3
$\begingroup$

I have found a result concerning projectivity of a certain ring extension: Lemma 2.64.

This says the following:

Let $A$ be an integral domain or a noetherian ring, $B$ an $A$-algebra, $C$ an $A$-subalgebra of $B$. Assume that $B$ is a f.g. flat $A$-module and $C$ is a f.g. free $A$-module. Then $B$ is a f.g. projective $C$-module.

I do not know French, so trying to read the proof (with the help of google translate) is difficult to me.

The problem is that I suspect (but of course may be wrong) that there exists a counterexample to that lemma: $A=\mathbb{Z}$, $C=\mathbb{Z}+2\mathbb{Z}i$, $B=\mathbb{Z}+\mathbb{Z}i$. (I am not sure whether $B$ is $C$-projective or not).

Can someone please either confirm the proof of Lemma 2.64 and disprove my counterexample, or find an error in the proof of Lemma 2.64.

Thank you very much.

Edit: Since user26857 confirmed that my counterexample is indeed true, hence Lemma 2.64 must contain an error,

I wonder what conditions one can add to Lemma 2.64 in order to guarantee the projectiveness (or only flatness; this also seems interesting) of $B$ over $C$?

BTW, I have now noticed another counterexample to Lemma 2.64: $k[x^2] \subset k[x^2,x^3] \subset k[x]$. Here $B$ and $C$ are f.g. free $A$-modules, but $B$ is NOT flat over $C$; this example appears in van den Essen's book "Polynomial Automorphisms and the Jacobian Conjecture" on page 294, Exercise 4. This counterexample to Lemma 2.64 is somewhat similar to the previous one ($B$ is integral over $C$, $C$ is not integrally closed, the field of fractions of $C$ equals that of $B$. Therefore, one sees that, for example, adding integrality of $B$ over $C$ is not enough to guarantee flatness).

(0) A trivial suggestion: All rings are integral domains, $C$ is integrally closed, $B$ is integral over $C$, and the field of fractions of $C$ equals that of $B$. Then $C=B$.

(1) Maybe one plausible answer (to guaranteeing flatness of $B$ over $C$) is adding the condition that all three rings are integral domains having the same field of fractions. This time $\mathbb{Z} \subset \mathbb{Z}[2i] \subset \mathbb{Z}[i]$ is not relevant since the field of fractions $\mathbb{Z}$ is $\mathbb{Q}$, while the field of fractions of the other two rings is $\mathbb{Q}(i)$.

(2) There are other options as well, for example, in a specific situation (described in a paper of Adjamagbo in van den Essen's book "Automorphisms of Affine Spaces"), normality of $C$ implies flatness of $B$ over $C$, or separability of $C$ over $A$ implies flatness of $B$ over $C$.

(3) See also, Nagata's flatness criterion.

Any plausible additional condition which guarantees projectiveness/flatness will be welcomed.

$\endgroup$
  • $\begingroup$ If $B$ is $C$-projective, then it is flat, hence faithfully flat (since it is integral), and therefore $C=B$, a contradiction. (Notice that $B$ and $C$ have the same field of fractions, namely $\mathbb Q(i)$.) $\endgroup$ – user26857 Jun 18 '15 at 0:39
  • $\begingroup$ That's a nice and easy argument I should have though of myself (since I am quite familiar with it; it appears, for example, in Matsumura's CRT, Exercise 7.2). $\endgroup$ – user237522 Jun 18 '15 at 0:51
1
$\begingroup$

Seydi's claim in the proof of Lemma 2.64 $$B\otimes_AB\simeq B\otimes_CB\otimes_AC$$ is wrong.

Let $A=\mathbb Z$, $C=\mathbb Z[2i]$, and $B=\mathbb Z[i]$. Then the claimed isomorphism becomes $$\mathbb Z[i]\otimes_{\mathbb Z}\mathbb Z[i]\simeq\mathbb Z[i]\otimes_{\mathbb Z[2i]}\mathbb Z[i]\otimes_{\mathbb Z}\mathbb Z[2i].$$ In the following we consider all objects involved in the above isomorphism as being $\mathbb Z$-modules. Then $$\mathbb Z[i]\otimes_{\mathbb Z}\mathbb Z[i]\simeq(\mathbb Z[i]\otimes_{\mathbb Z[2i]}\mathbb Z[i])^2$$ since $\mathbb Z[2i]$ is a free $\mathbb Z$-module of rank two. The left hand side is a free $\mathbb Z$-module of rank four, and therefore $\mathbb Z[i]\otimes_{\mathbb Z[2i]}\mathbb Z[i]$ must be a free $\mathbb Z$-module of rank two.
We have an exact sequence $$0\to J\to\mathbb Z[i]\otimes_{\mathbb Z[2i]}\mathbb Z[i]\to\mathbb Z[i]\to0,$$ where $J$ is generated by $x\otimes 1-1\otimes x$, $x\in\mathbb Z[i]$. It follows that $J$ is a free $\mathbb Z$-module of rank $0$, that is, $J=0$. But one can show, by using standard arguments, that $1\otimes i-i\otimes 1\ne0$.

$\endgroup$
  • $\begingroup$ @user237522 I have no idea why you bothered to read Seydi's paper which is full of mistakes as pointed out Adjamagbo in this paper that you are probably aware of it. $\endgroup$ – user26857 Jun 19 '15 at 16:53
  • $\begingroup$ Thank you very much! (I hoped that at least Lemma 2.64 was true). $\endgroup$ – user237522 Jun 21 '15 at 1:48

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.