**Proof.**
We will use Lemma 29.45.6 without further mention. Consider the commutative diagram

\[ \xymatrix{ B \ar[r] & B_{red} \\ A \ar[u] \ar[r] & A_{red} \ar[u] } \]

For any factorization $A \to C \to B$ of $A \to B$ as in (2), we see that $A_{red} \to C_{red} \to B_{red}$ is a factorization of $A_{red} \to B_{red}$ as in (2). It follows that if the lemma holds for $A_{red} \to B_{red}$ and produces the $A_{red}$-subalgebra $B'_{red} \subset B_{red}$, then setting $B' \subset B$ equal to the inverse image of $B'_{red}$ solves the lemma for $A \to B$. This reduces us to the case discussed in the next paragraph.

Assume $A$ and $B$ are reduced. In this case $A \subset B$ by Algebra, Lemma 10.30.6. Let $A \to C \to B$ be a factorization as in (2). Then we may apply Proposition 29.46.8 to $A \subset C$ to see that every element of $C$ is contained in an extension $A[c_1, \ldots , c_ n] \subset C$ such that for $i = 1, \ldots , n$ we have

$c_ i^2, c_ i^3 \in A[c_1, \ldots , c_{i - 1}]$, or

there exists a prime number $p$ with $pc_ i, c_ i^ p \in A[c_1, \ldots , c_{i - 1}]$.

Thus property (2) holds if we define $B' \subset B$ to be the subset of elements $b \in B$ which are contained in an extension $ A[b_1, \ldots , b_ n] \subset B $ such that (*) holds: for $i = 1, \ldots , n$ we have

$b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$, or

there exists a prime number $p$ with $pb_ i, b_ i^ p \in A[b_1, \ldots , b_{i - 1}]$.

There are only two things to check: (a) $B'$ is an $A$-subalgebra, and (b) $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism. Part (a) follows because given $n \geq 0$ and $b_1, \ldots , b_ n \in B$ satisfying (*) and $m \geq 0$ and $b'_1, \ldots , b'_ m \in B$ satisfying (*), the integer $n + m$ and $b_1, \ldots , b_ n, b'_1, \ldots , b'_ m \in B$ also satisfies (*). Finally, part (b) holds by Proposition 29.46.8 and our construction of $B'$.
$\square$

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