Part II Dimension
8.2 Lying Over, Going Up and Going Down
8.2 Lying Over, Going Up and Going Down 109
R∩I P
I
Q R
S
Figure 8.2.Lying over and going up
M → {Q0∈Spec(S0)|R0∩Q0=P0}.
Substituting all objects by their primed versions, we may therefore assume thatI={0}. By Proposition7.11, we have to show that the fiber ringS[P] is not the zero ring (implying (a)), has Krull dimension 0 (implying (b)), and has a finite spectrum ifS is finitely generated (implying (c)).
By way of contradiction, assume thatS[P]={0}. By the definition ofS[P], this is equivalent to the existence ofu∈R\P with u∈(P)S. Forming the localizationSP := (R\P)−1S, we obtain 1∈(PP)S
P, so 1 =
n
X
i=1
siai with si∈SP, ai ∈PP.
FormSe:=RP[s1, . . . , sn]⊆SP. Then the above equation implies (PP)
Se=S,e which we may write asPPSe=S. Sincee Seis an integral extension ofRP, it is finitely generated as anRP-module by Theorem 8.4. Applying Nakayama’s Lemma 7.3 yields Se ={0}. Since RP is embedded into S, this contradictse the fact that local rings are never zero. So we conclude thatS[P] is non-zero.
The homomorphism
K:= Quot (R/P)→S[P], a+P
b+P 7→a+ (P)S
b+ (P)S
makesS[P] into aK-algebra. The hypothesis thatS is integral overRtrans- lates into the fact that S[P] is algebraic overK. So ifQ∈Spec S[P]
, then the quotient ringS[P]/Qis algebraic overKas well, and Lemma1.1(a) yields that S[P]/Q is a field. This shows that dim S[P]
= 0.
Finally, if S is finitely generated as an R-algebra, then S[P] is an affine K-algebra, so Theorem 5.11yields that Specmax S[P]
is finite. Since S[P]
has dimension 0, Spec S[P]
= Specmax S[P]
, so we are done. ut Let R⊆S be an integral extension of rings. IfP0 $P1 $· · · $Pn is a chain of prime ideals Pi ∈ Spec(R), we can use Theorem8.12 to construct
8.2 Lying Over, Going Up and Going Down 111 a chain Q0 ⊆ · · · ⊆ Qn of prime ideals in Spec(S) with R∩Qi = Pi. In particular, all inclusions of theQi are proper. So dim(S)≥n, which implies
dim(R)≤dim(S). (8.2)
On the other hand, if Q∈ Spec(S) is a prime ideal and Q0 $Q1 $· · · $ Qn ⊆ Q is a chain of prime ideals in Spec(S), then Pi := R∩Qi yields a chain in Spec(S), and it follows from Theorem8.12(b) that the inclusions of thePi are proper. So withP :=R∩Qwe obtain
ht(Q)≤ht(P). (8.3)
This implies
dim(S)≤dim(R). (8.4)
By putting (8.2) and (8.4) together, we obtain
Corollary 8.13. Let R⊆S be an integral extension of rings. Then dim(R) = dim(S).
We now pose the question whether the reverse inequality of (8.3) also holds, i.e., whether (8.3) is in fact an equality. For proving this, we need to start with a chain of prime ideals in Spec(R) which are all contained in P, and construct an equally long chain of prime ideals in Spec(S) which are all contained inQ. The way to do this is to work our way downwards from Q.
But what we need for being able to do this is the going down property which was discussed in Section 7.2(see on page95). We have shown:
Corollary 8.14. LetR⊆Sbe an integral extension of rings such that going down holds for the inclusion R ,→S. Then forQ∈Spec(S)andP :=R∩Q we have
ht(P) = ht(Q).
Unfortunately, going down does not always hold for integral ring exten- sions, as Exercise 8.8 shows. We have proved that a sufficient condition for going down is freeness (see Lemma 7.16). However, freeness is rarely found for integral extensions. We will exhibit another sufficient condition for going down (see Theorem8.17). For proving this, we need two lemmas. The effort is worth it, since the reverse inequality of (8.3) is of crucial importance for proving some important results about affine algebras, such as Theorem8.22 and its corollaries. The first lemma is a result from field theory. The proof uses some standard results from field theory, which we will quote from Lang [33].
Lemma 8.15 (Elements fixed by field-automorphisms). Let N be a field of characteristic p ≥ 0 and let K ⊆ N a subfield such that N is finite and normal over K (see Lang [33, Ch. VII, Theorem 3.3] for the definition of a normal field extension). Let G:= AutK(N) be the group of automorphisms of N fixing K element-wise. Then for everyα∈NG in the fixed field of G,
there existsn∈N0 such thatαpn ∈K. IfN is separable overK, then n= 0, soα∈K.
Proof. In the separable case, the lemma follows directly from Galois theory.
The proof we give works for the separable case, too.
