# Hartshorne Deformation theory Exercise7.1

This is a exercise in "Deformation Theory [Hartshorne]".

Let C be a local Artin ring with residue field k.

Let X be a scheme flat over C , and let$X_0=X\times _Ck$.

If F is a coherent sheaf on X which is flat over C,

and $F_0=F⊗_{O_X}O_{X_0}$ is locally free on $X_0$,show that F is locally free on X.

what 'local Artin ring' implies in this question?

It suffices to show that if $X$ is affine and $F_0$ is free, then $F$ is free. Since $F_0$ is free over $X_0$, we can find an isomorphism $\mathscr{O}_{X_0}^r\rightarrow F_0$.

Arbitrarily lift this to a map $\mathscr{O}_X^r\rightarrow F$. We claim that this lift is an isomorphism. Let $K$ be the kernel and $M$ be the cokernel.

First, we see that, since $\mathscr{O}_X^r\rightarrow F$ is a surjection after applying $\cdot\otimes_{\mathscr{O}_X}\mathscr{O}_{X_0}$, $M\otimes_{\mathscr{O}_X}\mathscr{O}_{X_0}=0\Leftrightarrow M=mM$, where $m$ is the maximal ideal of $C$. Since $C$ is artinian, $m^N=0$ for $N>>0$, so $M=0$.

Now, we take $0\rightarrow K\rightarrow \mathscr{O}_X^r\rightarrow F\rightarrow0$ and tensor to get $0\rightarrow K\otimes_{\mathscr{O}_X}\mathscr{O}_{X_0}\rightarrow \mathscr{O}_{X_0}^r\rightarrow F_0\rightarrow 0$, which is left exact since $F$ is flat over $C$.

Therefore, we again see that $K\otimes_C k=0$, which means $K=mK$, so $K=0$ from the same argument as above.

(So we used $C$ is artinian to apply a version of Nakayama's lemma in the case where our coherent sheaves over $X$ aren't finitely generated over $C$.)

• (I copied the same trick Daniele A gave when I asked math.stackexchange.com/questions/1698112/…, if you want something clearer) – DCT Apr 17 '16 at 13:36
• thank you for the really nice proof! But I still not understand why the lift of the first isomorphsm exists? – 8Frog Apr 17 '16 at 20:31
• Okay, I made a small edit to the beginning of the proof, so we only need to consider the affine case. Then, the map $\mathscr{O}_{X_0}^r\rightarrow F_0$ is given by $r$ elements $f_1,\ldots,f_r$ of $F_0$. Since $F\rightarrow F_0$ is surjective, we can lift this map to $\mathscr{O}^r\rightarrow F$ by lifting those elements $\tilde{f}_1,\ldots,\tilde{f}_r$ to $F$ such that $\tilde{f}_i$ maps to $f_i$ under the surjection. – DCT Apr 18 '16 at 4:48
• Before the edit, I think I tried to claim that we could lift the map $\mathscr{O}_{X_0}^r\rightarrow F_0$ to $\mathscr{O}_X^r\rightarrow F$ in general (even if $X$ is not affine), and I doubt this is true. We have the long exact sequence $H^{0}(F)\rightarrow H^0(F_0)\rightarrow H^1(mF)$, where $m$ is the maximal ideal of $C$, and the fact that $H^1$ can be nontrivial is evidence to me that we shouldn't expect this to be true. (If this wasn't what you were worried about then, don't worry about this.) – DCT Apr 18 '16 at 4:53