Proposition 14

Let $A$ be a completely integrally closed domain. Then every polynomial ring $A[X_1, \ldots, X_n]$ (resp. every ring of formal power series $A[[X_1, \ldots, X_n]]$) is completely integrally closed.

By induction on $n$, it is sufficient to prove that $A[X]$ (resp. $A[[X]]$) is completely integrally closed. Then let $P$ be an element of the field of fractions of $A[X]$ (resp. $A[[X]]$) and suppose that there exists a non-zero element $Q \in A[X]$ (resp. $Q \in A[[X]]$) such that $Q P^m \in A[X]$ (resp. $Q P'' \in A[[X]]$) for every integer $m \geq 0$. If $K$ is the field of fractions of $A$, $A[X]$ (resp. $A[[X]]$) is a subring of $K[X]$ (resp. $K[[X]]$) and $K[X]$ (resp. $K[[X]]$) is a principal ideal domain (Algebra, Chapter VII, § 1, no. 1) and hence integrally closed (no. 4, Proposition 10) and Noetherian (Algebra, Chapter VIII, § 2, no. 3) and therefore completely integrally closed; then we have already seen that

$P \in K[X]$ (resp. $P \in K[[X]]$). Let $P = \sum_{k=0}^{\infty} a_k X^k$ ($a_k \in K$) and $Q = \sum_{k=0}^{\infty} b_k X^k$ ($b_k \in A$) and we argue by reductio ad absurdum by supposing that the $a_i$ do not all belong to $A$; then there is a least index $i$ such that $a_i \notin A$; if we write

it follows immediately from the hypothesis that also $Q(P - P_1)^m \in A[X]$ (resp. $Q(P - P_1)^m \in A[[X]]$) for all $m \geq 0$. Let $j$ be the least integer such that $b_j \neq 0$; clearly in $Q(P - P_1)^m$ the term of least degree with a coefficient $\neq 0$ is $b_j a_i^m X^{j + mi}$ and hence $b_j a_i^m \in A$ for all $m \geq 0$; but as $A$ is completely integrally closed this implies $a_i \in A$, contrary to the hypothesis.