The Stacks project

Proof. Choose a diagram as in Lemma 37.18.2. Set $\mathcal{F}' = (g')^*\mathcal{F}$. Divisors, Lemma 31.7.3 implies that $y' \in \text{Ass}_{X'/S'}(\mathcal{F}')$. By our choice of $y'$ we also see that $\dim (Y'_{t'}) = \dim (Y_ t)$, see for example Algebra, Lemma 10.116.7. By Algebra, Lemma 10.125.9 we see that $Y'_{s'}$ is equidimensional of dimension equal to $\dim (Y_ t)$. Since $\mathcal{F}$ is flat at $x$ over $S$ we see that $\mathcal{F}'$ is flat at $x'$ over $S'$, see Morphisms, Lemma 29.25.7.

Suppose that we can show $x' \in \text{Ass}_{X'/S}(\mathcal{F}')$. Then Divisors, Lemma 31.7.3 implies that $x \in \text{Ass}_{X/S}(\mathcal{F})$ and that the irreducible component $C'$ of $Y'_{s'}$ containing $x'$ is an irreducible component of $C \times _ s s'$ where $C \subset Y_ s$ is the irreducible component containing $x$. Whence $\dim (C) = \dim (C') = \dim (Y_ t)$ (see above) and the proof is complete. This reduces us to the case discussed in the next paragraph.

Assume $S = \mathop{\mathrm{Spec}}(A)$ where $A$ is a valuation ring and $t$ and $s$ are the generic and closed points of $S$. We will assume $x \not\in \text{Ass}_{X/S}(\mathcal{F})$ in order to get a contradiction. In other words, we assume $x \not\in \text{Ass}_{X_ s}(\mathcal{F}_ s)$ where $\mathcal{F}_ s$ is the pullback of $\mathcal{F}$ to $X_ s$. Consider the ring map

and the module $N = \mathcal{F}_ x$ over $B = \mathcal{O}_{X, x}$. Then $B/\mathfrak m_ A B = \mathcal{O}_{X_ s, x}$ and $N/\mathfrak m_ A N$ is the stalk of $\mathcal{F}_ s$ at the point $x$. Denote $\mathfrak q \subset B$ the prime ideal corresponding to the point $y$, see Schemes, Lemma 26.13.2. Since $x$ is a generic point of $Y_ s$ we see that the radical of $\mathfrak q + \mathfrak m_ A B$ is $\mathfrak m_ B$. Then $\text{Ass}_{B/\mathfrak m_ A B}(N/\mathfrak m_ A N)$ is a finite set of prime ideals (Algebra, Lemma 10.63.5) which doesn't contain the maximal ideal of $B/\mathfrak m_ A B$ since $x \not\in \text{Ass}_{X/S}(\mathcal{F})$. Thus the image of of $\mathfrak q$ in $B/\mathfrak m_ A B$ is not contained in any of those prime ideals. Hence by prime avoidance (Algebra, Lemma 10.15.2) we can find an element $g \in \mathfrak q$ whose image in $B/\mathfrak m_ A B$ is a nonzerodivisor on $N/\mathfrak m_ A N$ (this uses the description of zerodivisors in Algebra, Lemma 10.63.9). Since $N = \mathcal{F}_ x$ is $A$-flat by Lemma 37.18.1 we see that

is injective. In particular, if $K = \text{Frac}(A)$ is the fraction field of $A$, then we see that

is injective. Observe that $\mathfrak q$ corresponds to a prime ideal of $B \otimes _ A K$. Denote $\mathcal{F}_ t$ the restriction of $\mathcal{F}$ to the generic fibre $X_ t$. We have $(B \otimes _ A K)_{\mathfrak q} = \mathcal{O}_{X_ t, y}$ and $(N \otimes _ A K)_\mathfrak q$ is the stalk at $y$ of $\mathcal{F}_ t$. Hence we find that $g \in \mathfrak m_ y \subset \mathcal{O}_{X_ t, y}$ is a nonzerodivisor on the stalk $(\mathcal{F}_ t)_ y$ which contradicts our assumption that $y \in \text{Ass}_{X/S}(\mathcal{F})$. $\square$