**Proof.**
We first prove the second statement. Write $f = \alpha _0 + \alpha _1 x + \ldots + \alpha _ r x^ r$ with $\alpha _ r \not= 0$. By assumption there exists $h = b_0 + b_1 x + \ldots + b_ s x^ s \in R[x]$, $b_ s \not= 0$ such that $f^ n h \in R[x]$ for all $n \geq 0$. This implies that $b_ s \alpha _ r^ n \in R$ for all $n \geq 0$. Hence $\alpha _ r$ is almost integral over $R$. Since the set of almost integral elements form a subring (Lemma 10.36.4) we deduce that $f - \alpha _ r x^ r = \alpha _0 + \alpha _1 x + \ldots + \alpha _{r - 1} x^{r - 1}$ is almost integral over $R[x]$. By induction on $r$ we win.

In order to prove the first statement we will use absolute Noetherian reduction. Namely, write $\alpha _ i = a_ i / b_ i$ and let $P(t) = t^ d + \sum _{j < d} f_ j t^ j$ be a polynomial with coefficients $f_ j \in R[x]$ such that $P(f) = 0$. Let $f_ j = \sum f_{ji}x^ i$. Consider the subring $R_0 \subset R$ generated by the finite list of elements $a_ i, b_ i, f_{ji}$ of $R$. It is a domain; let $K_0$ be its field of fractions. Since $R_0$ is a finite type $\mathbf{Z}$-algebra it is Noetherian, see Lemma 10.30.3. It is still the case that $f \in K_0[x]$ is integral over $R_0[x]$, because all the identities in $R$ among the elements $a_ i, b_ i, f_{ji}$ also hold in $R_0$. By Lemma 10.36.4 the element $f$ is almost integral over $R_0[x]$. By the second statement of the lemma, the elements $\alpha _ i$ are almost integral over $R_0$. And since $R_0$ is Noetherian, they are integral over $R_0$, see Lemma 10.36.4. Of course, then they are integral over $R$.
$\square$

## Comments (1)

Comment #744 by Wei Xu on