**Proof.**
Assume (1). Using transfinite recursion we construct for each ordinal $\alpha $ an $A$-subalgebra $B_\alpha \subset B$ as follows. Set $B_0 = A$. If $\alpha $ is a limit ordinal, then we set $B_\alpha = \mathop{\mathrm{colim}}\nolimits _{\beta < \alpha } B_\beta $. If $\alpha = \beta + 1$, then either $B_\beta = B$ in which case we set $B_\alpha = B_\beta $ or $B_\beta \not= B$, in which case we apply Lemma 29.46.5 to choose a $b_\alpha \in B$, $b_\alpha \not\in B_\beta $ with $b_\alpha ^2, b_\alpha ^3 \in B_\beta $ and we set $B_\alpha = B_\beta [b_\alpha ] \subset B$. Clearly, $B = \mathop{\mathrm{colim}}\nolimits B_\alpha $ (in fact $B = B_\alpha $ for some ordinal $\alpha $ as one sees by looking at cardinalities). We will prove, by transfinite induction, that (2) holds for $A \to B_\alpha $ for every ordinal $\alpha $. It is clear for $\alpha = 0$. Assume the statement holds for every $\beta < \alpha $ and let $E \subset B_\alpha $ be a finite subset. If $\alpha $ is a limit ordinal, then $B_\alpha = \bigcup _{\beta < \alpha } B_\beta $ and we see that $E \subset B_\beta $ for some $\beta < \alpha $ which proves the result in this case. If $\alpha = \beta + 1$, then $B_\alpha = B_\beta [b_\alpha ]$. Thus any $e \in E$ can be written as a polynomial $e = \sum d_{e, i}b_\alpha ^ i$ with $d_{e, i} \in B_\beta $. Let $D \subset B_\beta $ be the set $D = \{ d_{e, i}\} \cup \{ b_\alpha ^2, b_\alpha ^3\} $. By induction assumption there exists an $A$-subalgebra $A[b_1, \ldots , b_ n] \subset B_\beta $ as in the statement of the lemma containing $D$. Then $A[b_1, \ldots , b_ n, b_\alpha ] \subset B_\alpha $ is an $A$-subalgebra of $B_\alpha $ as in the statement of the lemma containing $E$.

Assume (2). Write $B = \mathop{\mathrm{colim}}\nolimits B_\lambda $ as the colimit of its finite $A$-subalgebras. By Lemma 29.46.2 it suffices to show that $\mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism inducing isomorphisms on residue fields. Compositions of universally closed morphisms are universally closed and the same thing for morphisms which induce isomorphisms on residue fields. Thus it suffices to show that if $A \subset B$ and $B$ is generated by a single element $b$ with $b^2, b^3 \in A$, then (1) holds. Such an extension is integral and hence $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is universally closed and surjective (Lemma 29.44.7 and Algebra, Lemma 10.36.17). Note that $(b^2)^3 = (b^3)^2$ in $A$. For any ring map $\varphi : A \to K$ to a field $K$ we see that there exists a $\lambda \in K$ with $\varphi (b^2) = \lambda ^2$ and $\varphi (b^3) = \lambda ^3$. Namely, $\lambda = 0$ if $\varphi (b^2) = 0$ and $\lambda = \varphi (b^3)/\varphi (b^2)$ if not. Thus $B \otimes _ A K$ is a quotient of $K[x]/(x^2 - \lambda ^2, x^3 - \lambda ^3)$. This ring has exactly one prime with residue field $K$. This implies that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is bijective and induces isomorphisms on residue fields. Combined with universal closedness this shows (1) is true, see Lemmas 29.45.5 and 29.10.2.
$\square$

## Comments (0)

There are also: