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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.

ronno
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  • Natural follow-up question: What do we have to assume on $A \to B$ so that extension of scalars preserves noetherian modules? – Martin Brandenburg Aug 16 '14 at 08:59
  • I think this question does not show enough (any!) research effort but sadly I cannot downvote my own question. I have not deleted it since it has an answer with upvotes. – ronno May 08 '23 at 11:21

3 Answers3

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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.

Bruno Joyal
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Andrew Dudzik
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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: $$v\otimes b=\sum_i\sum_jx_j\cdot(v_i\otimes b_{ij}).$$

asking
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    I think you can also see this by: $M$ Noetherian $\implies$ $M$ finitely generated $\implies$ $M \otimes B$ is finitely generated as a $B$-module $\implies$ $M \otimes B$ is Noetherian as a $B$-module. Of course the last implication uses/needs that $B$ is Noetherian. – ronno May 08 '23 at 11:19
  • Nice! Way more elegant. :) – asking May 08 '23 at 11:37
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To give a requirement on $A\rightarrow B$ instead of $B$, a sufficient condition for a morphism $f:A\rightarrow B$ to have its induced extension of scalars functor to preserve Noetherian modules is for it to be flat and of finite type.

let $M$ be a Noetherian $A$ module and $B$ be flat over $A$. Observe that $M\otimes_AB$ is a finitely generated $B$ Module (since $M$ is f.g.), thus $M\otimes_AB$ is a finitely generated $B/Ann(M\otimes_AB)$ module by the induced action.

We will now show that $B/Ann(M\otimes_AB)$ is a Noetherian Ring. It is known that $A/\mathfrak{a}$ where $\mathfrak{a}=Ann(M)$ is a Noetherian ring. Given that $B$ is a finitely generated $A$-algebra and $A/\mathfrak{a}$ is Noetherian, we see that $B\otimes_AA/\mathfrak{a}$ is also Noetherian. (Hence why we require $f$ be of finite-type)

See also: When is a tensor product of two commutative rings noetherian?

Since $B$ is a flat ring, $B\otimes_AA/\mathfrak{a}=B/(B\otimes_A\mathfrak{a})$. Focusing on the quotient $B\otimes_A\mathfrak{a}=B\otimes_AAnn(M)$, we want to show the result : Given $M$ noetherian (hence finitely generated) and $B$ flat over $A$; $$B\otimes_AAnn(M)=Ann(M\otimes_AB)$$ See also: Annihilator of extension of scalars vs. the extension the annihilatar

Thus we see that $M\otimes_AB$ is a Noetherian $B/Ann(M\otimes_AB)$ module implying it is also a Noetherian $B$ module with the same $B$ action as the one induced by the extension of scalar.

Alex
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