The article is written with the assumption that all rings are commutative and have a unit.

The Eakin-Nagata Theorem is the following

Let $A\subset B$ be rings, with $B$ Noetherian, and assume that $B$ is finitely generated as an $A$-module. Then $A$ is Noetherian as well.

In order to prove this main result we will first prove some other results and reach this as a special case of Formanek’s Theorem. These are the exercises 24-28 from chapter 2 in the book Commutative Algebra by Ferretti.

Lemma (Prime Maximal Submodule)

Let $M$ be an $A$-module, $N$ a submodule of $M$ maximal with respect to the condition of not being finitely generated. Let $P=\text{Ann}(M/N)$ then $P$ is prime.

Proof

Suppose $P$ is not prime. Take $a, b \in A$ such that $ab \in P$ but $a, b \notin P$.

Define \(L = \{m \in M : am \in N\}\). Since $a \notin \text{Ann}(M/N)$, the containment $N \subsetneq N + aM$ is proper. By the maximality of $N$, $N + aM$ is finitely generated, say $N + aM = (n_1 + am_1, \dots, n_k + am_k)$.

Since $b \notin \text{Ann}(M/N)$, there is some $m \in M$ such that $bm \notin N$. However, $a(bm) = (ab)m \in N$, so $bm \in L$. Thus $N \subsetneq L$ is proper, and by maximality, $L$ is finitely generated.

Take $n \in N$. Then $n = \sum_{i=1}^k c_i(n_i + am_i)$ for some $c_i \in A$. Rearranging, we get\(n - \sum_{i=1}^k c_i n_i = a \left( \sum_{i=1}^k c_i m_i \right)\)The left side is in $N$, so $\sum c_i m_i \in L$. This implies $N \subseteq (n_1, \dots, n_k) + aL$. Since the other containment is obvious, $N = (n_1, \dots, n_k) + aL$.

Because both $(n_1, \dots, n_k)$ and $L$ are finitely generated, $N$ is finitely generated, which is a contradiction. Thus $P$ is prime.

Jothilingam’s Theorem

Let $A$ be a ring, $M$ a finitely generated module such that $P\cdot M$ is finitely generated for all prime ideals $P$ of $A$. Then $M$ is Noetherian.

Proof From A Simple Proof of Cohen’s Theorem by Naghipour.

Suppose $M$ is not Noetherian. Let $\mathcal{M}$ be the family of submodules of $M$ that are not finitely generated. By Zorn’s Lemma, $\mathcal{M}$ has a maximal element $N$. Then $P = \text{Ann}(M/N)$ is prime by the Prime Maximal Submodule Lemma.

Let $M = (m_1, \dots, m_k)$. Then $P = \bigcap_{i=1}^k \text{Ann}(\overline{m}_i)$, where $\overline{m}_i$ is the image of $m_i$ in $M/N$. Since $P$ is prime, $P = \text{Ann}(\overline{m}_i)$ for some $i$.

Consider the submodule $N + (m_i)$. Since $m_i \notin N$, we have a strict containment $N \subsetneq N + (m_i)$. By the maximality of $N$, $N + (m_i)$ is finitely generated: \(N + (m_i) = (n_1 + a_1 m_i, \dots, n_j + a_j m_i)\)

where $n_l \in N$ and $a_l \in A$.

Take $n \in N \subset N + (m_i)$ and write $n = \sum_{l=1}^j b_l(n_l + a_l m_i)$. Rearranging gives:

\[n - \sum_{l=1}^j b_l n_l = \left( \sum_{l=1}^j b_l a_l \right) m_i\]

This yields the containment $N \subseteq (n_1, \dots, n_j) + Pm_i$ (the other containment is obvious). Thus we have:

\[N = (n_1, \dots, n_j) + Pm_i \subseteq (n_1, \dots, n_j) + PM \subseteq N\]

By hypothesis, $PM$ is finitely generated, which implies $N$ is finitely generated.

