Ring Homomorphisms and Noetherian...?

Question

I am confused by the following corollary that appears in this book (just search "Corollary 7.8.11" for the appropriate page):

Corollary 7.8.11 Let $f : A \to B$ be a ring homomorphism and suppose that $B$ is finitely generated as a left $A$-module. Then if $A$ is left Noetherian or left Artinian, so is $B$

and am equally confused by this rather terse proof:

Every $B$-module is an $A$-module via $f$. So chain of ideals in $B$ is a chain $A$-submodules of a finitely generated $A$-module $B$.

First of all, in the statement of the collary, it says "Then if $A$ is left Noetherian or left Artinian, so is $B$" So is $B$ what? Are we trying to show $B$ is a left Noetherian ring or module?

"Every $B$-module is..." What's the purpose of this remark? Why do we care about $B$-modules? The statement of the corollary doesn't seem to indicate that we need to consider $B$-modules.Sure, if you have ring homomorphism $f : R \to S$ between rings $R$ and $S$, and $M$ is an $S$-module, then $M$ has a natural $R$-module structure given by $r \cdot m = f(r)m$. But in our case $B$ is already given some $A$-module structure already (indeed, it's finitely generated $A$-module). I don't see what role the module structure induced by $f$ is playing in the proof.

I'm hoping that someone will help me flesh out this proof. If it saves you any trouble, I'm really only interested in the Noetherian case, as the Artinian case is similar.

Answer 1

We are trying to show that $B$ is a left Noetherian ring. The fact that $B$ is a left Noetherian $A$-module is immediate (since $B$ is a finitely generated $A$-module and $A$ is a left Noetherian ring) -- we will need this fact in the proof.

To show $B$ is a left Noetherian ring is equivalent to showing $B$ is a left Noetherian $B$-module. This explains the remark about $B$-modules: we care here about $B$ as a left module over itself. Every $B$-submodule of $B$ (i.e., left ideal of $B$) is also an $A$-submodule by the same remark; thus every chain of left ideals is a chain of $A$-submodules. Since $B$ is a left Noetherian $A$-module, the ACC applies to each such chain; thus we have shown the ACC applies to each chain of left ideals, so $B$ is a left Noetherian ring.

Answer 2

The statement means "if $A$ is left Noetherian then so is $B$ (left Noetherian) and if $A$ is left Artinian then so is $B$ (left Artinian)".

The ideals of $B$ are $B$-modules and that's why we care about $B$-modules, as we want to show that every chain of ideals of $B$ has the ascending chain condition (for the Noetherian case that you are interested in). Because these ideals are $B$-modules they are also $A$-submodules of the finitely generated $A$-module $B$. As $A$ is Noetherian, by the proposition leading to the corollary, the ideals of $B$ satisfy the ascending chain condition, which is what we needed to prove to show that $B$ is Noetherian.

Answer 3

• They mean $B$ is a noetherian ring. $B$ being noetherian as an $A$-module results from a theorem you probably have seen.
• This remarks reduces the problem of chains of $B$-ideals to a problem of chains of $A$-submodules of $B$.