4
$\begingroup$

Let $A \subseteq B \subseteq C$ be commutative rings such that: $(1)$ $B$ is a projective $A$-module $(2)$ $C$ is a finitely presented $B$-module and $(3)$ $C$ is a flat $A$-module.

Does it follow that $C$ is a projective $A$-module??

Intuitively this kind of scenario might occur when we mange to "normalize" a flat finitely presented algebra, in which case the question asks if the projectiveness of the normalization would imply the projectiveness of the original algebra.

Recall that a module is projective iff it is Mittag-Leffler, flat, and it decomposes as a direct sum of countably generated modules. So the meat of the question is whether the hypotheses are enough to transfer the property of being Mittag-Leffler as an $A$-module from $B$ to $C$.

Being Mittag-Leffler is, as far as I know, not a transitive property.
On the other hand, being finitely presented is a transitive property, so in case $B$ is finitely generated the conclusion is immediately yes.

$\endgroup$
6
  • 3
    $\begingroup$ What about $A=\mathbb Z$, $B=\mathbb Z[T]$ and $C=B\times\mathbb Z[1/p]$ where $B\to C$ is induced by $T\mapsto(T,1/p)$? $\endgroup$
    – Z. M
    Commented May 7, 2021 at 0:26
  • $\begingroup$ @Z. M Yes that does the job. "Normalizing" in this case basically knocks out the localization component, which is precisely the obstruction to being projective $\endgroup$ Commented May 7, 2021 at 0:46
  • $\begingroup$ @Z.M Can you think of a counterexample with $C$ faithfully flat over $A$? $\endgroup$ Commented May 7, 2021 at 0:50
  • $\begingroup$ Just take $C=C_0\times B$ where $C_0$ is any counterexample. $\endgroup$ Commented May 7, 2021 at 6:10
  • $\begingroup$ I think that my example is already faithfully flat (it is surjective on Spec)? Anyway, I think that maybe you are interested in an example when $C$ is a domain (that seems more challenging)? $\endgroup$
    – Z. M
    Commented May 7, 2021 at 12:37

0

You must log in to answer this question.

Browse other questions tagged .