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This question is a byproduct of the following question.

Let $R$ be a commutative ring with unit. Using exterior algebra computations, one may show that if an $R$-module $M$ satisfies $M^n\simeq R^n$, then $M$ is a projective module of rank one satisfying $M^{\otimes n}\simeq R$. In other words, the isomorphism class of such a module $M$ is a $n$-torsion element of the Picard group of $R$.

I would be extremely surprised if we had the reverse implication. However, I cannot come up with a counterexample (the reverse implication is true for Dedekind rings).

So the question is:

Question. What is an example of a projective $R$-module of rank one $M$ such that $M^{\otimes n}\simeq R$ but $M^n\not\simeq R^n$ ?

If you allow me to be picky, I would rather prefer (if possible) $R$ to be a noetherian integral domain (so that the counterexample is not too artificial), and I would be happy with a counterexample for $n=2$.

GreginGre
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  • I can see that in your case ($R$ Noetherian of dimension $d$, $M$ as in your question with $M^{\otimes 2}$ isomorphic to $R$), then $M^{\oplus 2^d}$ is free. This of course is neither here nor there. – Mohan May 06 '23 at 15:19
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    I have now examples where $M\oplus M$ as in your question is not only not free, but even not stably free. Take $X$ to be the complement of a smooth quadric in $\mathbb{P}^n$ with $n\geq 4$ and $M$ be the restriction of $\mathcal{O}(1)$. – Mohan May 07 '23 at 02:27
  • @Mohan: Nice ! would you be kind enough to write some details in an answer? – GreginGre May 07 '23 at 07:48

1 Answers1

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Let $P=\mathbb{P}^n$, $Y\subset P$ a smooth hypersurface of degree $d\geq 2$ and let $X=P-Y$. Let $A^i$ denote the Chow group of codimension $i$ cycles.

The main thing is when do we have a surjection $A^i(P)\to A^i(Y)$ given by intersection, $\alpha\mapsto \alpha\cdot Y$ for all $i\leq r$? Depending on this $r$ (these are generally found using Lefschetz theorems), one gets examples of what you want.

So, let us focus on $n\geq 4$ and $d=2$. Then $A^1(P)\to A^1(Y)$ is an isomorphism, by Lefschetz.

One has the standard exact sequence of Chow groups, $$A^{\bullet}(Y)\to A^{\bullet}(P)\to A^{\bullet}(X)\to 0.$$

Since the composite $A^1(P)\to A^1(Y)\to A^2(P)$ is just multiplication by 2, we see by our surjection, we have $A^1(X)=A^2(X)=\mathbb{Z}/2\mathbb{Z}$, generated by the class of linear spaces.

Now let $M$ be the restriction of $\mathcal{O}_P(1)=L$ to $X$. Then $M^{\otimes 2}$ is the restriction of $\mathcal{O}_P(2)$ and since we are removing a quadric, we see that $M$ is 2-torsion. On the other hand, the second chern class of $M\oplus M$ (restriction of $c_1(L)\cdot c_1(L)=$ codimension two linear space) is non-zero in $A^{\bullet}(X)$. Thus $M\oplus M$ is not stably free.

Mohan
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