# $B$ is a finitely generated module over $A$. Then if $A$ is a Noetherian ring (or Artinian ring), then $B$ is a Noetherian (Artinian) ring.

Let $A$ be a subring of $B$. Suppose $B$ is a finitely generated module over $A$. Prove that if $A$ is a Noetherian ring (or Artinian ring), then $B$ is a Noetherian (Artinian) ring.

I am quite confused. My idea:

there exists a surjective $A$-module homomorphism $\phi: A^k\rightarrow B$. Suppose $A$ a Noetherian ring but $B$ not a Noetherian ring. Let ideals of $B$ be $J_1\subset J_2\subset\cdots$. Then $\phi_{-1}(J_1)\subset \phi^{-1}(J_2)\subset\cdots$ are $A$-submodules of $A^k$, moreover, they are also ideals of $A^k$. Since $A$ Noetherian, $A^k$ is Noetherian as a $A$-module. (But we cannot obtain $A^k$ Noetherian as a ring.) How to continue?

## 1 Answer

Here's a nice way to handle both Noetherian and Artinian cases at once: first notice that if $B$ is Noetherian (resp. Artinian) as an $A$-module, then $B$ is Noetherian (resp. Artinian) as a ring (since any chain of $B$-ideals is a chain of $A$-submodules of $B$). It thus suffices to prove the following:

Proposition: If $A$ is Noetherian (resp. Artinian), then any finite $A$-module $B$ is again Noetherian (resp. Artinian).

Proof: Induct on the number of generators of $B$. If $B$ is cyclic, then $B$ is a quotient of $A$, hence is Noetherian (resp. Artinian). In general, if $B = Ax_1 + \ldots + Ax_n$, then there is an exact sequence

$$0 \to Ax_n \to B \to B/Ax_n \to 0$$

Now $B/Ax_n$ needs only $n-1$ generators, so by induction $Ax_n, B/Ax_n$ are Noetherian (resp. Artinian), but then so is $B$.

By the way, the converse is also true: if $A \subseteq B$ as rings, and $B$ is a finite $A$-module, then $A$ is Noetherian iff $B$ is Noetherian (as rings): this is Eakin's theorem.