7. Localization. (d 1, m 1 ) (d 2, m 2 ) d 3 D : d 3 d 1 m 2 = d 3 d 2 m 1. (ii) If (d 1, m 1 ) (d 1, m 1 ) and (d 2, m 2 ) (d 2, m 2 ) then
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1 7. Localization To prove Theorem 6.1 it becomes necessary to be able to a enominators to rings (an to moules), even when the rings have zero-ivisors. It is a tool use all the time in commutative algebra, so we shall give some more results than strictly necessary for proving the Cohen-Seienberg theorems. In the beginning there are a lot of straightforwar verification to make, but they are all easy. Let R be a ring an M an R-moule. A multilpicatively close subset D of R is a subset of R containing 1 an such that if, D then also D. The most important example is the complement of a prime ieal D R\p, where p is a prime ieal, since if two elements are not in p then the prouct is not in p either. We moify the way we constructe the fiel Q starting from the integers Z. We first efine an equivalence relation on the cartesian prouct set D M (an in particular on D R). Lemma 7.1. Let M be a R-moule an D R a multiplicatively close subset. (i) The following efines an equivalence relation on the cartesian prouct D M: (, m 1 ) (, m 2 ) D : m 2 m 1. (ii) If (, m 1 ) ( 1, m 1 ) an (, m 2 ) ( 2, m 2 ) then (, m 2 + m 1 ) ( 1 2, 1m 2 + 2m 1). (iii) If (, r 1 ) ( 1, r 1 ) in D R an (, m 2 ) ( 2, m 2 ) in D M, then in D M. (, r 1 m 2 ) ( 1 2, r 1m 2) Proof. (i) Since 1 D we have 1m 1m it follows that (, m) (, m). If (, m 1 ) (, m 2 ) there exists a D such that m 2 m 1 so also (, m 2 ) (, m 1 ). If (, m 1 ) (, m 2 ) (, m 3 ) there exist 4, 5 D such that 4 m 2 4 m 1 an 5 m 3 5 m 2. So 4 5 m m m m m m 1, an (, m 1 ) (, m 3 ) since 4 5 D. So is inee an equivalence relation. (ii) There exist 1, 2 D such that 1 m m 1 an 2 m m 2. So ( m 2 + m 1 ) m m m m ( 1m 2 + 2m 1) (iii) There exist 1, 2 D such that 1 r r 1, 2 m m 2. So 1 2 r 1m 2 ( 1 r 1)( 2 m 2) ( 1 1r 1 )( 2 m 2) r 1 m 2.
2 22 We write m for the equivalence class of (, m), an D 1 M for the set of equivalence classes on D M. The lemma shows that the aition m 1 + m 2 : m 2 + m 1 is well efine on S 1 M, an that the multiplication r 1 m2 : r 1m 2 of an r 1 D 1 R an an m 2 D 1 M is also well efine. Lemma 7.2. Let M be a R-moule an D R a multiplicatively close subset. (i) D 1 R with the aition an internal multiplication just efine is a ring with unit an It is the zero ring if an only if 0 D. (ii) D 1 M with the aition an external multiplication just efine is a D 1 R-moule. D 1 M 0 if an only if for all m M there exists a D such that m 0. (iii) Suppose f : M N is an R-moule homomorphism. Then D 1 f : D 1 M D 1 N efine by (D 1 f)( m f(m) ) : is a D 1 R-moule homomorphism. (iv) Suppose f : R S is a ring homomorphism. Then f(d) S is a multiplicatively close subset an D 1 f : D 1 R f(d) 1 S efine by (D 1 f)( r f(r) ) : f() is a ring homomorphism. (v) The map ν : R D 1 R : r r 1 is a ring homomorphism. The kernel of ν is {r R; D : r 0}. In particular, the map is injective when R is an integral omain. (vi) We can consier D 1 M as an R-moule by r m : rm. Then the map ν : M D 1 M : m m 1 is an R-moule homomorphism. Proof. (i) an (ii) We shall constantly use that R is a ring an M an R-moule. The verifications are a bit long but straightforwar. Associativity of aition: ( m 1 + m 2 ) + m 3 m 2 + m 1 + m 3 m 3 + ( m 2 + m 1 ) ( ) m 3 + m 2 + m 1 ( m 2 + m 3 ) + m 1 ( ) m 1 + m 2 + m 3 m 1 + ( m 2 + m 3 ). Commutativity of aition: m 1 + m 2 m 2 + m 1 m 1 + m 2 m 2 + m 1. We have m + m m + m since (, 0) (1, 0). So (D 1 M, +) is a commutative group.
3 23 The associativity: The istributivity laws: ( r 1 r 2 ) m 3 r 1r 2 m3 (r 1r 2 )m 3 ( ) r 1(r 2 m 3 ) ( ) r 1 ( r 2 m 3 ). r 1 ( m 2 + m 3 ) r 1 3m 2 + m 3 r 1( m 2 + m 3 ) r 1 m 3 + r 1 m 2 r 1 m 3 + r 1 m 2 r 1m 2 + r 1m 3 r 1 m 2 + r 1 m 3. ( r 1 + r 2 ) m 3 r 2 + r 1 m 3 (r 2 + r 1 )m 3 Commutativity of multiplication in D 1 R: r 2 m 3 + r 1 m 3 r 2 m 3 + r 1 m 3 r 1m 3 + r 2m 3 r 1 m 3 + r 2 m 3. r 1 r 2 r 1r 2 r 2r 1 r 2 r 1. The unit is : 1 m 1 1m 1 m. We have 0 1 in D 1 R if an only if there exists a D such that 1 0 0, i.e., if 0 D. By efinition m 0 if an only if there exists a D such that m 0 0. (iii) Let m 1, m 2 D 1 M an r D 1 R. Then (D 1 f)( r m 1 + m 2 ) (D 1 f)( rm 1 + m 2 ) f(rm 1 + m 2 ) ) rf(m 1 ) + f(m 2 ) ) r f(m 1 ) + f(m 2) r (D 1 f)( m 1 ) + (D 1 f)( m 2 ). So D 1 f is inee a D 1 R-moule homomorphism. (iv) We skip the proof. (v) an (vi) This follows from m 1 + m 1 1m +1m 1 1 m+m 1 an r m 1 r 1 rm 1 rm 1 1 rm 1. If r R is in the kernel, then r 1 0, i.e., there exist a D such that r 0. Proposition 7.1. Let R be a ring with a multiplicatively close subset D. Let f : R S be a ring homomorphism such that for all D the image f() S is a unit. Then f factors uniquely through ν : R D 1 R: there is a unique ring homomorphism g : D 1 R S such that f g ν.
4 24 Proof. Existence: Let r 1 r 2, i.e, there exists a D such that r 1 r 2. So f( )f( )f(r 1 ) f( )f( )f(r 2 ), an since f() is a unit in S for every D, we get f( ) 1 f(r 1 ) f( ) 1 f(r 2 ). This shows that the map g( r ) : f() 1 f(r) is well efine. We check that g is a ring homomorphism. Let r 1, r 2 then g( r 1 ) + g( r 2 ) f( ) 1 f(r 1 ) + f( ) 1 f(r 2 ) f( ) 1 (f( )f(r 1 ) + f( )f(r 2 )) g( r 1 + r 2 ); g( r 1 ) g( r 2 ) f( ) 1 f(r 1 ) f( ) 1 f(r 2 ) f( ) 1 f(r 1 r 2 ) g( r 1 r2 ) an g(1) g( 1 1 ) f(1) 1 f(1) 1. Unicity: Suppose g : g : D 1 R S exists such that f g ν. Then So g( 1 ) g(( ) 1 ) ( g( )) 1 ( g ν()) 1 f() 1. g( r ) g(r 1 ) g(1 ) f(r)f() 1 g( r ), i.e. g g. Example 7.1. For any prime ieal p of R its complement D : R\p is a multiplicatively close subset We write M p : D 1 M in this case. For any f R the set D {1, f, f 2,..., f i,...} is a multiplicatively close subset of R. In this case we write M f : D 1 M. The ring R f is isomorphic to the ring R[X]/(Xf 1) use in the proof of the last version of the Nullstellensatz. The subset D R of non zero-ivisors is also a multiplicative subset. In that case we write D 1 R Q(R), the maximal fraction ring of R. If R is a omain, Q(R) is the quotient fiel of R Exactness of localization. A sequence of R-moule homomorphisms... δ i+2 δ i+1 δ M i+1 M i δ i 1 i Mi 1... is calle exact if for all i we have Im(δ i+1 ) Ker δ i. Localizing keeps this property intact, we say that localizing is an exact operation. This is a very important property of localizing. A short exact sequence is an exact sequence of the form 0 M f M g M 0; i.e., f is injecive, g is surjective an the image of f equals the kernel of g. Proposition 7.2. The localize sequence of an exact sequence of R-moules... D 1 δ i+2 D 1 D M 1 δ i+1 i+1 D 1 D M 1 δ i i D 1 D M 1 δ i 1 i 1... is an exact sequence of D 1 R-moules. In particular, localization preserves injectivity an surjectivity. Proof. Fix i. Let m i+1 i+1 D 1 M i+1, then (D 1 δ i D 1 δ i+1 )( m i+1 i+1 ) δ i δ i+1 (m i+1 ) i+1 0
5 since δ i+1 (m i+1 ) Im δ i+1 Ker δ i. So Im D 1 δ i+1 Ker D 1 δ i. Conversely, let m i i Ker D 1 δ i, δ i.e. i (m i ) i 0, i.e. there is a D such that δ i (m i ) 0 δ i (m i ). So m i Ker δ i Im δ i+1, say m i δ i+1 (m i+1 ) an (D 1 δ i+1 )( m i+1 i ) m i i m i i Im(D 1 δ i+1 ). Corollary 7.1. (i) If N is a submoule of M, then D 1 N is a submoule of D 1 M an D 1 (M/N) is isomorphic to D 1 M/D 1 N. (ii) If M 1 an M 2 are two R-moules, then D 1 (M 1 M 2 ) D 1 M 1 D 1 M 2. If N 1, N 2 are two submoules of M then D 1 (N 1 N 2 ) D 1 N 1 D 1 N 2 an D 1 (N 1 +N 2 ) D 1 N 1 +D 1 N 2. (iii) Every D 1 R-submoule of D 1 M is of the form D 1 N. It follows that if M is a Noetherian R-moule, then D 1 M is a Noetherian D 1 R. Proof. (i) Localizing the short exact sequence we get the short exact sequence 0 N f M 0 D 1 N f D 1 M g M/N 0, g D 1 M/N 0, from which follows that D 1 N is a submoule of D 1 M an D 1 (M/N) is isomorphic to D 1 M/D 1 N. (ii) The map D 1 (M 1 M 2 ) D 1 M 1 D 1 M 2 : (m 1, m 2 ) ( m 1, m 2 ) is an D 1 R-moule homomorphism, as is straightforwarly checke, with inverse Similarly it is shown that ( m 1, m 2 ) (m 1, m 2 ). D 1 (N 1 + N 2 ) D 1 N 1 + D 1 N 2 : (n 1 + n 2 ) is an D 1 R-moule isomorphism The exactness of the sequence gives the exactness of 0 N 1 N 2 M M/N 1 M/N 2, ( n 1 + n 2 ) 0 D 1 (N 1 N 2 ) D 1 M f D 1 M/D 1 N 1 D 1 M/D 1 N 2, where we use the isomorphism D 1 (M/N 1 M/N 2 ) D 1 M/D 1 N 1 D 1 M/D 1 N 2. Since the kernel of f is D 1 N 1 D 1 N 2, we get the isomorphism D 1 (N 1 N 2 ) D 1 N 1 D 1 N 2. We coul also have starte from the short exact sequence 0 N 1 N 2 N 1 N 2 N 1 + N 2 0. (iii) We use the map ν : M D 1 M. Let U be a D 1 R-submoule of M. Put N : ν 1 (U) for its preimage in M. It is an R-submoule of M. Then ν(n) D 1 N is a D 1 R-submoule of U. Let m U, then m 1 is also in U, hence m N an m D 1 N. So D 1 N U. If N is finitely generate by n 1,..., n r, then U is finitely generate by n 1 1,..., nr 1. So if M is Noetherian, then also D 1 M is Noetherian. 25
6 26 In particular, if I R is an ieal then D 1 I D 1 R is also an ieal, it is the ieal generate by the image ν(i). Every ieal J of D 1 R is of the form D 1 I for some ieal I R (we can take I ν 1 (J)). If I (r 1,..., r m ) then D 1 I ( r 1 1,..., rm 1 ). So if R is noetherian then also D 1 R is Noetherian Corresponence theorems. Fix a ring homomorphism f : R S. For any ieal I R write I e S, the extension of I, for the ieal generate by the image f(i). For any ieal J S write J c R, the contraction, for the preimage ieal f 1 (J). Clearly, if I 1 I i then also I1 e Ie 2, an if J 1 J 2 then J1 c J 2 c. Write C for the collection of all contracte ieals in R, an E for the collection of all extene ieals in S. These to sets are in natural bijection an inuce a corresponence theorem. Proposition 7.3 (Corresponence theorem for f : R S). (i) For any ieal I R we have I I ec an I ece I e. For any ieal J S we have J ce J an J J cec. (ii) The operations () e an () c give bijections between C an E, preserving the inclusion relation. (iii) If J 1, J 2 E, then J 1 J 2 E an J 1 + J 2 E. If I 1, I 2 C, then I 1 I 2 C an I 1 C. Proof. (i) Let r I, then f(r) I e, so r I ec. The ieal J ce is the ieal generate by f(j c ). If r J c then f(r) J, so the ieal generate by f(j c ) is containe in the ieal J. Since I I ec we get I e I ece, an since (I e ) ce I e we conclue I ece I e. That J c J cec is prove similarly. (ii) Every ieal in C is of the form J c for some ieal J S, an every ieal in E is of the form I e, for some ieal I R. Since I ece I e an J c J cec we see that contraction an extension are inverse operations. (iii) Exercise. For a given ring homomorphism we have to etermine what the sets E an C are to be able to profit from the corresponence theorem. If J S is a prime ieal, then J c is a prime ieal of R, but if I R is a prime ieal of R then I e is not a prime ieal in general. A primary ieal q is an ieal with the property that if xy q an x q then there is an n 1 such that y n q. For example, the primary ieals in Z are the ieals generate by the power of a prime number. Again, if J S is a primary ieal, then J c is a primary ieal of R. Proposition 7.4. Let a R be an ieal an ν a : R R/a the quotient map. Every ieal of R/a is an extene ieal an an ieal I of R is a contracte ieal if an only if I a. By restriction we also get a one-to-one corresponence between the set of prime ieals (resp. primary ieals, maximal ieals) in R containing a, an the prime ieals (resp. primary ieals, maximal ieals) in R/a. For localization we get the following version. Proposition 7.5. Let D be a multiplicatively close subset of R. Consier the ring homomorphism ν : R D 1 R. (i) Every ieal of D 1 R is an extene ieal. An ieal I R is a contracte ieal if an only if it has the property that if r I with D an r R then r I. In case q R is a primary ieal
7 with raical q p, then q is a contracte ieal if an only if D p. If I R is a primary ieal that is not a contracte ieal, then I e D 1 R. (ii) The operation () e inuces an inclusions preserving bijection between the set of prime ieals (resp. primary ieals) in C an the set of prime ieals (respectively primary ieals) of D 1 R. The inverse is given by the operation () c. (iii) Suppose I r i1 q i, where q i are primary ieals orere in such a way that q i D if an only if i s. Then I ec s i1 q i. Proof. (i) By Corollary 7.1(iii) every ieal of D 1 R is an extene ieal. We always have I ec I. Suppose I I ec an r I for some D an r R. So r 1 r D 1 I I e an so I ec I. Conversely, suppose I has the formulate property, an r I ec. Then for some a I an D we have r 1 a. This means there exists a D such that r a I. So by the assume property, r I. So I ec I. Suppose q is a primary ieal such that q D. If r q for some D an r R, then q so r q, by the property of being primary. So q is contracte. Suppose there exists a D q. Then there exists an n such that n q D, so 1 n q e n an q ec R q so q is not a contracte ieal. (ii) Suppose q is a primary ieal such that q D an r 1 r 2 q e. So there exist a q,, D such that a r 1 r 2. So r 1 r 2 q. Suppose ( r 1 ) n q e for all n 0. Then at least r1 n q for akk n. We conclue that r 2 q, so r 2 q e. So q e is a primary ieal. (iii) Exercise. Corollary 7.2. Suppose R is a ring, I an ieal an D R a multiplicative subset such that if a D, for all D an a I. There exists a prime ieal p of R containing I an with empty intersection with D. Proof. The proof uses the two corresponence theorems. The conition implies that D 1 I D 1 R, or D 1 R/D 1 I is not the zero ring. So by Krull s theorem, D 1 R/D 1 I contains a maximal ieal, that correspons to a prime ieal P of D 1 R containing D 1 I. It is of the form D 1 p for some prime ieal p with empty intersection with D. Since D 1 I D 1 p we also get I p. Corollary 7.3. (i) Suppose R is a ring an D R is a multiplicative subset. If R is catenary then D 1 R is catenary. If R is noetherian then D 1 R is noetherian. In particular, any localization of an affine ring is Noetherian an catenary. (ii) A ring is catenary if for every pair of prime ieals p 1 p 2, all maximal chains of prime ieals in R p2 /p 1p2 (R/p 1 ) p2 have the same length. The following will be use in the proof of Going-own in the next section. Lemma 7.3 (Lying-over). Let f : R S be any ring homomorphism. A prime ieal p R is a contracte ieal, i.e. p p ec if an only if it is the contraction of a prime ieal in S. It is not necessarily true that p e is a prime ieal. Proof. Let p be a contracte prime ieal, we want to show that it is also the contraction of some prime ieal of S. By assumption p p ec. Let D : f(r\p). It is a multiplicatively close subset of S isjoint from p e, since otherwise there woul be an r p such that f(r) p e or r p ec p, 27
8 28 contraiction. Let P be maximal such that P p e an P D. By the corresponence theorem for localization, D 1 P is a maximal ieal of D 1 R an so P is a prime ieal of S. Now p p ec M c. On the other han, if r P c but r p, then f(r) P an f(r) D. Which is a contraiction. So p P c is the contraction of a prime ieal. In other wors, Lying-over hols for a prime ieal if an only if it is a contracte ieal. Example 7.2. Consier Z Z[i]. The extene ieals in Z[i] are those principal ieals that are generate by a non-negative integer. All ieals of Z are contracte ieals. 2Z[i] (or 5Z[i]) is an extene ieal that is not a prime ieal (since (1 + i)(1 i) 2Z[i] but 1 + i an i are not in 2Z[i]), whereas (1 + i)z[i] (resp, (1 + 2i)Z[i]) is not an extene ieal, but it is a prime ieal having 2Z (resp 5Z) as contraction The ieal 3Z[i] is prime.
9 29 8. Proof of the Cohen-Seienberg theorem 8.1. Proof of Lying-over, Going-up an Incompatibility. Starting from an integral extension, we get other integral extension with useful corresponence theorems. Lemma 8.1. Let R S be an integral extension of rings, I S an ieal with contraction I e R. Then R/I c S/I is also an integral extension of rings. Lemma 8.2. Let R S be an integral extension of rings an D R a multiplicative subset. Then D 1 R D 1 S is also an integral extension of rings. Proof. Let s D 1 S. There is an integrality relation s n + r 1 s n r n, where r i R. We obtain an integrality relation of s over D 1 R: s n + r 1 s n rn. n Proof of Cohen-Seienberg theorem 6.1(i-iv). (i) The extension R/p S/P is also integral, an one of the the two omains is a fiel if an only if the other is a fiel. (ii)[incompatibility] We suppose Q P are two prime ieals of S with the same contraction p in R. Put D : {r R; r p}. We get an integral extension R p D 1 R D 1 S with unique maximal ieal D 1 p R p. Since Q D P D p D the ieals D 1 P D 1 Q of D 1 S are prime ieals, by the corresponence theorem of localisation. We have D 1 P D 1 R D 1 Q D 1 R D 1 p, a mximal ieal, so D 1 P an D 1 Q are both maximal ieals, by i. So from D 1 Q D 1 P we conclue D 1 Q P i.e. P e Q e. Hence by the corresponence theorem, P Q. (iii)[lying-over] Let p be a prime ieal of R, an put D {r R; r p}. Then we have an integral extension R p D 1 R D 1 S an D 1 p is the unique maximal ieal of R p. Let P be a prime ieal of S such that D 1 P is a maximal ieal of D 1 R. Then by (i) the intersection D 1 P D 1 R is a maximal ieal, hence D 1 p. So P R (D 1 P D 1 S) R (D 1 P D 1 R) R D 1 p R p, which shows that there is a prime ieal in S lying over p. (iv) We get an integral extension R/p m S/P m. By the corresponence theorem we can assume R an S are omains, m 0 an P 0 (0). By Lying-over there is a prime ieal P 1 in S lying over p 1. Of course, P 0 (0) P 1. We repeat this process to go up all the way Proof of Going-own theorem. The Going-own theorem is the trickiest to prove of the Cohen-Seienberg theorems. We nee the extra assumption that R an S are omains, an that R is integrally close in its fiel of fractions K : Q(R). We say that s is integral over an ieal I R if there exists an integrality relation s + r 1 s r 0 where all coefficients lie in I, i.e., r i I for 1 i. The collection of all elements integral over a given ieal I can be escribe. Lemma 8.3. Let R S be an integral extension of omains an suppose that R is integrally close in its fraction fiel K Q(R). Let I R be an ieal. (i) The collection of all elements in S that are integral over I equals the ieal I e, the raical of the extension to I in S. (ii) Let s S be integral over I. Then it is algebraic over K an the coefficients of its minimal polynomial lie in I, the raical of I.
