In this section, we will collect theorems about the spectrum of a ring: theorems that assert that the map on Spec is surjective, and theorems about lifting chains of primes. Given a ring homomorphism \(R \to S\text{,}\) we want to study the behavior of chains of primes: how chains in \(R\) behave under expansion to \(S\) and how chains in \(S\) behave under contraction to \(R\text{.}\)
Definition13.25.Fiber.
For a map of topological spaces \(f: X\to Y\) and \(y \in Y\text{,}\) the fiber over \(y\) is the subspace
where by abuse of notation we write \(R \setminus P\) for the image of \(R \setminus P\) in \(S\text{,}\) and \(P S\) for \(\psi(P) S\text{.}\)
EMPTY
Lemma13.27.Primes Contracting to p Fiber Ring.
Let \(\psi:R\to S\) be a ring homomorphism and \(P \in \Spec(R)\) be a prime ideal. The natural map \(S \to \kappa_{\psi}(P)\) induces a homeomorphism (in particular, an order-preserving bijection)
For any localization map or any quotient map, the induced map on Spec is a homeomorphism onto its image. The map on spectra induced by \(\pi\) can be identified with the inclusion of \(V(P S)\) into \(\Spec(S)\text{.}\) For the second map, \(g\text{,}\) we saw in that the map on spectra can be identified with the inclusion of the set of primes that do not intersect \(R \setminus P\text{,}\) i.e., those whose contraction is contained in \(P\text{.}\) Therefore, \((g \circ \pi)^*\) is injective, since it is the composition of two injective maps.
On the one hand, if a prime \(Q\) contains \(PS\text{,}\) then \(Q \cap R \supseteq PS \cap R \supseteq P\text{.}\) If moreover \(Q \cap R \subseteq P\text{,}\) we must have \(Q \cap R = P\text{.}\)
We have thus shown that the image of \((g \circ \pi)^*\) is the set of primes in \(S\) that contract to \(P\text{.}\) %So \(Q\) is in the image if and only if \(\psi^{-1}(Q) = P\text{.}\)
Lemma13.28.Image Criterion.
Let \(R \xrightarrow{\varphi} S\) be a ring homomorphism. For any \(P \in \Spec(R)\text{,}\)\(P \in \im(\varphi^*)\) if and only if
so localizing at \((R \setminus P)\text{,}\) we get an inclusion \(\kappa(P) \subseteq \kappa_{\varphi}(P)\text{.}\) Since \(\kappa(P)\) is nonzero, so is \(\kappa_{\varphi}(P).\) and thus its spectrum is nonempty. By , there is a prime mapping to \(P\text{.}\)
If \(P S \cap R \neq P\text{,}\) then \(P S \cap R \supsetneq P\text{.}\) If \(Q \cap R = P\text{,}\) then \(Q \supseteq P S\text{,}\) so \(Q \cap R \supsetneq P\text{.}\) So no prime contracts to \(P\text{.}\)
Example13.29..
Let \(R = \C[x^n] \subseteq S = \C[x]\text{.}\) The ideal \(x^n R\) is prime, while \(x^n S\) is not even radical. Nevertheless, \(x^n S \cap R = (x) \cap R = x^n R\text{.}\)
Example13.30.Inclusion Triangle.
Consider the inclusion \(R:=k[xy,xz,yz]\to S:=k[x,y,z]\) and the prime \(P = (xy)\) in \(R\text{.}\) Notice that \((xz)(yz) \in P S \cap R\text{,}\) but not in \(P\text{,}\) so \(P S \cap R \supsetneq P\text{,}\) and thus \(P \notin \im(\varphi^*)\text{.}\) We can check this more directly, by noting that any prime \(Q\) in \(S\) contracting to \(P\) would contain \(P S = (x) \cap (y)\text{,}\) so \(Q \supseteq (x)\) or \(Q \supseteq (y)\text{.}\) But \((x) \cap R = (xy,xz) \supsetneq P\) and \((y) \cap R = (xy,yz) \supsetneq P\text{,}\) so no prime in \(S\) contracts to \(P\text{.}\)
Corollary13.31.Direct Summand and Surj Spec Map.
