Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.