10 30 Proof. (i) Let s + r 1 s r 0 where all coefficients lie in I be an integrality relation for s S, where all coefficients lie in I. So s r 1 s 1 r 2 s 2... r I e an s I e. In the other irection, let s I e, so there exists an equation s n i0 a is i, where a i I an s i S. Since S is integral over R, the subring M : R[s 1,..., s n ] of S is finitely generate over R an s M IM. Let m 1,..., m r be a generating set for M as R-moule. Then there are b ij I such that s m i j b ijm j. Let F be the characteristic polynomial of the r r-matrix (a ij ). By Cayley-Hamilton s theorem, see the proof of Proposition 4.1 an the remark following it, F (s ) acts trivially on M, an since 1 M we get F (s ) F (s )1 0. Since every a ij is in I, the coefficients (except the principal) of F lie in I. So s satisfies an integrality relation over I; the same equation can serve as integrality relation of s over I. (ii) Let s S be integral over I then it is algebraic over K. Let F K[X] be the minimal polynomial of s over K, an f(x) R[X] be a monic polynomial with non-principal coefficients in I such that f(s) 0. We shall work with another integral extension. Let L be an algebraic closure of the quotient fiel of S, an let S be the integral closure of R in L. In particular it contains S. Since L is algebraically close there are s 1,..., s n such that F n i1 (X s i) L[X]. Since F ivies f in K[X], for every i we have f( s i ) 0, so each s i is integral over I. Since the nonprincipal coefficients of F are homogeneous polynomials in the s i s it follows that the non-principal coefficients of F are elements of K that are integral over I. Since R is integrally close in K, it follows that all these coefficients lie in R an are integral over I. By (i), it follows that these elements lie in the raical of I. Proof of Going-own theorem 6.1(v). We start with a special case, the general case will follow iteratively. Let R S be an integral extension of omains, where R is integrally close in it s fiel of fractions K. Let p 1 p 2 be two prime ieal of R an P 2 a prime ieal of S such that P 2 R p 2. We want to show there exists a prime ieal P 1, containe in P 2 such that P 1 R p 1. By the corresponence theorem for localization the prime ieals containe in P 2 are in one-to-one corresponence with the prime ieals in S P2. So what we are looking for is a prime ieal, say Q, in S P2 such that Q R p 1. By Proposition 7.3, such a prime ieal exists if an only if p 1 is a contracte ieal in R for the extension R S P2, i.e. p 1 S P2 R p 1. Let r s p 1S P2 R with s p 1 S an S\P 2 an suppose r p 1. We inten o erive a contraiction. By the previous lemma s is integral over p 1 an its minimal polynomial over the fiel of fractions of R is of the form X n + r 1 X n r n, where each r i p 1. Now X n + r 1 r X n 1 + r 2 X n rn r 2 r has as root r 1 s S\P n 2 an the coefficients lie in K. It is the minimal polynomial of over K. By the previous lemma it follows that r i We suppose that r p 1, it follows from the primeness of p 1 an r i ri r i all i. It follows that R an r r i i p 1. p 1 for r i p 1 that r i r i n ( r 1 r n 1 + r 2 r 2 n r n r n ) p 1S p 2 S P 2, hence P 2. Which is the esire contraiction. This finishes the proof of the Going-own theorem.
