The Stacks project

Proof. Since $C \to A$ is integral (Lemma 39.23.4) and injective we see that $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(C)$ is surjective, see Algebra, Lemma 10.36.17. Thus $|U| \to |M|$ is surjective.

Let $k$ be an algebraically closed field and let $C \to k$ be a ring map. Since surjective morphisms are preserved under base change (Morphisms, Lemma 29.9.4) we see that $A \otimes _ C k$ is not zero. Now $k \subset A \otimes _ C k$ is a nonzero integral extension. Hence any residue field of $A \otimes _ C k$ is an algebraic extension of $k$, hence equal to $k$. Thus we see that $U(k) \to M(k)$ is surjective.

Let $a_0, a_1 : A \to k$ be two ring maps. If there exists a ring map $b : B \to k$ such that $a_0 = b \circ t^\sharp $ and $a_1 = b \circ s^\sharp $ then we see that $a_0|_ C = a_1|_ C$ by definition. Thus the map $U(k) \to M(k)$ equalizes the two maps $R(k) \to U(k)$. Conversely, suppose that $a_0|_ C = a_1|_ C$. Let us name this algebra map $c : C \to k$. Consider the diagram

If we can construct a dotted arrow making the diagram commute, then the proof of part (2) of the lemma is complete. Since $s : A \to B$ is finite there exist finitely many ring maps $b_1, \ldots , b_ n : B \to k$ such that $b_ i \circ s^\sharp = a_1$. If the dotted arrow does not exist, then we see that none of the $a'_ i = b_ i \circ t^\sharp $, $i = 1, \ldots , n$ is equal to $a_0$. Hence the maximal ideals

of $A \otimes _ C k$ are distinct from $\mathfrak m = \mathop{\mathrm{Ker}}(a_0 \otimes 1 : A \otimes _ C k \to k)$. By Algebra, Lemma 10.15.2 we would get an element $f \in A \otimes _ C k$ with $f \in \mathfrak m$, but $f \not\in \mathfrak m_ i'$ for $i = 1, \ldots , n$. Consider the norm

By Lemma 39.23.2 this lies in the invariants $C^1 \subset A \otimes _ C k$ of the base change groupoid (base change via the map $c : C \to k$). On the one hand, $a_1(g) \in k^*$ since the value of $t^\sharp (f)$ at all the points (which correspond to $b_1, \ldots , b_ n$) lying over $a_1$ is invertible (insert future reference on property determinant here). On the other hand, since $f \in \mathfrak m$, we see that $f$ is not a unit, hence $t^\sharp (f)$ is not a unit (as $t^\sharp \otimes 1$ is faithfully flat), hence its norm is not a unit (insert future reference on property determinant here). We conclude that $C^1$ contains an element which is not nilpotent and not a unit. We will now show that this leads to a contradiction. Namely, apply Lemma 39.23.5 to the map $c : C \to C' = k$, then we see that the map of $k$ into the invariants $C^1$ is injective and moreover, that for any element $x \in C^1$ there exists an integer $n > 0$ such that $x^ n \in k$. Hence every element of $C^1$ is either a unit or nilpotent.

We still have to finish the proof of (1). We already know that $|U| \to |M|$ is surjective. It is clear that $|U| \to |M|$ is $|R|$-invariant. Finally, suppose $u_0, u_1 \in U$ maps to the same point $m \in M$. Then the induced field extensions $\kappa (u_0)/\kappa (m)$ and $\kappa (u_1)/\kappa (m)$ are algebraic (as $A$ is integral over $C$ as used above). Hence if $k$ is an algebraic closure of $\kappa (m)$ then we can find $\kappa (m)$-embeddings $\overline{u}_0 : \kappa (u_0) \to k$ and $\overline{u}_1 : \kappa (u_1) \to k$. These determine $k$-valued points $\overline{u}_0, \overline{u}_1 \in U(k)$ mapping to the same point of $M(k)$. By part (2) we see that there exists a point $\overline{r} \in R(k)$ with $s(\overline{r}) = \overline{u}_0$ and $t(\overline{r}) = \overline{u}_1$. The image $r \in R$ of $\overline{r}$ is a point with $s(r) = u_0$ and $t(r) = u_1$ as desired. $\square$