Proof. Let $R$ be a UFD. Let $x$ be an element of the fraction field of $R$ which is integral over $R$. Say $x^ d - a_1 x^{d - 1} - \ldots - a_ d = 0$ with $a_ i \in R$. We can write $x = u p_1^{e_1} \ldots p_ r^{e_ r}$ with $u$ a unit, $e_ i \in \mathbf{Z}$, and $p_1, \ldots , p_ r$ irreducible elements which are not associates. To prove the lemma we have to show $e_ i \geq 0$. If not, say $e_1 < 0$, then for $N \gg 0$ we get

$u^ d p_2^{de_2 + N} \ldots p_ r^{de_ r + N} = p_1^{-de_1}p_2^ N \ldots p_ r^ N( \sum \nolimits _{i = 1, \ldots , d} a_ i x^{d - i} ) \in (p_1)$

which contradicts uniqueness of factorization in $R$. $\square$

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