Puzzles around finiteness and projectivity of modules

Stream: maths

[Junyan Xu]

It's well known (and in mathlib) that the dual Pᵛ of a finite projective R-module P is also finite projective. When trying to generalize Module.finite_dual_iff from free modules to projective modules, I investigated all possible implications between (conjunctions of) the four conditions Finite R P, Finite R Pᵛ, Projective R P and Projective R Pᵛ; 3 of the 12 binary implications appear harder than the others, and I've only figured out one of them so far. These look like some good puzzles to test one's intuition, so I'm posting them as a poll. If you are able to construct counterexamples or give proofs, they are of course appreciated.

[Junyan Xu]

/poll Which of the following always hold?

Projective R P ∧ Finite R Pᵛ → Finite R P
Projective R P ∧ Finite R Pᵛ ↛ Finite R P
Projective R P ∧ Finite R Pᵛ → Projective R Pᵛ
Projective R P ∧ Finite R Pᵛ ↛ Projective R Pᵛ
Projective R Pᵛ ∧ Finite R P → Finite R Pᵛ
Projective R Pᵛ ∧ Finite R P ↛ Finite R Pᵛ

[Kevin Buzzard]

Shouldn't we have options which are the negations of current options?

[Junyan Xu]

What do you mean exactly? Each of the three questions comes with two options (implication followed by non-implication). Maybe it's not displayed correctly on mobile?
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[Kevin Buzzard]

Oh sorry for the noise! Yes it was a mobile display issue (I couldn't see the slashes through the implication symbols)

[Junyan Xu]

It looks like it might get displayed as the following in some fonts (could be tiny)
image.png
image.png

[Mitchell Lee]

All three statements Projective R P ∧ Finite R Pᵛ → Finite R P, Projective R P ∧ Finite R Pᵛ → Projective R Pᵛ, Projective R Pᵛ ∧ Finite R P → Finite R Pᵛ hold if $R$ is noetherian. I suspect they do not hold generally, but I do not have a specific counterexample.

Lemma.

Let $P$ be a projective module over a commutative ring $R$. Then the evaluation map $\Psi \colon P \to (P^{\vee})^\vee$ is injective.

Proof

Since $P$ is a direct summand of a free module, it is enough to check that the statement holds if $P = \bigoplus_{i \in I} R$ is a free module. In this case, we have $P^{\vee} = \prod_{i \in I} R$. Let $x \in P$ and suppose that $\Psi(x) = 0$. Then for all $i \in I$, we have that $x$ evaluates to zero on the element $e^i \in \prod_{i \in I} R$ whose $j$-th component is $\delta_{ij}$. This implies that the $i$-th component of $x$ is $0$; since this holds for all $i$, we have $x = 0$.

Lemma.

Let $M$ be a module over a noetherian commutative ring $R$. If $M$ is finite, then $M^\vee$ is finite.

Proof

We have a surjection $R^m \twoheadrightarrow M$ for some natural number $m$. Dualizing yields an injection $M^\vee \hookrightarrow (R^m)^\vee$. Since $(R^m)^\vee = R^m$ is finite and $R$ is noetherian, the lemma is proved.

Theorem.

Let $P$ be a projective module over a noetherian commutative ring $R$. If $P^\vee$ is finite, then $P$ is finite.

Proof.

By the first lemma, the evaluation map $\Psi \colon P \to (P^{\vee})^\vee$ is injective. By the second lemma, $(P^\vee)^{\vee}$ is finite. Hence, $P$ is finite.

Theorem.

Let $P$ be a projective module over a noetherian commutative ring $R$. If $P^\vee$ is finite, then $P^\vee$ is projective.

Proof.

By the previous theorem, $P$ is finite, and it is well known that the dual of a projective finite module is projective.