# Subtleties in the construction of base change morphisms

Given a flat and projective morphism of noetherian schemes, $f: X \rightarrow Y$ and $F$, $G$ two coherent $O_X$-modules, flat over $Y$. Furthermore given a morphism $u: Y' \rightarrow Y$ of noetherian schemes and a quasi coherent $O_{Y'}$-module $\mathcal{M}$. Then we have $X':=X\times_YY'$ with $p_i$ the i-th projection.

1) Take a locally free resolution $P_{\cdot} \rightarrow F \rightarrow 0$. Now the article says: "Since $f$ and $F$ are flat over $Y$, $p_1^{\*}P_{\cdot}$ is a resolution of $p_1^{\*}F$". Don't we need $u$ to be flat for this? Why is this implied by the flatness of $f$ and $F$?

2) Assume $Y$ and $Y'$ are affine, say $Y=Spec(A)$ and $Y'=Spec(A')$. Given an $A'$-module $M$. For a descending induction, the article says there is an exact sequence:

$0\rightarrow p_1^{\*}G'(n)\otimes_{A'} M \rightarrow \bigoplus\limits_{i=1}^r O_{X'}(-k+n)\otimes_{A'} M \rightarrow p_1^{\*}G(n)\otimes_{A'} M\rightarrow 0$.

I see that we have $0\rightarrow G'(n) \rightarrow \bigoplus\limits_{i=1}^r O_{X}(-k+n)\rightarrow G(n)\rightarrow 0$ on X, now if $p_1$ would be flat (see (1)), we'd have $0\rightarrow p_1^{\*}G'(n) \rightarrow \bigoplus\limits_{i=1}^r O_{X'}(-k+n) \rightarrow p_1^{\*}G(n)\rightarrow 0$ on $X'$, but why is $\otimes_{A'} M$ exact in this case? Or is there any other way to construct such a sequence?

Background: I'm reading the article "Universal families of extensions" by Herbert Lange (J. Algebra 83 (1983), 101–112), where he constructs base change homomorphisms for relative $Ext$-sheaves on schemes. I get the basic idea of the construction, but now i'm trying to understand the details, and these two questions stayed open.

-

## 1 Answer

It is easy, $p_1^*G = G\otimes_A A'$ is flat over $A'$ since $G$ is flat over $A$. Hence $Tor_1^{A'}(p_1^*G,M) = 0$, hence your sequence is exact after tensoring with $M$. The same argument works for the first question as well.

-
Ah, of course. Thanks for your help. Again :-). These little details start to annoy me, since i should see this myself. – TonyS Nov 17 '10 at 21:33