6

Let $f:X \rightarrow Y $ be a morphism of locally Noetherian schemes. Let $Z$ be a closed subscheme of $X$ and suppose that there exists a point $y \in Y$ such that $Z_y=X_y$ as schemes. Show that if $Z$ is flat over $Y$ at $z \in X_y$ , then $Z$ is equal to $X$ in a open neighborhood of $z$.

The problem is from Liu's book and he tells me to use the following, but I can't seem to put it together
1. Use : Let $A \rightarrow B$ be a ring homomorphism and $J$ an ideal such that $B/J$ is flat over $A$. Then for any ideal $I$ of $A$, we have $IB \cap J = IJ$.
2. Use Nakayama's lemma.

user26857
  • 52,094
Heidar Svan
  • 411

1 Answers1

5

Take $A=O_{Y,y}$, $I$ the maximal ideal of $A$, $B=O_{X,z}$ and $J\subseteq B$ the image of the ideal defining the closed subscheme $Z$ in $X$.

The hypothesis $X_y=Z_y$ in a neighborhood of $z$ means the canonical map $$B/IB=B\otimes_A A/I\to (B/J)\otimes_A A/I=B/(IB+J)$$ is an isomorphism. The kernel in general is $(IB+J)/IB\simeq J/(IB\cap J)$. So $J=IB\cap J$. By (1), this implies $J=IJ$. Applying Nakayama to the $B$-module $J$, we get $J=0$. Hence $Z=X$ in a neighborhood of $z$.

  • Thanks! For the follow-up (4.3.8) to this exercise, do you have any hints? Should it really be an open immersion? (I thought $Z \cap f^{-1}(V)$ was closed in $f^{-1}(V)$? – Heidar Svan Mar 15 '13 at 15:17
  • If I should ask it as a separate question , please say so. I have been working for hours on it! – Heidar Svan Mar 15 '13 at 17:06
  • @HeidarSvan: $Z\cap f^{-1}(V)$ can be at the same time open and closed in $f^{-1}(V)$. As a hint for (b), a morphism flat and finite type is always an open map. If you still can't find the solution, please ask a separate question. –  Mar 15 '13 at 18:45