The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.37 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 10.37.1. 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 10.37.2. 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 10.37.3. 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 10.37.2. 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 10.35.7. $\square$

Lemma 10.37.4. 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 10.37.3. $\square$

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. $\square$

Lemma 10.37.6. 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 10.37.5 we see the $a_ i$ are integral over $R$. Since $R$ is normal this means they are in $R$. $\square$

Proposition 10.37.7. 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 10.40.1.

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 10.16.9 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 10.37.4 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 10.37.6 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 10.37.5 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 037E. Beware of the difference between the letter 'O' and the digit '0'.