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Let $A \to B$ be a ring homomorphism. Let $M_1 \supseteq M_2\supseteq \ldots$ be an infinite chain of $A$-modules ($M_i$ not necessarily finite free). Suppose that the limit $\cap_{i=1}^{\infty} M_i$ exists and is finite free. What can we say about $\cap M_i \otimes_A B$? (Suppose we can impose some nice conditions on $A$ and $B$)

Thanks!

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  • $\begingroup$ Tensoring with $B$ does not even induce a filtration on $M_1$ unless $B$ is $A$-flat. $\endgroup$
    – Fernando Muro
    Commented Apr 7, 2013 at 7:29
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    $\begingroup$ @Fernando, then let's impose this flatness condition...in fact, any nice condition you want...I am just wondering if there is any well-stated, well-known basic result in this direction? Thanks! $\endgroup$
    – ringq
    Commented Apr 7, 2013 at 7:34
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    $\begingroup$ This question seems awfully broad and vague. What is your motivation? Is there some example you have in mind? Do you want to know when taking the intersection commutes with tensor products? $\endgroup$
    – Eric Wofsey
    Commented Apr 7, 2013 at 8:23
  • $\begingroup$ Hi Eric. The commutativity is precisely the question! $\endgroup$
    – ringq
    Commented Apr 7, 2013 at 8:34
  • $\begingroup$ If $B$ is projective over $A$, it is easy to see that it commutes. I conjecture that the converse holds: If $B$ is not projective, there is some chain of modules for which it does not commute. Any sort of general description of $\bigcap M_i\otimes B$ is likely to involve $\lim^1$, which tends to be pretty nasty to compute (when it's nontrivial). $\endgroup$
    – Eric Wofsey
    Commented Apr 8, 2013 at 2:54

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Let $A=k[X]$, let $M_i$ be the $A$-ideal generated by $X^i$, and let $B=k(X)$. Then $\cap M_i=0$ is certainly finite and free, but

$$0=(\cap M_i)\otimes B\neq \cap(M_i\otimes B)=B$$

which is a counterexample to what you're looking for (as clarified in your response to Eric Wofsey's comment), and it's hard to imagine rings nicer than $A$ and $B$.

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  • $\begingroup$ In Eric Wofsey's comment there was a reference to $\varprojlim ^1$: in that direction, let me expand a bit saying that the condition that $\cap M)i$ stabilizes (i.e. $M_j=\cap M_i$ for some sufficiently big $j$), although seeming terribly strong is most probably the unique "universal" condition to impose in order to make the tensor product commute with the intersection: it coincides with being "Mittag-Leffler" which (with some noetherianity assumption and finite generation of $A/B$) is "the" requirement to make $\varprojlim$ and $\otimes$ commute. $\endgroup$
    – Filippo Alberto Edoardo
    Commented Apr 8, 2013 at 3:26

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