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.


  • $\begingroup$ 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! $\endgroup$ Jan 15, 2013 at 2:55
  • $\begingroup$ @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. $\endgroup$ Jan 15, 2013 at 2:59
  • 4
    $\begingroup$ 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. $\endgroup$ Jan 15, 2013 at 3:18
  • $\begingroup$ (And in my comment, there should be a hypothesis that $M$ is finitely generated.) $\endgroup$ Jan 15, 2013 at 3:19

3 Answers 3


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.

  • $\begingroup$ This is a very nice example. $\endgroup$ Jan 15, 2013 at 10:55
  • $\begingroup$ 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. $\endgroup$ Jan 16, 2013 at 18:16
  • $\begingroup$ @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 $K = \mathbf{R}$). $\endgroup$
    – user30180
    Jan 17, 2013 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 $(iV,Z)$.

  • $\begingroup$ @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). $\endgroup$
    – user30180
    Jan 15, 2013 at 4:32
  • $\begingroup$ The proof Steven has in mind for the non-freeness of $M$ does depend muchly on $\mathbb R$ being $\mathbb R$, really. $\endgroup$ Jan 15, 2013 at 4:43
  • $\begingroup$ 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. $\endgroup$
    – user30180
    Jan 15, 2013 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.

  • $\begingroup$ We were all looking at the question at the same time :-) $\endgroup$ Jan 15, 2013 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.