# Sheaf without embedded associated points

Hi. I want to know if the following is true:

Let $f:X\rightarrow S$ be a flat morphism of complex spaces without embedded components. Let $F$ be a $O_{X}$-coherent $S$-flat sheaf. Then the following are equivalent:

(1) There is some point $s$ in $S$ s.t $F_{s}$ (the usual schematic or analytic restriction on the fiber) is $S_{1}$.

(2) $F$ is $S_{1}$.

Thank you.

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This is not true.

## (2) $\not\Rightarrow$ (1)

Let $f=\mathrm{id}_X$, then $F_s$ is supported at a point, so no matter what, $F_s$ is not $S_1$. So, you just have to choose an $X$ and and $F$ that's $S_1$.

## (1) $\not\Rightarrow$ (2)

Let $S=\mathrm{Spec} k[\varepsilon]/\varepsilon^2$ and $X=S\times T$ for some reduced $T$. Then $\mathscr O_X$ is not $S_1$ but $(\mathscr O_X)_{(\varepsilon)}\simeq \mathscr O_T$ is $S_1$.

In general, you should expect that restrictions to fibers of morphisms are $S_n$ for a smaller $n$ than they are to start with. More precisely the following is true:

Let $f:X\to S$ be a flat morphism such that $S$ is an integral regular $1$-dimensional scheme and let $\mathscr F$ be a coherent sheaf on $X$. Then $\mathscr F$ is $S_n$ if and only if $\mathscr F_s$ is $S_{n-1}$ for every $s\in S$ closed point.

Remark: This is true for more general morphisms, but you can probably figure that out if you need to.

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Thank you, Sandor. – kaddar Oct 24 '11 at 13:48