The Stacks project

Proof. Let $\mathop{\mathrm{Spec}}(A) \subset U$ be an affine neighbourhood of $u$ such that $u$ corresponds to a closed point of $U$, see Morphisms, Lemma 29.16.4. Let $\mathop{\mathrm{Spec}}(B) \subset R$ be an affine neighbourhood of $e(u)$ which maps via $j$ into the open $\mathop{\mathrm{Spec}}(A) \times _ S \mathop{\mathrm{Spec}}(A) \subset U \times _ S U$. Let $\mathfrak m \subset A$ be the maximal ideal corresponding to $u$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $e(u)$. Pictures:

Note that the two induced maps $s, t : \kappa (\mathfrak m) \to \kappa (\mathfrak q)$ are equal and isomorphisms as $s \circ e = t \circ e = \text{id}_ U$. In particular we see that $\mathfrak q$ is a maximal ideal as well. The ring maps $s, t : A \to B$ are of finite presentation and flat. By assumption the ring

is Cohen-Macaulay of dimension $d_2$. The equality of dimension holds by Morphisms, Lemma 29.28.1.

Let $R''$ be the restriction of $R$ to $u = \mathop{\mathrm{Spec}}(\kappa (u))$ via the morphism $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$. As $u \to U$ is locally of finite type, we see that $(\mathop{\mathrm{Spec}}(\kappa (u)), R'', s'', t'', c'')$ is a groupoid scheme with $s'', t''$ locally of finite type, see Lemma 40.9.1. By Lemma 40.10.10 this implies that $\dim (G'') = \dim (R'')$. We also have $\dim (R'') = \dim _{e''}(R'') = \dim (\mathcal{O}_{R'', e''})$, see Lemma 40.10.9. By Groupoids, Lemma 39.18.4 we have $G'' = G_ u$. Hence we conclude that $\dim (\mathcal{O}_{R'', e''}) = d_1$.

As a scheme $R''$ is

Hence an affine open neighbourhood of $e''$ is the spectrum of the ring

We conclude that

and so now we know that this ring has dimension $d_1$.

We claim this implies we can find an element $f \in \mathfrak m$ such that

Namely, suppose $\mathfrak n_ j \supset s(\mathfrak m)B_{\mathfrak q}$, $j = 1, \ldots , m$ correspond to the minimal primes of the local ring $B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q}$. There are finitely many as this ring is Noetherian (since it is essentially of finite type over a field – but also because a Cohen-Macaulay ring is Noetherian). By the Cohen-Macaulay condition we have $\dim (B_{\mathfrak q}/\mathfrak n_ j) = d_2$, for example by Algebra, Lemma 10.104.4. Note that $\dim (B_{\mathfrak q}/(\mathfrak n_ j + t(\mathfrak m)B_{\mathfrak q})) \leq d_1$ as it is a quotient of the ring $\mathcal{O}_{R'', e''} = B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q} + t(\mathfrak m)B_{\mathfrak q}$ which has dimension $d_1$. As $d_1 < d_2$ this implies that $\mathfrak m \not\subset t^{-1}(\mathfrak n_ i)$. By prime avoidance, see Algebra, Lemma 10.15.2, we can find $f \in \mathfrak m$ with $t(f) \not\in \mathfrak n_ j$ for $j = 1, \ldots , m$. For this choice of $f$ we have the displayed inequality above, see Algebra, Lemma 10.60.13.

Set $A' = A/fA$ and $U' = \mathop{\mathrm{Spec}}(A')$. Then it is clear that $U' \to U$ is an immersion, locally of finite presentation and that $u \in U'$. Thus (1), (2) and (3) of the lemma hold. The morphism

factors through $\mathop{\mathrm{Spec}}(A)$ and corresponds to the ring map

Now, we see $t(f)$ is not a zerodivisor on $B_{\mathfrak q}/s(\mathfrak m)B_{\mathfrak q}$ as this is a Cohen-Macaulay ring of positive dimension and $f$ is not contained in any minimal prime, see for example Algebra, Lemma 10.104.2. Hence by Algebra, Lemma 10.128.5 we conclude that $s : A_{\mathfrak m} \to B_{\mathfrak q}/t(f)B_{\mathfrak q}$ is flat with fibre ring $B_{\mathfrak q}/(s(\mathfrak m)B_{\mathfrak q} + t(f)B_{\mathfrak q})$ which is Cohen-Macaulay by Algebra, Lemma 10.104.2 again. This implies part (4) of the lemma. To see part (5) note that by Diagram (40.9.0.1) the fibre $F'_ u$ is equal to the fibre of $h$ over $u$. Hence $\dim _{e'(u)}(F'_ u) = \dim (B_{\mathfrak q}/(s(\mathfrak m)B_{\mathfrak q} + t(f)B_{\mathfrak q}))$ by Morphisms, Lemma 29.28.1 and the dimension of this ring is $d_2 - 1$ by Algebra, Lemma 10.104.2 once more. This proves the final assertion of the lemma and we win. $\square$