Proposition 4.3.9 in Liu: flatness by domination

Question

Proposition 4.3.9 in Liu says: Let $Y$ be a Dedeking scheme. Let $f:X\to Y$ be a morphism with $X$ reduced. Then $f$ is flat if and only if every irreducible component of $X$ dominates $Y$.

I don't understand details in the proof: the proof is going so

Suppose every irreducible component of $X$ dominates $Y$. Let $x\in X$ and $y=f(x)$

First the case $y$ generic point of $Y$ we deduced $\mathcal{O}_{X,x}$ flat over $\mathcal{O}_{Y,y}$: ok

Then suppose $y$ closed and $\pi$ uniformising parameter for $\mathcal{O}_{Y,y}$ (Discret Valuation Ring because $Y$ Dedeking).

Then $\pi$ does not belong to any minimal prime ideal of $\mathcal{O}_{X,x}$, that is with notation $f^\sharp_x:\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$, $f^\sharp_x(\pi)$ does not belong to any prime ideal of $\mathcal{O}_{X,x}$: I don't understand why (question °1)

As $X$ reduced then $\pi$ is not a zero divisor, then $\mathcal{O}_{X,x}$ flat over $\mathcal{O}_{Y,y}$: ok

Conclusion: ok (but I don't see the necessity of the generic point case... maybe just a remark)

The converse: one know from a preceeding lemma that if $f:X\to Y$ is flat, $Y$ irreducible and $X$ has only a finite number of irreducible components, then every one of them dominates $Y$.

Question n°2: But I don't see how we can use this lemma because we don't have the finitness of number of irreducible component for $X$ which is only reduced.

Answer 1

For your second question, this is mentioned in the second errata to the paperback edition of the book: http://www.math.u-bordeaux1.fr/~qliu/Book/errata-third-b1.pdf. One does need to assume that the number of irreducible components of $X$ is finite.

I believe I have an argument for your first question. If $V$ is an affine open of $Y$ containing $y$, then the morphism $f^{-1}(V)\rightarrow V$ still has the property that the irreducible components of $f^{-1}(V)$ dominate $V$, so for the purposes of proving flatness, one can assume that $Y=V$ is affine, $Y=\mathrm{Spec}(A)$. Similarly one can assume that $\pi$ lifts to a global section of $Y$ (which I will continue to denote by $\pi$ since $A$ injects into its localizations). Suppose that $f^\sharp(\pi)$ lies in a minimal prime $\mathfrak{p}$ of $\mathscr{O}_{X,x}$, and let $U=\mathrm{Spec}(B)$ be an affine open around $x$ in $X$. The minimal prime $\mathfrak{p}$ of $\mathscr{O}_{X,x}$ corresponds to a minimal prime $\mathfrak{p}^\prime$ of $B$, and we have $f^\sharp(\pi)\in\mathfrak{p}^\prime$. Let $x^\prime$ be the point of $X$ corresponding to $\mathfrak{p}^\prime$, which is a generic point of $X$. From $f^\sharp(\pi)\in\mathfrak{p}^\prime$, we have $f(x^\prime)\in V(\pi A)$. But then the closure of $\{f(x^\prime)\}$ is contained in $V(\pi A)$, and this closure contains $f(Z)$, where $Z=\overline{\{x^\prime\}}$ is the irreducible component corresponding to $x^\prime$. Since $\pi$ is not nilpotent, $V(\pi A)$ is a proper closed subset of $A$. This is a contradiction.