Lemma Ascending Chain & Ideals

Let $M$ be a finitely generated $A$ module, \(E_M:= \{I\cdot M: I\subset A\text{ and }I \text{ is an ideal}\}.\) Then submodules of $E_M$ are finitely generated if and only if $E_M$ has the ascending chain condition.

Proof

$(\rightarrow)$

Let $I_1 M \subseteq I_2 M \subseteq \dots$ be an ascending chain in $E_M$. Let $N = \bigcup I_i M$. Define $L = {a \in A : aM \subseteq N}$. Clearly $I_j \subseteq L$ for all $j$, so $I_j M \subseteq LM$. Thus $N = \bigcup I_j M \subseteq LM$. Since $LM \subseteq N$ by definition of $L$, we have $N = LM \in E_M$. By hypothesis, $N$ is finitely generated. Since the generators of $N$ must eventually appear in some $I_k M$, the chain is stationary.

$(\leftarrow)$

Assume $E_M$ has the ACC. For any $IM \in E_M$, consider the family:

\[\mathcal{F} = \{JM : J \subseteq I \text{ is a finitely generated ideal}\}\]

Since $0 \in \mathcal{F}$, it is non-empty. By the ACC on $E_M$, $\mathcal{F}$ contains a maximal element $J_0 M$. Suppose $J_0 M \subsetneq IM$. Then there exists some $\sum_{}a_i m_i \in IM \setminus J_0 M$ for $a_i \in I$. Let $J_1 = J_0 + (a_1,\dots , a_k )$. Then $J_1$ is finitely generated and $J_1 \subseteq I$, so $J_1 M \in \mathcal{F}$. However, $J_0 M \subsetneq J_1 M$, contradicting maximality. Thus $IM = J_0 M$. Since $J_0$ and $M$ are finitely generated, $IM$ is finitely generated.

Formanek’s Theorem

Let $A$ be a ring, $M$ a finitely generated, faithful $A$-module. Let $E_M$ be as defined above. If $E_M$ satisfies the ascending chain condition then $M$ is Noetherian, and $A$ is Noetherian as well.

Proof

Let $P$ be a prime ideal of $A$. Since $PM \in E_M$, the Ascending Chain Lemma implies that $PM$ is finitely generated. By Jothilingam’s Theorem, since $M$ is finitely generated and $PM$ is finitely generated for all prime ideals $P$, $M$ is Noetherian.

Let $M = (m_1, \dots, m_n )$. Since $M$ is Noetherian, the direct sum $M^n$ is also a Noetherian $A$-module. Define the $A$-module homomorphism:

\[\phi : A \to M^n \quad \text{where} \quad \phi(a) = (am_1, \dots, am_n)\]

The kernel of this map is:

\[\ker \phi = \bigcap_{i=1}^n \text{Ann}(m_i) = \text{Ann}_A(M)\]

Because $M$ is faithful, $\text{Ann}_A(M) = 0$, making $\phi$ injective. Thus, $A \cong \phi(A)$ as $A$-modules. Since $\phi(A)$ is a submodule of the Noetherian module $M^n$, it is itself Noetherian. Since $A$ is Noetherian as a module over itself, it follows that $A$ is a Noetherian ring.

The main result is

Eakin-Nagata Theorem

Let $A\subset B$ be rings, with $B$ Noetherian, and assume that $B$ is finitely generated as $A$-module. Then $A$ is Noetherian as well.

Proof

First, $\text{Ann}_A(B) = 0$. Take $a \in \text{Ann}_A(B)$, then $a \cdot 1 = 0$, giving $a = 0$. Thus $B$ is faithful as an $A$-module.

Define $E_B = {IB : I \subset A \text{ is an ideal}}$. For any ideal $I \subset A$, $IB$ is an ideal of $B$. Since $B$ is Noetherian, ascending chains in $B$ stabilize, so $E_B$ satisfies the ascending chain condition. By Formanek’s Theorem, $A$ is Noetherian.