When Do Finitely Generated Prime Ideals Imply a Noetherian Ring?

The phrase “Noetherian prime ideal” can mean different things, so it helps to state the precise result. For a commutative ring with identity, the ring is Noetherian exactly when every prime ideal is finitely generated. This is often called Cohen’s theorem.

A prime ideal is an ideal, not usually a ring in its own right. The theorem concerns whether it is finitely generated as an ideal of the ambient ring.

The statement

A ring $R$ is Noetherian if every ideal of $R$ is finitely generated. Cohen’s theorem says that it is enough to check prime ideals:

$$R\text{ is Noetherian} \quad\Longleftrightarrow\quad \text{every prime ideal of }R\text{ is finitely generated}.$$

The forward direction is immediate: prime ideals are ideals, so the definition of a Noetherian ring applies to them. The converse is the substantive part.

Why the converse holds

Suppose every prime ideal of $R$ is finitely generated, but suppose, for contradiction, that $R$ is not Noetherian. Then some ideal is not finitely generated. Among all such ideals, choose one, $I$, that is maximal under inclusion.

Such a maximal choice is possible: the union of a chain of non-finitely generated ideals cannot be finitely generated. If that union had finitely many generators, all of them would lie in one ideal in the chain, forcing the union to equal that finitely generated ideal.

We show that $I$ is prime. It is proper because $R$ is finitely generated as an ideal. Suppose $ab\in I$ while $a\notin I$ and $b\notin I$. The ideal $I+(a)$ strictly contains $I$, so maximality of $I$ implies that $I+(a)$ is finitely generated. The colon ideal

$$(I:a)=\{r\in R:ra\in I\}$$

also strictly contains $I$, since $b\in(I:a)$ but $b\notin I$. Thus $(I:a)$ is finitely generated as well.

Write $I+(a)=(a,x_1,\ldots,x_n)$ with each $x_i\in I$. For any $y\in I$, this expression gives

$$y=ra+\sum_{i=1}^n r_i x_i$$

for some $r,r_i\in R$. Since $y$ and every $x_i$ belong to $I$, we have $ra\in I$, so $r\in(I:a)$. Therefore

$$I=a(I:a)+(x_1,\ldots,x_n).$$

The colon ideal is finitely generated, and the $x_i$ are a finite list. This equation makes $I$ finitely generated, contradicting its choice. Hence no such $a,b$ exist, and $I$ is prime.

But the hypothesis says every prime ideal is finitely generated. That contradicts the fact that $I$ was not finitely generated. Therefore $R$ must be Noetherian.

A concrete example

Let $R=k[x_1,x_2,x_3,\ldots]$, the polynomial ring in countably many variables over a field $k$. The ideal

$$\mathfrak m=(x_1,x_2,x_3,\ldots)$$

is prime because $R/\mathfrak m\cong k$ is an integral domain. It is not finitely generated: any finite list of elements of $\mathfrak m$ involves only finitely many variables, say variables with indices at most $N$, where we may take $N\ge 1$. Each generator then belongs to $(x_1,\ldots,x_N)R$, so the ideal they generate is contained in $(x_1,\ldots,x_N)R$. But $x_{N+1}\notin(x_1,\ldots,x_N)R$, as its image in $R/(x_1,\ldots,x_N)R\cong k[x_{N+1},x_{N+2},\ldots]$ is nonzero. Thus the finite list cannot generate $\mathfrak m$. Hence $R$ is not Noetherian, consistent with Cohen’s theorem.

What this does—and does not—say

If $R$ is Noetherian and $\mathfrak p$ is prime, then $\mathfrak p$ is finitely generated as an ideal of $R$. Also, the quotient $R/\mathfrak p$ is Noetherian: its ideals correspond to ideals of $R$ containing $\mathfrak p$, and those ideals are finitely generated.

These are different statements. Saying that $R/\mathfrak p$ is Noetherian does not mean that $\mathfrak p$ is a ring. And Cohen’s theorem does not require checking that prime ideals, considered as rings, are Noetherian.

Common mistakes

• Checking only maximal ideals: Cohen’s theorem requires every prime ideal, not only maximal ideals.
• Confusing finitely generated with prime: A prime ideal need not be finitely generated in an arbitrary ring; the theorem characterizes when all prime ideals are.
• Ignoring the ring assumptions: The proof above uses the usual commutative, identity-containing ring setting. For noncommutative rings, definitions of prime ideals and Noetherianity vary, and this exact statement should not be applied without specifying conventions.