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

Then f_* (E) is locally free.

Zhenya  in litteris
<e2557eb2-74a8-42d2-a387-ebb22e188...@26g2000hsk.googlegroups.com>
scripsit:

Then f_*(E) is flat because E is flat and f is (see Hartshorne, *Algebraic
Geometry*, proposition III.9.2(c)).

But f_*(E) is coherent because E is and f is projective
(op. cit. III.8.8(b)).

Now this implies that f_*(E) is locally free (op. cit. III.9.2(e)).

On 16.05.2008 11:15, David Madore wrote:

Note that f_*(E(n)) is locally free for n>>0 where E(n) is the n-th
twist of E with respect to a f-ample line bundle L on Y if X is integral
(cf. http://groups.google.com/group/sci.math/msg/9c4a3dfe05992be0, too).
This appears to be the best result which is possible in the light of
Hartshorne, Thm. III.9.9 and the equivalent assertions in its proof.

... but I am keen on seeing some elaboration on (*). ;)

Jannick Asmus  in litteris <g0js9i$g4...@dizzy.math.ohio-state.edu>
scripsit:

In truth, I find it very annoying that this question is not mentioned
in standard texts on algebraic geometry and commutative algebra.  If
you have the affine case and quasi-coherent sheaves in mind, the
direct image of a quasi-coherent sheaf by a morphism Spec(B)->Spec(A)
corresponds to viewing a B-module as the induced A-module; so it's
pretty natural to think that if a morphism is flat then direct image
by it will take flat sheaves to flat sheaves (see below).  A decent
book on the subject should contain some explanation as to the relation
between "f_*(E) is flat" and "E is f-flat": either my expectations are
too high or there are no decent textbooks on the subject. :-)

However, I can see more clearly the reason why finding a
counterexample is not obvious.  Let f:Y->X be a morphism of finite
type between noetherian schemes, and assume X = Spec A is affine
(which we can do, because our problem is local on the target).
Consider a finite covering of Y by affine subschemes V_i = Spec(B_i),
each B_i being an A-algebra (of finite type).  If E is a
quasi-coherent sheaf on Y, the restriction E|V_i of E to each V_i is
some (M_i)~ with M_i a B_i-module; the direct image f_*(E|V_i) is
([A]M_i)~ where [A]M_i is M_i seen as an A-module; if we assume that E
and f are flat (or more generally that E is flat over X), then the
[A]M_i are flat, so the f_*(E|V_i) are (this is essentially what
Hartshorne says around II.5 and III.9).  Now this does not imply that
f_*(E) is flat as I was too quick to believe, but it is true that
f_*(E) injects in the direct sum sum of the f_*(E|V_i) which are flat
(so f_*(E) is, e.g., torsion free).  Now it is not true in general
that a subsheaf (submodule) of a flat sheaf (module) is flat, but it
is true if, say, X is a smooth curve.  So my attempts to find a
counterexample with X = A^1 were doomed.

I need to think further about this...

On 17.05.2008 16:15, David Madore wrote:

The direct image is thoroughly examined in terms of the cohomology
groups H^i(X,F) or higher direct images R^i(f_*)(F). This might be a bit
disguised, but basically direct images are considered by cohomological
tools noted before.

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

==== Dictionary of linear algebra [with O(d) a line bundle on P1]:

 a pencil is a pair of MxN matrices (A,B), or a family xA + yB;

 isomorphism of pencils is (A,B) |--> (QAR,QBR) with invertible Q,R;

 a pencil is "a Jordan block" if isomorphic to one with A or B equal to
   id, and the other one a Jordan block;

 a pencil is "decreasing Kronecker" if isomorphic to (d/dx,d/dy) acting
   Pol_n(x,y) --> Pol_{n-1}(x,y); "increasing Kronecker" is transposed;

 over algebraically closed fields, any pencil decomposes to a direct sum
   of Jordan blocks, and increasing or decreasing Kronecker blocks; the
   blocks appearing in the decomposition are uniquely defined;

 Ker and Coker(xA + yB) induce sheaves on P1 (consisting of ratios x:y);
   Jordan blocks correspond to skyscrapers in Coker, increasing
   Kronecker blocks to O(n) in Coker, decreasing to O(-n) in Ker;

 a family of pencils has pencils with Jordan blocks on a closed subset;

 a generic pencil with M=N breaks into 1x1 Jordan blocks; with M>N into
   M-N decreasing Kronecker blocks; n's of these blocks differ at most
   by one (which determines the list uniquely);

 a generic family of pencils has principal degeneration on:

   M=N: 2x2 Jordan block appears on hypersurface Disc(det(xA+yB))=0;

   M=N+1: 1x1 Jordan block appears on hypersurface HyperDet(A,B)=0
     for N+1 x N x 2 flavor of GKZ hyperdeterminant;

   M>N: 1x1 Jordan block appears in codimension M-N;

   M=N+2, M=2m+1, m>1: a sum of two Kronecker blocks with n=m,m+1
     degenerates into one with n=m-1,m+2 on a subfamily of codimension 3;

   M=N+2, M=2m, m>1: a sum of two Kronecker blocks with n=m degenerates
     into a sum with n=m-1,m+1 on a subfamily of codimension 2;

      [I vaguely rememeber seeing calculations of stabilizers published
       which would imply the last two subcases, but do not recollect
       where; I needed to calculate stabs explicitly...  When m=1,
       codim decreases by 1.

       The M=N+1 case was observed by I.M.Gelfand/Valery???]

 To translate to the cases N>M, transpose.

=============================== The last two sub-statements suggest:

Conjecture: a generic family of rk=2 vector bundles on P1 of degree
  2m+1 degenerates into O(m-1) (+) O(m+2) on codimension 3; of degree
  2m degenerates into O(m-1) (+) O(m+1) on codimension 2.

     [These families are generically O(m) (+) O(m+1) and O(m) (+) O(m).]

================================ Comparing with your example:

Conjecture: consider a family of rk=2 vector bundles on P1 parameterized
  by S which is isomorphic to O(n) (+) O(n') outside of subvariety S'
  of S, and degenerates into O(N) (+) O(N') on S'; N<n,n'.  Then the
  direct image to S is not flat if codim S' >= 2 and n,n' >= -1, N < -1.

[*]   [I do not know what happens if one drops n,n' >= -1, N < -1.]

[Heuristically, I would expect that if direct image is flat, so is the
 higher direct image (to "preserve Euler characteristic"). If so, then
 since higher direct image vanishes at generic point, it vanishes
 everywhere; hence the fiber of direct image is global sections of the
 vector bundle.  But dimension of global sections jumps up on S'.]

=================================  Going back to your example:

You have N=M+2, M=5, and take Coker; generically, it would be
O(2) (+) O(3).  You twist by O(-2); generically, the result would
be O(0) (+) O(1), which in codimension 3 would degenerate into
O(-1) (+) O(2) (on a closed subset it may also acquire Jordan blocks;
one can ignore this).  This is not in context of the conjecture above,
but of [*], which for me is an unchartered territory.

Following my argument, the minimal example to check would be a pencil
with M=3, N=M+2: take a direct sum of Kronecker blocks with n=0,3, and
take a generic 2-parameter deformation (it is enough to require that
for a deformed pencil, Ker A and Ker B do not intersect).  So, I would
take Ker(xA + yB) with matrices A,B depending on a,b:

  1 0 0 0 0     0 1 0 0 0
  0 1 0 0 a     0 0 1 0 0
  0 0 1 0 b     0 0 0 1 0

- and I would twist by O(1) - while to closely follow your example,
the twist would be by O(2).  (Since I prefer Ker to Coker, the picture
is kinda dual to yours.)  So O(-2) (+) O(1) would deform into O(-1) (+) O(0).

Should not one be able to do it without computer algebra?  :-(

  Apparently, it was a kind of circular reasoning...

============================================ Calculation:

Twist by O(1): to calculate ( Ker(xA+yB) )(1), one needs to find
linear functions V(x,y) = x v1 + y v2 of (x,y) such that (xA + yB)V = 0.
When (a,b) != (0,0), one gets that V is proportional to V0 with

   v1 = -b^2 e4 + a (e5 - a e2 - b e3)
   v2 =  a^2 e1 + b e5.

So the direct image is {P(a,b) (x v1 + y v2)} with a regular scalar
function P; so it is iso to the structure sheaf.

Twist by O(2): to calculate ( Ker(xA+yB) )(2), one needs to find
quadratic functions Q(x,y) of (x,y) such that (xA + yB)V = 0.  When
(a,b) != (0,0), one gets x V0(x,y), y V0(x,y), and

  Q0 = x^2 (e5 - a e2 - b e3) + xy (a e1 + b e2) + y^2 b e2;

all 3 quadratic functions are linearly independent (at least when a,b
!= 0), so the direct image is {x P1(a,b) V0 + y P2(a,b) V0 + P3(a,b) Q0}.

So AFAICS, the direct image is 3 copies of the structure sheaf...