# $A \oplus M$ is Noetherian iff $A$ is Noetherian and $M$ is a f.g. $A$-module

Let $A$ be a commutative ring and $M$ an $A$-module. Define a multiplication on the additive group $B:=A \oplus M$ by $$(a,x)(b,y) := (ab,bx+ay) \ \ \ (a,b \in A,\ x,y \in M).$$ An easy computation verifies that $B$ is a ring under the above multiplication and $B_1:=A_1 \oplus M$ is an ideal of $B$ where $A_1$ is any ideal of $A$.

My question is:

Prove that $B$ is a Noetherian ring $\iff$ $A$ is a Noetherian ring and $M$ is a finitely generated $A$-module.

My attempt: Suppose that $B$ is a Noetherian ring. We know in particular that $\{ 0 \} \oplus M$ is an ideal of $B$ hence $M$ is finitely generated. For any ideal $A_1$ of $A$, $B_1:=A_1 \oplus M$ is an ideal of $B$. This means that $B_1$ is finitely generated so is $A_1$. We conclude that $A$ is Noetherian. Is this proof correct?

Conversely, if $A$ is a Noetherian ring and $M$ is a finitely generated $A$-module then in fact $M$ is Noetherian $A$-module (by quite long arguments but I'm sure it's correct). So for any ideal $B_1$ of $B$ how can I prove it's finitely generated? Of course if $B_1$ is of the form either $A_1 \oplus M$ for some ideal $A_1$ of $A$ or $A \oplus M_1$ for some submodule $M_1$ of $M$ then we are done.

Any help would be much appreciated.

Since $A\simeq B/(\{0\}\oplus M)$ we have an exact sequence of $B$-modules (hence of $A$-modules, too):
$$0\to M\to B\to A\to 0.$$
"$\Rightarrow$" $B$ is a noetherian ring, and $A$ is q quotient of $B$, so $A$ is also a noetherian ring. Moreover, the ideal $\{0\}\oplus M$ of $B$ is finitely generated hence $M$ is a finitely generated $A$-module.
"$\Leftarrow$" From the given exact sequence we get that $B$ is a noetherian $A$-module, therefore a noetherian ring.