Proposition 10

Every principal ideal domain is integrally closed.

Let $A$ be a principal ideal domain, $K$ its field of fractions and $x$ an element of $K$. There exist two relatively prime elements $a, b$ of $A$ such that $x = ab^{-1}$ (Algebra, Chapter VII, § 1, no. 2, Proposition 1 and Chapter VI, § 1, no. 11, Proposition 9 (DIV)). If $x$ is integral over $A$, it is a root of a polynomial

X^n + c_1 X^{n-1} + \ldots + c_n \text{ of } A[X]. Then $a^n = b(-c_1 a^{n-1} - \ldots - c_n b^{n-1})$, which proves that $b$ divides $a''$. Since $a$ and $b$ are relatively prime, this implies that $b$ is invertible in $A$ (Algebra, Chapter VI, § 1, no. 12, Corollary 1 to Proposition 11 (DIV)); hence $x \in A$.