$ and $Spec(R)$ is empty. $\mathcal{C} \cap Spec(R) = \emptyset$.

Now we reason by reductio ad absurdum.

Supossition : $\mathcal{C}\neq\emptyset$.

We go to undestand that if $\mathcal{C}\neq \emptyset$ then there exist some maximal ideal into $(\mathcal{C},\subseteq)$.

• For any ascendent (in the inclusion sense) chain of ideals $S\subset\mathcal{C}$ where $S={\lbrace S_i \rbrace}_{i \in I}$ for some set of indexes $I$, we have that $\bigcup S$ is a proper ideal of $R$.
• And $\bigcup S \in \mathcal{C}$, because if it is not, the finite set of generators of $\bigcup S$ was into some ideal of the chain, we say $S_{i_0}$ with $i_0 \in I$, and this ideal of the chain was finitely-generated as $R$-mod. Absurd. Then $\bigcup S$ is not finitely-generated as $R$-mod, then $\bigcup S \in \mathcal{C}$, and it is maximal for its chain.

Then $\mathcal{C}$ is an inductive set. There exists maximal ideals in $\mathcal{C}$ in the sense of inclusion of sets.

Let $\mathfrak{m}$ a maximal ideal in $\mathcal{C}$.

Now we go to search (and finding) a pair of ideals related between them and with maximal ideal $\mathfrak{m}$ such that they are finitely-generated as $R$-mod, and then, using the fact that any ideal of $\mathcal{C}$ is not prime we will obtain a contradiction. The searched contradiction is that $\left(\mathfrak{m} \in \mathcal{C}\right) \wedge \left(\mathfrak{m} \notin \mathcal{C}\right)$.

As $\mathfrak{m}$ is in $\mathcal{C}$, it is not in $Spec(R)$, and if the proper ideal $\mathfrak{a}$ is such that $\mathfrak{m} \subsetneq \mathfrak{a}$ then $\mathfrak{a}$ is finitely-generated as $R$-mod.

Let $x,y\in R\setminus \mathfrak{m}$ and $xy\in \mathfrak{m}$. They exist because the mentioned maximal ideal is neccesarily non-prime.

• Now we realize that $\mathfrak{m} \subsetneq \left( \mathfrak{m} : {\langle x \rangle}_R \right)$ because $y \notin \mathfrak{m}$ and $y \in \left( \mathfrak{m} : {\langle x \rangle}_R \right)$. Then, there exists a finite subset $V$ of $\left( \mathfrak{m} : {\langle x \rangle}_R \right)$ such that ${\langle V \rangle}_R = \left( \mathfrak{m} : {\langle x \rangle}_R \right)$.
• In the other hand we see that ${\mathfrak{m}} \subsetneq {\mathfrak{m}}+{\langle x \rangle}_R$. Because of it, there exists a finite subset of ${\mathfrak{m}}$, I say $U$, such that ${\langle U \cup \lbrace x \rbrace \rangle}_R = {\mathfrak{m}}+{\langle x \rangle}_R$.

We go to prove that there exists a finite subset $U \subset \mathfrak{m}$ such that ${\langle U \cup \lbrace x \rbrace \rangle}_R = \mathfrak{m}+{\langle x \rangle}_R$. First we write a generator system for $\mathfrak{m}+{\langle x \rangle}_R = {\langle \lbrace w_1 , w_2 , \ldots , w_l \rbrace \rangle}_R$.

We can write this generators by the way of the sum of ideals where they are: $\exists {\lbrace u_1 , u_2 , \ldots , u_l \rbrace} \subset \mathfrak{m}$ and $\exists {\lbrace a_1 , a_2 , \ldots , a_l \rbrace} \subset R$ such that $w_i = u_i + a_i x \; i = 1, 2 , \ldots , l$. Now we obtain ${u_i = w_i - a_i x \; i= 1 , 2 , \ldots , l}$ which if we take a minimal subset generator (join to $\lbrace x \rbrace$), we have got the set $U=\lbrace u_1 , u_2 , \ldots , u_n \rbrace$ sucht that $U \subset \mathfrak{m}$ and ${\langle U \cup \lbrace x \rbrace \rangle}_R = \mathfrak{m}+{\langle x \rangle}_R$.

Now:

$U = \lbrace u_1 , u_2 , \ldots , u_n \rbrace \subset \mathfrak{m} \;\; V = \lbrace v_1 , v_2 , \ldots , v_m \rbrace$

$z \in \mathfrak{m} \implies z \in \mathfrak{m}+{\langle x \rangle}_R$ then we have $z = r_1 u_1 + r_2 u_2 + \ldots + r_n u_n + r_{n+1} x$.

As $\left( \;z = r_1 u_1 + r_2 u_2 + \ldots + r_n u_n + r_{n+1} x \in \mathfrak{m} \right)$ $\; \wedge \;$ $\left( w := r_1 u_1 + r_2 u_2 + \ldots + r_n u_n \in \mathfrak{m} \; \right) \; \implies \; z-w = r_{n+1} x \in \mathfrak{m}$ which again means that $r_{n+1} \in \left(\mathfrak{m}{:} {\langle x \rangle}_R \right)$.

Now $r_{n+1} = s_1 v_1 + s_2 v_2 + \ldots + s_m v_m$ and substituting the anterior expression in the expresion of $z$ we obtain:

$z = r_1 u_1 + r_2 u_2 + \ldots + r_n u_n + s_1 v_1 x + s_2 v_2 x + \ldots + s_m v_m x$.

In this way we define $G := { \lbrace { u_1 , u_2 , \ldots , u_n , v_1 x, v_2 x , \ldots , v_m x } \rbrace }$. And above we have just proved that $\mathfrak{m} \subseteq {\langle G \rangle}_R$ because any element of $\mathfrak{m}$ can be expressed as a linear combination of the elements of $G$.

But that $G \subset \mathfrak{m}$ is easy to check, because the elements of $U$ are all of them into $\mathfrak{m}$ and the remaining are elements of $\left(\mathfrak{m}{:}{\langle x \rangle}_R\right)$ times $x$. By definition they are elements of $\mathfrak{m}$. Then ${\langle G \rangle}_R = \mathfrak{m}$.

Then $G$ is a finite system generator for $\mathfrak{m}$ with $n+m$ elements of $\mathfrak{m}$. Then $\mathfrak{m} \notin \mathcal{C}$ which is the contradiction expected.

Then the supossition $\mathcal{C} \neq \emptyset$ is false. All the ideals of $R$ are finitely-generated as $R$-mod.