
Lemma 10.37.5. Let $K$ be a field. Let $n, m \in \mathbf{N}$ and $a_0, \ldots , a_{n - 1}, b_0, \ldots , b_{m - 1} \in K$. If the polynomial $x^ n + a_{n - 1}x^{n - 1} + \ldots + a_0$ divides the polynomial $x^ m + b_{m - 1} x^{m - 1} + \ldots + b_0$ in $K[x]$ then

1. $a_0, \ldots , a_{n - 1}$ are integral over any subring $R_0$ of $K$ containing the elements $b_0, \ldots , b_{m - 1}$, and

2. each $a_ i$ lies in $\sqrt{(b_0, \ldots , b_{m-1})R}$ for any subring $R \subset K$ containing the elements $a_0, \ldots , a_{n - 1}, b_0, \ldots , b_{m - 1}$.

Proof. Let $L/K$ be a field extension such that we can write $x^ m + b_{m - 1} x^{m - 1} + \ldots + b_0 = \prod _{i = 1}^ m (x - \beta _ i)$ with $\beta _ i \in L$. See Fields, Section 9.16. Each $\beta _ i$ is integral over $R_0$. Since each $a_ i$ is a homogeneous polynomial in $\beta _1, \ldots , \beta _ m$ we deduce the same for the $a_ i$ (use Lemma 10.35.7).

Choose $c_0, \ldots , c_{m - n - 1} \in K$ such that

$\begin{matrix} x^ m + b_{m - 1} x^{m - 1} + \ldots + b_0 = \\ (x^ n + a_{n - 1}x^{n - 1} + \ldots + a_0) (x^{m - n} + c_{m - n - 1}x^{m - n - 1}+ \ldots + c_0). \end{matrix}$

By part (1) the elements $c_ i$ are integral over $R$. Consider the integral extension

$R \subset R' = R[c_0, \ldots , c_{m - n - 1}] \subset K$

By Lemmas 10.35.17 and 10.29.3 we see that $R \cap \sqrt{(b_0, \ldots , b_{m - 1})R'} = \sqrt{(b_0, \ldots , b_{m - 1})R}$. Thus we may replace $R$ by $R'$ and assume $c_ i \in R$. Dividing out the radical $\sqrt{(b_0, \ldots , b_{m - 1})}$ we get a reduced ring $\overline{R}$. We have to show that the images $\overline{a}_ i \in \overline{R}$ are zero. And in $\overline{R}[x]$ we have the relation

$\begin{matrix} x^ m = x^ m + \overline{b}_{m - 1} x^{m - 1} + \ldots + \overline{b}_0 = \\ (x^ n + \overline{a}_{n - 1}x^{n - 1} + \ldots + \overline{a}_0) (x^{m - n} + \overline{c}_{m - n - 1}x^{m - n - 1}+ \ldots + \overline{c}_0). \end{matrix}$

It is easy to see that this implies $\overline{a}_ i = 0$ for all $i$. Indeed by Lemma 10.24.1 the localization of $\overline{R}$ at a minimal prime $\mathfrak {p}$ is a field and $\overline{R}_{\mathfrak p}[x]$ a UFD. Thus $f = x^ n + \sum \overline{a}_ i x^ i$ is associated to $x^ n$ and since $f$ is monic $f = x^ n$ in $\overline{R}_{\mathfrak p}[x]$. Then there exists an $s \in \overline{R}$, $s \not\in \mathfrak p$ such that $s(f - x^ n) = 0$. Therefore all $\overline{a}_ i$ lie in $\mathfrak p$ and we conclude by Lemma 10.24.2. $\square$

Comment #2182 by David Savitt on

At the reference to 10.29.3, it seems to me that you're using something slightly stronger than what's literally available to reference at 10.29.3(2); you want "for any ideal $I \subset R$ the inverse image of $\sqrt{IS}$ in $R$ is $\sqrt{I}$" there instead.

Also all the $b_m$'s in this argument should be $b_{m-1}$'s.

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