If M is not a free A-module, can tensoring with a bigger field make it free?

I need a simple commutative algebra lemma for a paper, but can't find a reference. Maybe I don't know the right keywords. Here's the setup. $F:K$ is a field extension, $A$ is an algebra over $K$, and $M$ an $A$-module. If $M \otimes_K F$ is a free $A \otimes_K F$ module, must $M$ have been a free $A$-module?

I can do the particular case I need: $K = \mathbb{F}_2$, $F$ a finite extension, $A = 1 \oplus N$ for $N$ nilpotent, $M \otimes_K F$ free of rank one. But I suspect that something more general holds, and it would be nice to quote rather than reprove.

Thanks!

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if you don't know how to prove it, one can guess the reader of your paper might also have a little trouble and then providing a proof would be nice to her! – Mariano Suárez-Alvarez Jan 15 '13 at 2:55
@Mariano: I'll certainly outline my proof, I just wanted to see if there was a simple trick (which would be great), or a more general framework, which I would indicate. If my hands-on way is the only way, then I'll give more details than if there is another reference. – Joshua Batson Jan 15 '13 at 2:59
In your situation, $A$ is local--thus the result holds as projectivity descends through faithfully flat base change. I don't know about the general situation. – Daniel Litt Jan 15 '13 at 3:18
(And in my comment, there should be a hypothesis that $M$ is finitely generated.) – Daniel Litt Jan 15 '13 at 3:19

Yes, there are such examples even with invertible $M$ and smooth $K$-algebras of dimension 1, with $K$ any field that is not algebraically closed.

Choose such a $K$, so we may and do also choose a nontrivial primitive finite extension $F$ of $K$. Consider the projective line over $K$ and remove a closed point $\xi$ such that $F = K(\xi)$. This is an affine open $U = {\rm{Spec}}(A)$ for a Dedekind $A$, and let $M$ be the maximal ideal of $A$ corresponding to an $K$-point. This is invertible as an $A$-module (as for any Dedekind domain) but cannot be principal. Indeed, if $a \in A$ were a generator of $M$ then its divisor on $\mathbf{P}^1_K$ has restriction to $U$ that is a single point of degree 1 yet the condition ${\rm{deg}}_K({\rm{div}}(a)) = 0$ forces the "negative" part of the divisor to have degree $-1$ over $K$, contradicting that this negative part is supported at the closed point $\xi$ with $K$-degree larger than 1.

Clearly $A \otimes_K F$ is the coordinate ring of the complement in $\mathbf{P}^1_F$ of ${\rm{Spec}}(F \otimes_K F)$, which contains an $F$-point, so $A \otimes_K F$ is the coordinate ring of a non-empty affine open in the open complement $\mathbf{A}^1_F = {\rm{Spec}}(F[t])$ of an $F$-point in $\mathbf{P}^1_F$. Thus, $A \otimes_K F$ is the localization of the PID $F[t]$ at some nonzero element, so $A \otimes_K F$ is a PID and hence its nonzero ideal $M \otimes_K F$ (corresponding to a single $F$-point) is principal.

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This is a very nice example. – Piotr Pstrągowski Jan 15 '13 at 10:55
I had a little trouble understanding this. Would an instance be $F:K = \mathbb{R}:\mathbb{C}$, with $A = \mathbb{R}[x,y]/x^2+y^2-1$ and $M = \langle x-1,y \rangle$? Then $M_\mathbb{C} = \langle z-1 \rangle$ (where $z = x+iy$) is principal, though $M$ is not. – Joshua Batson Jan 16 '13 at 18:16
@Joshua Batson: Yes, the "circle" was even my original example. In the course of writing it up as an answer I planned to say at the end that I didn't use any too essential about $\mathbf{R}$, but I then decided that such a remark would be more convincing if I actually explained the construction in the general case. Perhaps in so doing I made it look too exotic, but indeed the circle example is the simplest special case (note you could use $K = \mathbf{Q}$ and $F = \mathbf{Q}(i)$ as well, maybe even "simpler" than with $K = \mathbf{R}$). – user30180 Jan 17 '13 at 3:52

The implication does not hold.

Let $K$ be the real numbers, $F$ the complex numbers, $A=K[X,Y,Z]/(X^2+Y^2+Z^2-1)$, and $M$ the kernel of the map $A^3\rightarrow A$ given by the unimodular row $(X,Y,Z)$.

Then $M$ cannot be free, by the same argument I gave in my answer to this question.

But $M$ becomes free after tensoring with the complex numbers. To see this, write $U=(X+iY)/2$, $V=(X-iY)/2$. Then it suffices to show that $(X,Y,Z)=(U+V,-iU+iV,Z)$ can be transformed via elementary operations to $(1,0,0)$ (so that its kernel is isomorphic to the kernel of $(1,0,0)$, which is evidently free).

To construct such a series of elementary transformations, first transform $(U+V,-iU+iV,Z)\rightarrow (2U,-iU+iV,Z)\rightarrow (2U,iV,Z)$ and then note that $2U$ is equal to 1 mod $(iV,Z)$.

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@Steven Landsburg: This is the same example that initially occurred to me (albeit not quite as explicitly as you have worked it out), but in the course of writing it up I decided to adapt the argument to work over more general fields (since ultimately there's nothing too special about $\mathbf{R}$ in the method). – user30180 Jan 15 '13 at 4:32
The proof Steven has in mind for the non-freeness of $M$ does depend muchly on $\mathbb R$ being $\mathbb R$, really. – Mariano Suárez-Alvarez Jan 15 '13 at 4:43
Oops, I misread his question as being about the affine open in the projective conic $x^2 + y^2 + z^2$ over $\mathbf{R}$ obtained by deleting a single closed point. So my comment above is indeed incorrect; thanks for the correction, Mariano. – user30180 Jan 15 '13 at 5:13

Example 4.7 in [Guralnick, Robert; Jaffe, David B.; Raskind, Wayne; Wiegand, Roger. On the Picard group: torsion and the kernel induced by a faithfully flat map. J. Algebra 183 (1996), no. 2, 420--455. MR1399035 (97c:14002)] is an example of an algebra $A$ —A Dedekind domain in fact— such that after an extension $K/k$ the map $Pic(A)\to Pic(A_K)$ has non-finitely generated kernel. Each element in the kernel gives a counterexample.

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We were all looking at the question at the same time :-) – Mariano Suárez-Alvarez Jan 15 '13 at 4:19