The Stacks project

Proof. Choose $b_1, \ldots , b_ n \in B$ generating $B$ as an $A$-module. For each $i$ choose a monic $P_ i \in A[x]$ such that $P_ i(b_ i) = 0$ in $B$, see Algebra, Lemma 10.36.3. Thus $B$ is a quotient of the finite free $A$-algebra $B' = A[x_1, \ldots , x_ n]/(P_1(x_1), \ldots , P_ n(x_ n))$. Let $J \subset B'$ be the kernel of the surjection $B' \to B$. Then $J =(f_1, \ldots , f_ m)$ is finitely generated as $B$ is a finitely generated $A$-algebra, see Algebra, Lemma 10.6.2. Choose an $A$-basis $b'_1, \ldots , b'_ N$ of $B'$. Consider the algebra

where $I$ is the ideal generated by the coefficients in $A[z_1, \ldots , z_ n, y_1, \ldots , y_ m]$ of the basis elements $b'_1, \ldots , b'_ N$ of the expression

in $B'[z_1, \ldots , z_ N, y_1, \ldots , y_ m]$. By construction the element $\sum z_ j b'_ j$ maps to an idempotent $e_{univ}$ in the algebra $B \otimes _ A A_{univ}$. Moreover, if $e \in B \otimes _ A A'$ is an idempotent, then we can lift $e$ to an element of the form $\sum b'_ j \otimes a'_ j$ in $B' \otimes _ A A'$ and we can find $a''_ k \in A'$ such that

is zero in $B' \otimes _ A A'$. Hence we get an $A$-algebra map $A_{univ} \to A$ sending $z_ j$ to $a'_ j$ and $y_ k$ to $a''_ k$ mapping $e_{univ}$ to $e$. This finishes the proof. $\square$