If \(R\subseteq S\) is a direct summand, the map \(\Spec(S) \to \Spec(R)\) is surjective.
Proof.
By , we know \(I S \cap R = I\) for all ideals in this case, so says the map on Spec is surjective.
Definition13.32.Integral Element.
Let \(R\) be a ring, \(S\) an \(R\)-algebra, and \(I\) an ideal. An element \(r\) of \(R\) is integral over \(I\) if it satisfies an equation of the form
In both cases, we say that \(r\) and \(s\) satisfy an equation of integral dependence over \(I.\)
FIX
Remark13.33.
Notice that \(\overline{I}^S \cap R = \overline{I}\) is immediate from the definition.
Example13.34.Rees Algebra.
Let \(R\subseteq S\text{,}\)\(I\) be an ideal of \(S\text{,}\) and \(t\) be an indeterminate. Consider the rings \(R[It]\subseteq R[t] \subseteq S[t].\) Here \(R[It]\) is the subalgebra of \(R[t]\) generated by elements of the form \(at\) for all \(a \in I\text{.}\) Notice that we can give this a structure of a graded ring by setting all elements in \(R\) to have degree \(0\) and \(t\) to have degree \(1\text{,}\) so
This is usually called the Rees algebra of \(I\text{.}\) - \(\overline{I}^S =\{ s \in S \ | \ st\in S[t] \ \text{is integral over the ring} \ R[It]\}\text{.}\) - \(\overline{I}^S\) is an ideal of \(S\text{.}\)
Lemma13.35.Extension-Contraction Lemma for Int. Ext..
Let \(R \subseteq S\) be integral, and \(I\) be an ideal of \(R\text{.}\) Then \(IS \subseteq\overline{I}^S\text{,}\) and hence \(I S \cap R \subseteq \overline{I}\text{.}\)
Proof.
Let \(x\in IS\text{.}\) We can write \(x = a_1 s_1 + \cdots + a_t s_t\) for some \(a_i\in I\text{.}\) Moreover, taking \(S'=R[s_1,\dots,s_t]\text{,}\) we also have \(x\in IS'\text{.}\) We will show that \(x \in \overline{I}^{S'}\text{,}\) so \(x \in \overline{I}^{S}\) follows as a corollary. So we might as well replace \(S\) with \(S'\text{,}\) so that \(R \subseteq S\) is also integral and module-finite. By , the extension is also module-finite.
Let \(S= R b_1 + \cdots + R b_n\text{.}\) We can write
with \(a_{ij}\in I\text{.}\) We can write these equations in the form \(x v = Av\text{,}\) where \(v=(b_1,\dots,b_u)\text{,}\) and \(A=[a_{ij}]\text{.}\) By the determinantal trick, , we have \(\det(xI-A)v=0\text{.}\) Since we can assume \(b_1=1\text{,}\) we have \(\det(xI-A)=0\text{.}\) The fact that this is the type of equation we want follows from the monomial expansion of the determinant: any monomial is a product of \(n\) terms where some of them are copies of \(x\text{,}\) and the rest are elements of \(I\text{.}\) Since this is a product of \(n\) terms, a term in \(x^i\) has a coefficient coming from a product of \(n-i\) elements of \(I\text{.}\)
So this shows that \(IS \subseteq\overline{I}^S\text{.}\) Now notice that \(\overline{I}^S \cap R = \overline{I}\) is immediate from the definition, as noted in .
Theorem13.36.Lying Over.
If \(R\subseteq S\) is an integral extension, then \(P S \cap R = P\) for every \(P \in \Spec(R)\text{,}\) and the induced map \(\Spec(S) \longrightarrow \Spec(R)\) is surjective.
Proof.
Let \(I\) be any ideal in \(R\text{.}\) Given any \(r \in \overline{I}\text{,}\) we have
for some \(n\) and some \(a_i \in I^i\) for all \(i\text{,}\) so
\begin{equation*}
r^n = - a_1 r^{n-1} - \cdots - a_{n-1} r - a_n \in I.
\end{equation*}
Thus \(\overline{I} \subseteq \sqrt{I}\text{.}\) Therefore, if \(P\) is a prime in \(R\text{,}\) by we have \(P S \cap R \subseteq \overline{P}\text{,}\) and
\begin{equation*}
P S \cap R \subseteq \overline{P} \subseteq \sqrt{P} = P.
