The Stacks project

10.38 Going down for integral over normal

We first play around a little bit with the notion of elements integral over an ideal, and then we prove the theorem referred to in the section title.

Definition 10.38.1. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. We say an element $g \in S$ is integral over $I$ if there exists a monic polynomial $P = x^ d + \sum _{j < d} a_ j x^ j$ with coefficients $a_ j \in I^{d-j}$ such that $P^\varphi (g) = 0$ in $S$.

This is mostly used when $\varphi = \text{id}_ R : R \to R$. In this case the set $I'$ of elements integral over $I$ is called the integral closure of $I$. We will see that $I'$ is an ideal of $R$ (and of course $I \subset I'$).

Lemma 10.38.2. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $A = \sum I^ nt^ n \subset R[t]$ be the subring of the polynomial ring generated by $R \oplus It \subset R[t]$. An element $s \in S$ is integral over $I$ if and only if the element $st \in S[t]$ is integral over $A$.

Proof. Suppose $st$ is integral over $A$. Let $P = x^ d + \sum _{j < d} a_ j x^ j$ be a monic polynomial with coefficients in $A$ such that $P^\varphi (st) = 0$. Let $a_ j' \in A$ be the degree $d-j$ part of $a_ i$, in other words $a_ j' = a_ j'' t^{d-j}$ with $a_ j'' \in I^{d-j}$. For degree reasons we still have $(st)^ d + \sum _{j < d} \varphi (a_ j'') t^{d-j} (st)^ j = 0$. Hence we see that $s$ is integral over $I$.

Suppose that $s$ is integral over $I$. Say $P = x^ d + \sum _{j < d} a_ j x^ j$ with $a_ j \in I^{d-j}$. Then we immediately find a polynomial $Q = x^ d + \sum _{j < d} (a_ j t^{d-j}) x^ j$ with coefficients in $A$ which proves that $st$ is integral over $A$. $\square$

Lemma 10.38.3. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. The set of elements of $S$ which are integral over $I$ form a $R$-submodule of $S$. Furthermore, if $s \in S$ is integral over $R$, and $s'$ is integral over $I$, then $ss'$ is integral over $I$.

Proof. Closure under addition is clear from the characterization of Lemma 10.38.2. Any element $s \in S$ which is integral over $R$ corresponds to the degree $0$ element $s$ of $S[x]$ which is integral over $A$ (because $R \subset A$). Hence we see that multiplication by $s$ on $S[x]$ preserves the property of being integral over $A$, by Lemma 10.36.7. $\square$

Lemma 10.38.4. Suppose $\varphi : R \to S$ is integral. Suppose $I \subset R$ is an ideal. Then every element of $IS$ is integral over $I$.

Proof. Immediate from Lemma 10.38.3. $\square$

Lemma 10.38.5. 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 10.36.7).

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 10.36.17 and 10.30.3 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 10.25.1 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 10.25.2. $\square$

Lemma 10.38.6. Let $R \subset S$ be an inclusion of domains. Assume $R$ is normal. Let $g \in S$ be integral over $R$. Then the minimal polynomial of $g$ has coefficients in $R$.

Proof. Let $P = x^ m + b_{m-1} x^{m-1} + \ldots + b_0$ be a polynomial with coefficients in $R$ such that $P(g) = 0$. Let $Q = x^ n + a_{n-1}x^{n-1} + \ldots + a_0$ be the minimal polynomial for $g$ over the fraction field $K$ of $R$. Then $Q$ divides $P$ in $K[x]$. By Lemma 10.38.5 we see the $a_ i$ are integral over $R$. Since $R$ is normal this means they are in $R$. $\square$

Proposition 10.38.7. Let $R \subset S$ be an inclusion of domains. Assume $R$ is normal and $S$ integral over $R$. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q'$ be a prime of $S$ with $\mathfrak p' = R \cap \mathfrak q'$. Then there exists a prime $\mathfrak q$ with $\mathfrak q \subset \mathfrak q'$ such that $\mathfrak p = R \cap \mathfrak q$. In other words: the going down property holds for $R \to S$, see Definition 10.41.1.

Proof. Let $\mathfrak p$, $\mathfrak p'$ and $\mathfrak q'$ be as in the statement. We have to show there is a prime $\mathfrak q$, with $\mathfrak q \subset \mathfrak q'$ and $R \cap \mathfrak q = \mathfrak p$. This is the same as finding a prime of $S_{\mathfrak q'}$ mapping to $\mathfrak p$. According to Lemma 10.17.9 we have to show that $\mathfrak p S_{\mathfrak q'} \cap R = \mathfrak p$. Pick $z \in \mathfrak p S_{\mathfrak q'} \cap R$. We may write $z = y/g$ with $y \in \mathfrak pS$ and $g \in S$, $g \not\in \mathfrak q'$. Written differently we have $zg = y$.

By Lemma 10.38.4 there exists a monic polynomial $P = x^ m + b_{m-1} x^{m-1} + \ldots + b_0$ with $b_ i \in \mathfrak p$ such that $P(y) = 0$.

By Lemma 10.38.6 the minimal polynomial of $g$ over $K$ has coefficients in $R$. Write it as $Q = x^ n + a_{n-1} x^{n-1} + \ldots + a_0$. Note that not all $a_ i$, $i = n-1, \ldots , 0$ are in $\mathfrak p$ since that would imply $g^ n = \sum _{j < n} a_ j g^ j \in \mathfrak pS \subset \mathfrak p'S \subset \mathfrak q'$ which is a contradiction.

Since $y = zg$ we see immediately from the above that $Q' = x^ n + za_{n-1} x^{n-1} + \ldots + z^{n}a_0$ is the minimal polynomial for $y$. Hence $Q'$ divides $P$ and by Lemma 10.38.5 we see that $z^ ja_{n - j} \in \sqrt{(b_0, \ldots , b_{m-1})} \subset \mathfrak p$, $j = 1, \ldots , n$. Because not all $a_ i$, $i = n-1, \ldots , 0$ are in $\mathfrak p$ we conclude $z \in \mathfrak p$ as desired. $\square$

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