Let g = irr(α, K) ∈ K[x] be the minimal polynomial of α over K. Let N be the algebraic closure of N, and let β ∈ N be a zero of g. Since K[α] ∼= K[x]/(g) ∼=K[β] with an isomorphism sending α to β, we have a homomorphism σ: K[α] → N of K-algebras with σ(α) = β. By Lang [33, Ch. VII, Theorem 2.8], this extends to a homomorphism σ: N → N. The normality ofN implies σ∈G(see Lang [33, Ch. VII, Theorem 3.3]). Since σ(α) =β, the hypothesis of the lemma implies β =α. Soαis the only zero ofg, and we obtaing= (x−α)mwithm∈N. Writem=k·pn withp-k. If N is separable overK, theng has to be separable, som= 1 andn= 0. We have
g= (x−α)m= (xpn−αpn)k =xkpn−k·αpn·x(k−1)pn+ (lower terms),
sog∈K[x] impliesapn∈K. ut
Lemma 8.16. Let N be a field and K⊆N a subfield such that N is finite and normal over K. LetR ⊆K be a subring that is integrally closed in K, and let S ⊆N be the integral closure of R in N. Then for two prime ideals Q,Qe ∈Spec(S) with R∩Q=R∩Q, there existse σ∈ G:= AutK(N) with Qe=σ(Q).
Proof. Leta∈Q. Then the producte Q
σ∈Gσ(a) lies inNG, so by Lemma8.15 there existsn∈N0with
b:= Y
σ∈G
σ(a)pn ∈K, (8.5)
where p = char(K) and n = 0 if p = 0. Since a is integral over R and all σ∈GfixRelement-wise, allσ(a) are integral overRas well. Sobis integral over R, too, and (8.5) implies b ∈ R. Moreover, b is an S-multiple of a, so b∈R∩Qe=R∩Q⊆Q. SinceQis a prime ideal, it follows from (8.5) that there existsσ∈Gwithσ(a)∈Q. Since this holds for alla∈Q, we concludee
Qe⊆ [
σ∈G
σ(Q).
By the Prime Avoidance Lemma7.7, this implies that there existsσ∈Gwith Qe⊆σ(Q). SinceσfixesRelement-wise, we haveR∩σ(Q) =R∩Q=R∩Q,e so by Theorem8.12(b), the inclusionQe⊆σ(Q) cannot be strict. ut Theorem 8.17(Going down for integral extensions of normal rings). LetS be a ring and R⊆S a subring such that
8.2 Lying Over, Going Up and Going Down 113 (1) S is an integral domain,
(2) Ris normal,
(3) S is integral overR, and
(4) S is finitely generated as anR-algebra.
Then going down holds for the inclusionR ,→S. In particular, the conclusion of Corollary8.14 holds.
Proof. The proof is not difficult but a bit involved. Figure8.3shows what is going on. Given prime ideals P ∈Spec(R) and Q0 ∈Spec(S) with P ⊆Q0, we need to produce Q ∈Spec(S) with R∩Q = P and Q⊆ Q0. The field of fractions L := Quot(S) is a finite field extension of K := Quot(R). By Lang [33, Ch. VII, Theorem 3.3], there exists a finite, normal field extension N of K such thatL⊆N. Let T ⊆N be the integral closure of R in N, so S ⊆T. By Theorem8.12, there existZ, Ze 0∈Spec(T) such thatR∩Ze=P and S∩Z0 = Q0. We cannot assume that Ze is contained in Z0. However, applying Theorem 8.12 again, we see that there exists Ze0 ∈ Spec(T) such that R∩Ze0=R∩Q0 andZe⊆Ze0. We have
R∩Z0=R∩S∩Z0=R∩Q0=R∩Ze0.
So by Lemma8.16there existsσ∈AutK(N) withZ0 =σ(Ze0). SetZ :=σ(Z)e andQ:=S∩Z∈Spec(S). Then
R∩Q=R∩Z=R∩σ(Z) =e R∩Ze=P and
Q=S∩σ(Z)e ⊆S∩σ(Ze0) =S∩Z0 =Q0.
This finishes the proof. ut
P R∩Q0
R
Q Q
0
S
Z Z
0
σ Ze
σ Ze0
S S S T
Figure 8.3.Going down: givenP andQ0, constructQ
We finish this section by drawing some conclusions about geometric prop- erties of normalization.
Proposition 8.18 (Geometric properties of normalization). LetRbe an in- tegral domain with normalization R, and consider the morphisme f: Spec(R)e →Spec(R) induced from the inclusionR⊆R. Thene
(a) dim(R) = dim(R).e
(b) The morphism f is surjective.
(c) LetP ∈Spec(R)such thatRP is normal. Then the fiberf−1({P})con- sists of one point.
Proof. Parts (a) and (b) follow from Corollary8.13and Theorem8.12(a).
To prove (c), takeP ∈Spec(R) withRP normal. BothRP andReare con- tained in Quot(R). WithU :=R\P we haveU−1Re⊆Quot(R) = Quot(RP), andU−1Re is integral overRP, soU−1Re =RP by the normality of RP. Let Q∈Spec(R) be in the fiber ofe P, soR∩Q=P. By Theorem6.5it follows that U−1Q∈Spec(U−1R) = Spec(Re P), and Re∩U−1Q=Q, soR∩U−1Q=P.
But Theorem 6.5also says thatPP is the only prime ideal inRP whose in- tersection withRisP, soU−1Q=PP. It follows that Q=Re∩PP, showing
uniqueness. ut