\end{equation*}
Then \(P S \cap R = P\text{,}\) and by we conclude that \(P\) is in the image of the map on Spec.
Remark13.37.
Both assumptions that the extension is integral and that it is an inclusion are needed in [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Lying Over|Lying Over]]; we cannot reduce to the case of an integral inclusion. The point is that \(R \to S\) being an inclusion does not translate into a property of the induced map on Spec.
Example13.38..
We saw in that the map induced on Spec by the inclusion \(k[xy,xz,yz] \subseteq k[x,y,z]\) is not surjective. So does not apply — indeed, this inclusion is not module-finite, and thus by it is not integral. For example, the infinite set \(\{ 1, x^n, y^n, z^n \mid n \geqslant 1\}\) is a minimal generating set for \(k[x,y,z]\) over \(k[xy,xz,yz]\text{.}\)
Example13.39..
Suppose \(f\) is a regular element on \(R\text{,}\) but not a unit. Since \(f\) is regular, the map \(R \longrightarrow R_f\) is an inclusion, but we claim it is not integral. If \(\frac{1}{f}\) is integral over \(R\text{,}\) there would be \(a_i \in R\) such that
and \(f\) must be a unit. So whenever \(f\) is a regular element but not a unit, \(R \longrightarrow R_f\) is an example of an inclusion that is not integral. Note that the image of the map on Spec is the complement of \(V(f)\text{,}\) so in particular the map is not surjective.
In contrast, the map \(R \longrightarrow R/(f)\) is integral, since it is module-finite, but it is not an inclusion. The map on Spec is again not surjective: its image is \(V(f)\text{.}\)
Remark13.40.
Let \(I\) be an ideal in \(S\text{.}\) Suppose \(R \to S\) is an integral extension. There is an induced map \(R/(I \cap R) \to S/I\text{,}\) and that map is integral: an equation of integral dependance for \(s \in S\) over \(R\) gives an equation for integral dependance of its class in \(S/I\) over \(R/(I \cap R)\text{.}\)
Lemma13.41.Integral Extension Maximal Contracts to Maximal.
If \(\varphi: R\to S\) is integral, \(Q \cap R\) is maximal if and only if \(Q\) is maximal in \(S\text{.}\)
Proof.
By , the induced map \(R/(Q \cap R) \subseteq S/Q\) is an integral extension of domains, and \(Q\) (respectively, \(Q \cap R\)) is maximal if and only if \(S/Q\) (respectively, \(R/(Q \cap R)\)) is a field. So this follows by .
Lemma13.42.Integral Preserved by Localization.
Let \(R \to S\) be integral, and let \(W\) be a multiplicatively closed subset of \(R\text{.}\) Then \(W^{-1}R \to W^{-1}S\) is integral.
Proof.
Since \(W\) is a multiplicatively closed subset of \(R\text{,}\) its image in \(S\) is also multiplicatively closed. Given \(x\in S\) and \(w\in W\text{,}\) then we have equations of the form
Since the map on spectra induced by \(R \longrightarrow R/\ker(f)\) is injective, we can replace \(R\) by the quotient and assume \(\varphi\) is an integral inclusion.
So suppose \(R \subseteq S\) is integral, and let \(\fa = P \cap R = Q \cap R\text{.}\) By , localizing at \((R\setminus \fa)\) preserves integrality.
By localizing \(R\) at \((R \setminus \fa)\text{,}\) the image of \(\fa\) is a maximal ideal. So we can reduce to the situation where \(R \cap P = R \cap Q\) is a maximal ideal. By , \(P \subseteq Q\) are both maximal ideals. Therefore, \(P = Q\text{.}\)
Corollary13.44.Finite Contractions in Int Noeth Alg.
Let \(S\) be an integral \(R\)-algebra. If \(S\) is noetherian, then only finitely many primes contract to each \(P \in \Spec(R)\text{.}\)
Proof.
