Zhenya wrote:
Hi!
Does anybody know is this statement right or wrong?

f: Y-X - flat projective morphism
E - locally free sheaf on Y

Then f_* (E) is locally free.

Zhenya


If I am not mistaken in the following conclusions, the statement is
wrong. To find a counterexample I refer to the article

D. Eisenbud, F.-O. Schreyer, Relative Beilinson Monad and Direct Image
for Families of Coherent Sheaves,
arXiv:math.AG/0506391 v2 27 Sep 2006

The articles cites the following lemma, which is also in
Hartshorne(III,Lemma 12.3):

If f:X-Y, Y=spec A, is a projective morphism and F a Y-flat sheaf on X,
then there exists a complex AA^*

AA^*: 0 - A^h0 - A^h1 - A^h2 - ... - A^hn - 0

whose cohomologies are h^i(AA^*) = R^i f_* F(Y)

The abovementioned article gives a kind of converse of this lemma as
Theorem 0.4:

Let A be a noetherian ring and let AA^* a complex like the above. Then a
vector bundle F on P^n_A exists, such that h^i(AA^*)=R^i f_*(Y)

Therefore f_*F can correspond to an arbitrary kernel ker(A^h0 - A^h1)
and need not be a flat (=locally free) sheaf on Y.

I constructed a counterexample (more or less by trial and error) with
Macaulay2 by choosing A=k[a,b,c,d] and S=A[x,y], therefore X=Proj
S=P^1_A, and a certain matrix m : S^5(-1) - S^7. So there is a sequence

S^5(-1) - S^7 - M' - 0

Let M be M' ** S^(-2)

Let I3 the ideal in k[a,b,c,d], where there are points x,y, so that
the matrix m doesn't have the full rank 5, it can be computed by taking
the 5x5 minors of m in k[a,b,c,d,x,y] and eliminating x, setting y=1
intersect eliminate y, setting x=1. Above Y-V(I3) F=M^\tilde is a vector
bundle.

Using the command directImageComplex(M) one computes a complex of A
modules of the abovementioned kind, whose cohomologies are the (higher)
direct image sheaves of M^\tilde. In the case I considered, it took the form

0 - A^6 - A^3 - 0

with 0 - K - A^6 - A^3 - 0 exact and

phi
0 - A^1 ----- A^4 - K - 0

exact (and K=f_*(M^\tilde)(Y))

Let I4 be the ideal of k[a,b,c,d], where the map A^1 - A^4 vanishes. It
can be checked, that rad(I4) \not\supseteq rad(I3), so there are points
P in spec A, above which F=M^\tilde is a vector-bundle, but on tensoring
with k(P) the map phi ** k(P) vanishes, so Tor_1^(A_P)(k(p),K_P) 0,
so K_P = (f_* M^\tilde)_P is not projective.

If there is interest, I can post the script doing the calculation for
further experimenting.


Greetings

Jurgen


--
Jurgen Bohm www.aviduratas.de
"At a time when so many scholars in the world are calculating, is it not
desirable that some, who can, dream ?" R. Thom