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.
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.
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_ j, 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 s^ d + \sum _{j < d} \varphi (a_ j'') s^ j = 0 and 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
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
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 10.18.6 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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