If \(Q' \in \Spec(S)\) contracts to \(P\text{,}\) then \(Q' \supseteq P S\text{,}\) so in particular \(Q'\) contains some prime \(Q\) minimal over \(P S\text{.}\) Then
\begin{equation*}
P S \subseteq Q \subseteq Q' \implies P \subseteq Q \cap R \subseteq Q' \cap R = P,
\end{equation*}
so \(Q' \cap R = Q \cap R\text{.}\) By , \(Q = Q'\text{.}\) So all the primes contracting to \(P\) are in \(\Min(P S)\text{,}\) which is a finite set since \(R\) is noetherian, by .
Corollary13.45.Integral Dim Inequality.
If \(R \longrightarrow S\) is integral, then for any \(Q \in \Spec(S)\) we have
In particular, \(\dim(S) \leqslant \dim(R)\text{.}\)
Proof.
Given a chain of primes \(P_0 \subsetneq \cdots \subsetneq P_n = Q\) in \(\Spec(S)\text{,}\) we can contract to \(R\text{,}\) and by we get a chain of distinct primes in \(\Spec(R)\text{.}\)
Theorem13.46.Going Up.
If \(R \longrightarrow S\) is integral, then for every \(P \subsetneq P'\) in \(\Spec(R)\) and \(Q \in \Spec(S)\) with \(Q \cap R = P\text{,}\) there is some \(Q' \in \Spec(S)\) with \(Q \subsetneq Q'\) and \(Q' \cap R = P'\text{.}\)
Proof.
Consider the map \(R / P \to S / Q\text{.}\) This is integral, as we observed in . It is also injective, so applies. Thus, there is a prime \(\fa\) of \(S/Q\) that contracts to the prime \(P' / P\) in \(\Spec(R/P)\text{.}\) We can write \(\fa = Q' / Q\) for some \(Q' \in \Spec(S)\text{,}\) and we must have that \(Q'\) contracts to \(P'\text{.}\)
Corollary13.47.Integral Dim Equality.
If \(R\subseteq S\) is integral, then \(\dim(R)=\dim(S)\text{.}\)
Proof.
We have already shown that \(\dim(S) \leqslant \dim(R)\) in , so we just need to show that \(\dim(R) \leqslant \dim(S)\text{.}\) Fix a chain of primes \(P_0 \subsetneq \cdots \subsetneq P_n\) in \(\Spec(R)\text{.}\) By Lying Over, , there is a prime \(Q_0 \in \Spec(S)\) contracting to \(P_0\text{.}\) Then by Going up, , we have \(Q_0 \subsetneq Q_1\) with \(Q_1 \cap R = P_1\text{.}\) Continuing, we can build a chain of distinct primes in \(S\) of length \(n\text{.}\) So \(\dim(R) \leqslant \dim(S)\text{,}\) and equality follows.
Lemma13.48.Lemma Minimal Polynomial.
Let \(R\) be a integrally closed and let \(x\) be an element in some larger domain that is integral over \(R\text{.}\) Let \(k\) be the fraction field of \(R\text{,}\) and \(f(t)\in k[t]\) be the minimal polynomial of \(x\) over \(k\text{.}\) (a) If \(x\) is integral over \(R\text{,}\) then \(f(t) \in R[t]\subseteq k[t]\text{.}\) (b) If \(x\) is integral over a prime \(P\text{,}\) then \(f(t)\) has all of its nonleading coefficients in \(P\text{.}\)
Proof.
Let \(x\) be integral over \(R\text{.}\) Fix an algebraic closure of \(k\) containing \(x\text{,}\) and let \(x_1~=~x\text{,}\)\(x_2,\ldots,x_u\) be the roots of \(f\text{.}\) Since \(f(t)\) divides any polynomial with coefficients in \(k\) that \(x\) satisfies, it also divides a monic equation of integral dependance for \(x\) over \(R\text{.}\) Therefore, each \(x_i\) is a solution to such an equation of integral dependence, and thus must be integral over \(R\text{.}\)
Let \(S=R[x_1,\dots,x_u] \subseteq \overline{k}\text{.}\) This is a module-finite extension of \(R\text{,}\) so all of its elements are integral over \(R\text{.}\) The leading coefficient of \(f(t)\) is 1, and the remaining coefficients of \(f(t)\) are polynomials in the \(x_i\text{,}\) hence they lie in \(S\text{.}\) On the other hand, \(R\) is normal, so \(S \cap k = R\text{.}\) We conclude that all the coefficients of \(f\) are in \(R\text{,}\) and \(f \in R[t]\text{.}\)
Now let \(x\) be integral over \(P\text{.}\) By the same argument as above, all of the \(x_i\) are integral over \(P\text{.}\) Since each \(x_i \in \overline{P}^S\text{,}\) any polynomial in the \(x_i\) lies in \(\overline{P}^S\text{.}\) So the nonleading coefficients of \(f\) lie in \(\overline{P}^S \cap R = P\text{,}\) by [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Lying Over|Lying Over]].
