0

Let $A$ be a regular local ring, $k$ its residue field. Assume $k$ is perfect and $A$ is the localization of finitely generated algebra. Then $\Omega_{A/k}\otimes_A K \cong \Omega_{K/k}$.

I want to show this to be true. I can see how the result would lead straight from the existence of a split exact sequence $0\to A\to K\to k\to 0$. But I can't see how this would be exact.

coolpenguin
  • 163

1 Answers1

1

This isjust the consequence of a more general statement.

For a multiplicatively closed subset $S$ of $A$, $$ S^{-1}\Omega_A/k\simeq \Omega_{S^{-1}A/k}.$$

See Hartshorne, Chapter II, Proposition 8.2A.

Now for $S=A-0$, $\Omega_A/k\otimes _A K =S^{-1}\Omega_A/k$.

Evans Gambit
  • 2,043
  • I'm sorry but I still don't see it! You are connecting two different parts of the proposition. You still need to have $K=A\otimes_k k$, to get $\Omega_{K/k}=\Omega_{A/k}\otimes_A K$, correct? I don't see how $K=A\otimes_k k$ is true – coolpenguin Sep 21 '21 at 12:03
  • I am using that $S^{-1}A=K$ for $S=A-{0}$. If you recall the definiton of the faction field $K$ of $A$, it is exactly the localization of $A$ along the multiplicatively closed set $S=A-0$. – Evans Gambit Sep 21 '21 at 12:32
  • Yes, so $S^{-1}A=K$ implies $S^{-1}\Omega_{A/k}=\Omega_{S^{-1}A/k}$. How does this lead to $S^{-1}\Omega_{A/k}=\Omega_{A/k}\otimes_A K$? – coolpenguin Sep 21 '21 at 13:08
  • 1
    That is again a general statement: for an $A$-module $M$, $S^{-1}M\simeq S^{-1}A\otimes_A M$. – Evans Gambit Sep 21 '21 at 13:16
  • 1
    Ah yes, thank you! – coolpenguin Sep 21 '21 at 13:29