Proposition 13

Let $A$ be a Krull domain and $X_1, X_n, \ldots, X_n$ indeterminates. The ring $A[X_1, \ldots, X_n]$ is a Krull domain.

Arguing by induction on $n$, it is sufficient to show that, if $X$ is an indeterminate, $A[X]$ is a Krull domain. Let $K$ be the field of fractions of $A$. The field of fractions of $A[X]$ is $K(X)$. Let $I$ be the set of monic polynomials in $K[X]$ which are irreducible over $K$; for all $f \in I$, let $v_f$ be the valuation on $K(X)$ defined by $f$ (Chapter VI, §3, no. 3, Example 4). On the other hand, for every essential valuation $w$ of $A$, let $w$ be the extension of $w$ to $K(X)$ defined by

for $\sum_j a_j X^j \in K[X]$ (Chapter VI, §10, no. 1, Lemma 1). Clearly the $u$, and the $\bar{w}$ are discrete and normed and, for all $u \in K[X]$, $v_f(u) = 0$ (resp. $\bar{w}(u) = 0$) except for a finite number of valuations $v_f$ (resp. $\bar{w}$).

To show the proposition, it therefore suffices to show that $A[X]$ is the intersection of the rings of the valuations $v_f$ and $\bar{w}$. Now the intersection of the rings of the valuations $v_f$ is $K[X]$. On the other hand, for $\sum_j a_j X^j \in K[X]$, the relation $\bar{w}\left( \sum_j a_j X^j \right) \geq 0$ is equivalent to " $w(a_j) \geq 0$ for all $j$"; hence the relation " $\bar{w}\left( \sum_j a_j X^j \right) \geq 0$ for every valuation $\bar{w}'$" is equivalent to " $w(a_j) \geq 0$ for all $j$ and every essential valuation $w$ of $A$." This proves our assertion.