Theorem13.49.Going Down.
Suppose that \(R\) is a integrally closed, \(S\) is a domain, and \(R\subseteq S\) is integral. Then, for every \(P \subsetneq P'\) in \(\Spec(R)\) and \(Q'\) in \(\Spec(S)\) with \(Q' \cap R = P'\text{,}\) there is some \(Q \in \Spec(S)\) with \(Q \subsetneq Q'\) and \(Q \cap R = P\text{.}\)
Proof.
Let \(W=(S \setminus Q')(R \setminus P)\) be the multiplicative set in \(S\) consisting of products of elements in \(S \setminus Q'\) and \(R \setminus P\text{.}\) Note that each of the sets \(S \setminus Q'\) and \(R \setminus P\) contains \(1\text{,}\) so \(S \setminus Q'\) and \(R \setminus P\) are both contained in \(W\text{.}\) We will show that \(W \cap P S\) is empty. Once we do that, it will follow from that there is a prime ideal \(Q\) in \(S\) containing \(P S\) such that \(W \cap Q\) is empty. Since \(S \setminus Q' \subseteq W\text{,}\) our new prime \(Q\) must necessarily be contained in \(Q'\text{.}\)
Moreover, \(R \setminus P \subseteq W\text{,}\) so \((Q \cap R) \cap (R \setminus P)\) is empty, or equivalently, \(Q \cap R \subseteq P\text{.}\) Since \(Q \cap R \subseteq P\) and \(Q \cap R \supseteq P\text{,}\) we conclude that \(Q \cap R = P\text{.}\)
So our goal is to show that \(W \cap P S\) is empty. We proceed by contradiction, and assume there is some \(x \in P S \cap W\text{.}\) We can write \(x=rs\) for some \(r\in R \setminus P\) and \(s \in S \setminus Q'\text{.}\) Moreover, since \(x \in P S\text{,}\)\(x\) is integral over \(P\text{,}\) by .
Consider the minimal polynomial of \(x\) over \(\mathrm{frac}(R)\text{,}\) say
We claim that this is the minimal polynomial of \(s\text{.}\) If \(s\) satisfied a monic polynomial of degree \(d < n\text{,}\) multiplying by \(r^d\) would give us a polynomial of degree \(d\) that \(x\) satisfies, which is impossible. So indeed, this is the minimal polynomial of \(s\text{.}\)
Since \(s\in S\text{,}\) and thus integral over \(R\text{,}\) says that each \(\frac{a_i}{r^i} =: v_i\in R\text{.}\) Since \(r \notin P\) and \(r^i v_i =a_i \in P\text{,}\) we must have \(v_i\in P\text{.}\) The equation \(g(s)=0\) then shows that \(s \in \sqrt{P S}\text{.}\) Since \(Q' \in \Spec(S)\) contains \(P' S\) and hence \(P S\text{,}\) we have \(s \in \sqrt{P S} \subseteq Q'\text{.}\) This is the desired contradiction, since \(s \notin Q'\) by construction.
Corollary13.50.Height in Normal Domains.
If \(R\) is a integrally closed, \(S\) is a domain, and \(R \subseteq S\) is integral, then \(\ht(Q) = \ht(Q \cap R)\) for any \(Q \in \Spec(S)\text{.}\)
Proof.
We already know from that \(\ht(Q) \leqslant \ht(Q \cap R)\text{.}\) To show that \(\ht(Q) \geqslant \ht(Q \cap R)\text{,}\) given a saturated chain up to \(Q \cap R\text{,}\) we can apply to get a chain just as long that goes up to \(Q\text{.}\)
