Lemma 10.159.4. Let $A$ be a ring. Let $\kappa = \max (|A|, \aleph _0)$. Then every flat $A$-algebra $B$ is the filtered colimit of its flat $A$-subalgebras $B' \subset B$ of cardinality $|B'| \leq \kappa $. (Observe that $B'$ is faithfully flat over $A$ if $B$ is faithfully flat over $A$.)
Proof. If $B$ has cardinality $\leq \kappa $ then this is true. Let $E \subset B$ be an $A$-subalgebra with $|E| \leq \kappa $. We will show that $E$ is contained in a flat $A$-subalgebra $B'$ with $|B'| \leq \kappa $. The lemma follows because (a) every finite subset of $B$ is contained in an $A$-subalgebra of cardinality at most $\kappa $ and (b) every pair of $A$-subalgebras of $B$ of cardinality at most $\kappa $ is contained in an $A$-subalgebra of cardinality at most $\kappa $. Details omitted.
We will inductively construct a sequence of $A$-subalgebras
\[ E = E_0 \subset E_1 \subset E_2 \subset \ldots \]
each having cardinality $\leq \kappa $ and we will show that $B' = \bigcup E_ k$ is flat over $A$ to finish the proof.
The construction is as follows. Set $E_0 = E$. Given $E_ k$ for $k \geq 0$ we consider the set $S_ k$ of relations between elements of $E_ k$ with coefficients in $A$. Thus an element $s \in S_ k$ is given by an integer $n \geq 1$ and $a_1, \ldots , a_ n \in A$, and $e_1, \ldots , e_ n \in E_ k$ such that $\sum a_ i e_ i = 0$ in $E_ k$. The flatness of $A \to B$ implies by Lemma 10.39.11 that for every $s = (n, a_1, \ldots , a_ n, e_1, \ldots , e_ n) \in S_ k$ we may choose
\[ (m_ s, b_{s, 1}, \ldots , b_{s, m_ s}, a_{s, 11}, \ldots , a_{s, nm_ s}) \]
where $m_ s \geq 0$ is an integer, $b_{s, j} \in B$, $a_{s, ij} \in A$, and
\[ e_ i = \sum \nolimits _ j a_{s, ij} b_{s, j}, \forall i, \quad \text{and}\quad 0 = \sum \nolimits _ i a_ i a_{s, ij}, \forall j. \]
Given these choicse, we let $E_{k + 1} \subset B$ be the $A$-subalgebra generated by
$E_ k$ and
the elements $b_{s, 1}, \ldots , b_{s, m_ s}$ for every $s \in S_ k$.
Some set theory (omitted) shows that $E_{k + 1}$ has at most cardinality $\kappa $ (this uses that we inductively know $|E_ k| \leq \kappa $ and consequently the cardinality of $S_ k$ is also at most $\kappa $).
To show that $B' = \bigcup E_ k$ is flat over $A$ we consider a relation $\sum _{i = 1, \ldots , n} a_ i b'_ i = 0$ in $B'$ with coefficients in $A$. Choose $k$ large enough so that $b'_ i \in E_ k$ for $i = 1, \ldots , n$. Then $(n, a_1, \ldots , a_ n, b'_1, \ldots , b'_ n) \in S_ k$ and hence we see that the relation is trivial in $E_{k + 1}$ and a fortiori in $B'$. Thus $A \to B'$ is flat by Lemma 10.39.11. $\square$
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