Suppose $A$ is a commutative ring with unity, and $B$ is an $A$-algebra. If $M$ is a Noetherian $A$-module, is $M \otimes_A B$ Noetherian as a $B$-module? Note that there are no finiteness conditions on $A$ or $B$.
This seems a natural generalization of exercise 7.10 in Atiyah-Macdonald, which asks to show that $M[x]$ is Noetherian as an $A[x]$ module.
On
I’m very late to this, but while generalising A&M 7.10 I came up with the same question, and it seems like assuming $B$ is Noetherian is sufficient and necessary (by slade’s answer) for this to hold. I post it here for the sake of completeness (answering Martin’s comment).
Indeed, let $N$ be a submodule of $M\otimes B$, and let $$N’:=\{v\in M : v\otimes b\in N \text{ for some } b\in B\}.$$
$N’$ is an $A$-submodule of $M$: the only non-trivial part to check is that it’s closed under addition. Suppose $v,w\in N’$, then there are $b,b’\in B$ such that $v\otimes b,w\otimes b’\in N’$, so $v\otimes (bb’),w\otimes (b’b)\in N$, hence $(v+w)\otimes (bb’)\in N$, thus $v+w\in N’$. Therefore, as $M$ is $A$-Noetherian, there are $v_1,\cdots,v_n\in M$ that $A$-generate $N’$.
For each $i=1,\cdots,n$, let $J_i:=\{b : v_i\otimes b\in N\}$. $J_i$ are ideals in $B$ (clear), so generated by $b_{i1},\cdots,b_{i,n_i}\in B$. (Here we use $B$ is Noetherian).
Note that $N$ is $B$-generated by $\{v_i\otimes b_{ij} : i=1,\cdots,n, j=1,\cdots,n_i\}$, as:
$v\otimes b\in N$, then $v=\sum a_iv_i$ for $a_i\in A$, so $v\otimes b=\sum (v_i\otimes a_ib)$, and each $a_ib$ is in $I_i$, so $a_ib=\sum_j x_jb_{ij}$, therefore: $$v\otimes b=\sum_i\sum_jx_j\cdot(v_i\otimes b_{ij}).$$
If $A$ is noetherian and $B$ is not, then $A$ is a noetherian $A$-module, but $A\otimes_A B \cong B$ is not a noetherian $B$-module.