Lemma 15.14.6. For any ring $A$ there exists an extension $A \subset B$ such that

$B$ is a filtered colimit of finite free $A$-algebras,

$B$ is free as an $A$-module, and

$B$ is absolutely integrally closed.

Lemma 15.14.6. For any ring $A$ there exists an extension $A \subset B$ such that

$B$ is a filtered colimit of finite free $A$-algebras,

$B$ is free as an $A$-module, and

$B$ is absolutely integrally closed.

**Proof.**
Let $I$ be the set of monic polynomials over $A$. For $i \in I$ denote $x_ i$ a variable and $P_ i$ the corresponding monic polynomial in the variable $x_ i$. Then we set

\[ F(A) = A[x_ i; i \in I]/(P_ i; i \in I) \]

As the notation suggests $F$ is a functor from the category of rings to itself. Note that $A \subset F(A)$, that $F(A)$ is free as an $A$-module, and that $F(A)$ is a filtered colimit of finite free $A$-algebras. Then we take

\[ B = \mathop{\mathrm{colim}}\nolimits F^ n(A) \]

where the transition maps are the inclusions $F^ n(A) \subset F(F^ n(A)) = F^{n + 1}(A)$. Any monic polynomial with coefficients in $B$ actually has coefficients in $F^ n(A)$ for some $n$ and hence has a solution in $F^{n + 1}(A)$ by construction. This implies that $B$ is absolutely integrally closed by Lemma 15.14.2. We omit the proof of the other properties. $\square$

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