Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


Question 1: let $f:X\rightarrow S$ be a proper and surjective morphism of complex reduced spaces with $X$ pure dimensional. Let $F$ be a $S$-flat coherent sheaf on $X$. Is it true that the two condition a) and b) are equivalent:

a) There $s$ in $S$ for which $F_{s}$ is ${\cal O}_{X_{s}$-torsion free.

b) $F$ is ${\cal O}_{X}$-torsion free.

Remark: Of course, we can suppose that $Supp(F)=X$ and then $f$ is universally open.

Question 2: Suppose that $f$ is open of fiber dimension constant $n$ and let $U$ be an open polydisk in $C^{n}$, $g:X\rightarrow S\times U$ a relative local parametrization of $f$ which is finite, surjective and universally open morphism. Consider the relative canonical sheaf $\omega^{n}_{X/S}$ and the isomorphism $g_{*}\omega^{n}_{X/S}\simeq {\cal H}om(g_{*}({\cal O}_{X}), {\cal O}_{S\times U})$

then it is true that (i) $S$ reduced imply $\omega^{n}_{X/S}$ is torsion free on $X$.

(ii) $S$ normal imply $\omega^{n}_{X/S}$ is of depth > 1 on $X$.

Remark: (i) and (ii) are true for $S$-torsion freeness and $S$-depth.

Thank you very much.

share|cite|improve this question
For the first question the answer is NO even for $S$ being a point! – Sasha Oct 8 '10 at 10:59
Dear Sasha, can you give me an example of torsion free sheaf on complex reduced pure dimensional space which are no flat ? – kaddar Oct 8 '10 at 16:41
Flat over what? – Sasha Oct 9 '10 at 19:35
We have a partial answer to the first question: $f:X\rightarrow S$ proper and surjective morphisme of reduced complex with $X$ pure dimensional and $S$ irreducible. If $F$ is an $S$-flat coherent sheaf, we have the equivalence: a) There existe some point $s$ in $S$, for which $F_{s}$ is a torsion free $O_{X_{s}}$-module. b) $F$ is a torsion free $O_{X}$-coherent sheaf. The proof can be donne by intensive use of Siu-Trautman result on degeneracy set of coherent sheaf ... – kaddar Oct 11 '10 at 7:00

Your Answer


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

Browse other questions tagged or ask your own question.