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

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .