Is this true?

X : noetherian scheme
F : coherent sheaf of O_X-module
U in X : affine open subset
s in F(U) : a section such that s(U) != 0
s_y != o for some y in U
z in U : a generic point of U (*)

Then, s_z != 0

I made a proof by myself, but I'm not sure of it. Since if that is
true, a statement of Mumford's redbook is wrong. So, please point out
the wrong point of my proof.

My Proof)
FACT) Closed subsets are stable under specialization.
By the FACT, y and z are in the same irreducible component of
X.
So, let Y be the same irreducible component of X containing y
and z.
Suppose s_z = 0.
Then, there exists an open set V in U containing z such that
s(V) = 0.
But, since s_y != 0 and y is in V, this is a contradiction.
Thus, s_z != 0. (END)

I don't know where I made a mistake.

And, here's an another question.
If above statement is true, how about we replace the statement (*)
by
"z is a generilization of y."? Is that still hold?

I'll be happy if you give me any comments.

In article ,
(beginner) writes:
Is this true?

X : noetherian scheme
F : coherent sheaf of O_X-module
U in X : affine open subset
s in F(U) : a section such that s(U) != 0
s_y != o for some y in U
z in U : a generic point of U (*)

Then, s_z != 0

No, it is false. Consider, for instance, the affine line X = U = Spec(F[t])
over a field F. You get open subsets as Spec(F[t][1/h(t)] for various
nonconstant polynomials. The generic point is given by the ideal {0}.

Let M be a finitely generated F[t]-module; this gives us a coherent sheaf
on X whose sections on the open set Spec(F[t][1/h(t)]) are the tensor product of
M and F[t][1/h] over F[t].

In particular, now, take M = F[t]/(t-1)F[t]. This is nonzero, and it has
nonzero stalk at the point y given by the ideal (t-1)F[t]. But it becomes
trivial once we restrict to the open subsset where t-1 is invertible.

I made a proof by myself, but I'm not sure of it. Since if that is
true, a statement of Mumford's redbook is wrong. So, please point out
the wrong point of my proof.

The problem in the proof is the assumption that the point y has to
be in each open subset V of U.


William C. Waterhouse
Penn State



My Proof)
FACT) Closed subsets are stable under specialization.
By the FACT, y and z are in the same irreducible component of
X.
So, let Y be the same irreducible component of X containing y
and z.
Suppose s_z = 0.
Then, there exists an open set V in U containing z such that
s(V) = 0.
But, since s_y != 0 and y is in V, this is a contradiction.
Thus, s_z != 0. (END)

I don't know where I made a mistake.

And, here's an another question.
If above statement is true, how about we replace the statement (*)
by
"z is a generilization of y."? Is that still hold?

I'll be happy if you give me any comments.