4
$\begingroup$

My issue is in Theorem 8.8 and Lemma 8.9 of Chapter II concerning Kähler differentials. I have managed to "relax" Lemma 8.9 as follows :

Let $A$ be a noetherian local integral domain with residue field $k$ and quotient field $K$. If $M$ is a finitely generated $A$-module and if $r = \dim_k M \otimes_A k \le \dim_K M \otimes_A K$, then the inequality is an equality and $M$ is free of rank $r$.

The proof is exactly the same as outlined in Hartshorne, since the inequality is sufficient to complete the proof (the surjectivity of $K^r \to M \otimes_A K$ with $\dim_K M \otimes_A K \ge r$ shows that it has to be injective and the dimensions need to be equal).

My problem comes in the proof of Theorem 8.8 :

Let $B$ be a local ring containing a field $k$ isomorphic to its residue field (I noticed afterwards that they more precisely mean that the map $k \to B \to B/\mathfrak m$ is a field isomorphism). Assume furthermore that $k$ is perfect, and that $B$ is the localization of a finitely generated $k$-algebra. Then $\Omega_{B/k}$ is a free $B$-module of rank equal to $\dim B$ if and only if $B$ is a regular local ring.

It is assumed that $B$ is the localization of a finitely generated $k$-algebra, say $A$. Let us assume for simplicity that $A$ is a domain (since it is an admissible context in the hypotheses). The theorem claims that $\Omega_{B/k}$ is free of rank $\dim B$ if and only if $B$ is a regular local ring. The proof goes fine until they claim that $\dim B = \mathrm{tr.deg}(K/k)$ ; this equality is false in general since it doesn't hold when $B = A_{\mathfrak p}$ where $\mathfrak p \in \mathrm{Spec}(A)$ is not maximal, because $$ \dim(A_{\mathfrak p}) = \mathrm{ht}(\mathfrak p) < \dim A = \mathrm{tr.deg}(K/k) $$ (that last equality is using the Theorem (I,1.8A) that Hartshorne quoted directly after) and $K = Q(A) = Q(A_{\mathfrak p})$. But we still have $\dim B \le \mathrm{tr.deg}(K/k)$.

A mistake in a book, I could live with it and try to find the proof. But here's where I get confused. Using that inequality and the rest of the proof, we get $$ \dim_k \Omega_{B/k} \otimes_B k = \dim_k \mathfrak m/\mathfrak m^2 = \dim B \le \mathrm{tr.deg}(K/k) = \dim_K \Omega_{K/k} = \dim_K \Omega_{B/k} \otimes_B K, $$ from which it follows that $\dim_k \Omega_{B/k}\otimes_B k= \dim_K \Omega_{B/k} \otimes_B K$ by the Lemma, and in particular that $\dim B = \mathrm{tr.deg}(K/k)$. Which is now confusing since I know this to be generally false.

Question : Where is the flaw in this argument?

$\endgroup$
4
$\begingroup$

The proof goes fine until they claim that $\dim B = \mathrm{tr.deg}(K/k)$ ; this equality is false in general since it doesn't hold when $B = A_{\mathfrak p}$ where $\mathfrak p \in \mathrm{Spec}(A)$ is not maximal, because $\dim(A_{\mathfrak p}) = \mathrm{ht}(\mathfrak p) < \dim A = \mathrm{tr.deg}(K/k)$ (that last equality is using the Theorem (I,1.8A) that Hartshorne quoted directly after).

This isn't a mistake; remember, we are assuming that $k$ is the residue field of $B$. If $B=A_{\mathfrak p}$ and $\mathfrak{p}$ were not maximal, then the residue field of $B$ would be larger than the field $k$ over which $A$ is finitely generated. Indeed, if $B=A_{\mathfrak p}$ for some prime $\mathfrak{p}$ then we have natural inclusions $k\subseteq A/\mathfrak{p}\subseteq B/\mathfrak{m}=k$, so we must have $A/\mathfrak{p}=k$ and $\mathfrak{p}$ is maximal.

$\endgroup$
1
  • $\begingroup$ Aaaaaaaaaaaaaaaah. So I was overlooking an assumption! In some sense "$B$ is the local ring of a closed point". Thanks for pointing it out! $\endgroup$ Nov 10 '15 at 6:11

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.