Let $X\xrightarrow{f} Y\xrightarrow{g} Z$ be a stein factorisation.
It is known that the fibres of $f$ is connected.
Are the fibres of $f$ reduced?
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4$\begingroup$ Some fibers may be nonreduced. Let $Z$ equal $\mathbb{A}^2_k$, and let $x_0$ be a smooth $k$-point. Form the blowing up, $\nu_1:Z_1\to Z,$ at $0$ with exceptional divisor $E_1$. For a smooth point $x_1\in E_1$, denote by $\nu_2:Z_2\to Z_1$ the blowing up at $x_1$ with exceptional locus $E_2$. For the strict transform $\widetilde{E}_1\subset Z_2$, there is a unique intersection point $x_2\in \widetilde{E}_1\cap E_2$. Denote the blowing up at $x_2$ by $\nu_3:X\to Z_2$. The composite $f=\nu_1\circ \nu_2\circ \nu_3$ has connected fibers, but the fiber over $x_0$ is nonreduced. $\endgroup$– Jason StarrCommented Jan 19, 2018 at 12:36
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$\begingroup$ Typo correction: "... at $0$ with exceptional divisor $E_1$." --> "... at $x_0$ with exceptional divisor $E_1$." $\endgroup$– Jason StarrCommented Jan 19, 2018 at 14:16
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3$\begingroup$ Another (perhaps more elementary) example: take $Z=\mathrm{Spec}\,k[t]$, and let $X\subset \mathbb{P}^2_Z$ be the relative conic defined by $x^2=ty^2$. Then $Y=Z$ but the fiber at $t=0$ is not reduced. $\endgroup$– Laurent Moret-BaillyCommented Jan 19, 2018 at 14:21
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$\begingroup$ As a matter of fact, you can even restrict the preceding example to $t=0$: then $Z=\mathrm{Spec}\,k$ and $X$ is the conic $x^2=0$ in $\mathbb{P}^2$, which is its own Stein factorization since $\mathrm{H}^0(X,\mathscr{O}_X)=k$. $\endgroup$– Laurent Moret-BaillyCommented Jan 19, 2018 at 16:20
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1$\begingroup$ @SándorKovács: If, say, $Z=\mathrm{Spec}\,k$ and $X=\mathrm{Spec}\,(k[t]/t^2)$, then $Y=X$ anf $f$ is the identity. (I now realize that my phrase "is its own Stein factorization" is ambiguous). $\endgroup$– Laurent Moret-BaillyCommented Jan 20, 2018 at 8:59
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