11 8.3. Invariant theory an going own. The Going-own theorem was prove uner the hypothesis of integral closeness of the small ring. There are other situations where Going-own hols, even when this hypothesis oes not hol. This is the case in invariant theory of finite groups. Proposition 8.1. Let S be a ring an G a finite group of ring automorphisms. Put R : S G for the ring of invariants. (i) The extension R S is integral, so Lying-over, Incompatibility an Going-up hol. (ii) If P 1 an P 2 are prime ieal of S both lying over the same prime ieal of R, then there exists a σ G such that σ(p 1 ) P 2. Or in other wors, the prime ieals of R are in corresponence with the G-orbits of prime ieals in R. (iii) Going-own also hols (without any extra assumptions on S an R). (iv) If S is a omain integrally close in its fiel of fractions L, then R is also omain integrally close in its fiel of fractions K, an K L G. Proof. (i) Let s S, then F σ G (X σ(s)) is a polynomial such that σ(f ) F, when we exten the action to S[X] by trivial action on X. So F S[X] G S G [X] R[X] an F (s) 0, so s is integral over R. (ii) Let x P 1, then y σ G σ(y) is an invariant containe in P 1. Since P 1 R P 2 R, it follows that y P 2. Since P 2 is prime one of the factors σ(x) is in P 2, or x σ 1 (P 2 ). So P 1 is containe in the union σ G σ(p 2 ). By prime avoiance, see Lemma 8.4, we get that there exists a σ G such that P 1 σ(p 2 ). Since σ(p 2 ) R σ(p 2 R) P 2 R P 1 R, it follows from the incompatibility theorem that P 1 σ(p 2 ). (iii) Suppose p 1 p 2 are prime ieals in R an P 2 a prime ieal of S such that P 2 R p 2. By Lying-over an Going-up there are prime ieals Q 1 Q 2 such that Q i R p i. By (ii) there exists a σ G such that σ(q 2 ) P 2. Then P 1 σ(q 1 ) σ(q 2 ) P 2 an P 1 R σ(q 1 ) R p 1. So we prove Going-own in this special case. The general case follows iteratively. (iv) Write L for the fraction fiel of S an K for the fraction fiel of R. Then G also acts on L by fiel automorphisms σ ( ) a b : σ(a) σ(b) (check that the action is well efine at least). We certainly have K L G, the invariant fiel of the action. But we even have equality: suppose z a b LG. Write b 1 σ G σ(b) an a 1 a σ 1 G σ(b). Then z a 1 b 1, b 1 S G R an therefore a 1 zb 1 L G S S G R. Or z a 1 b 1 K. Let now z K be integral over R, then it is also integral over S, hence z S K S L G S G R. Remark. Consier the corresponence theorem for the integral extension R S G S. By Lyingover all the prime ieals of R are contracte ieals. If we have projection operator F : S R as in Proposition 3.1, then every ieal in R is a contracte ieal. This oes not nee the case without the presence of such a projection operator Prime avoiance. By efinition, if a prouct of two elements lies in a prime ieal, then at least one of the two elements lies in the prime ieal. The analogous property hol for ieals: if a prouct of ieals is containe in a prime ieal, then at least one of those ieals is containe in the 31
12 32 prime ieal. Another property is: if an ieal is containe in the union of two prime ieals, then it necessarily lies in at least one of the two. Lemma 8.4 (Prime avoiance lemma). Let R be a ring. (i) Let I 1,..., I n be ieals an p a prime ieal containing the intersection i I i (or containing only the proiuct ieal I 1 I 2 I n ). Then I i p for some i. If even p i I i then p I i for some i. (ii) Let p 1,..., p n be prime ieals an I an ieal containe in the union i p i. Then I p i for some i. Proof. (i) Suppose non of the I i s is containe in p. Then for every i we can pick an x I i such that x i p. The prouct x : x 1 x 2... x n i I i i I i p is then in the prime ieal p hence one of the x i is in p i. Contraiction. So there eixsts an i such that I i p. If even p i I i, then I i p also. (ii) We prove this result by inuction on n. If n 1 it is obviously true. Assume the result is true for n 1. If I is containe in the union of the prime ieals except one, we can use inuction to conclue. Suppose that for every i the ieal I is not containe in j i p j. So for every i we can pick an x i I such that x i p j, if j i. Since I is nevertheless in the union of all prime ieals, it follows that x i p i. Consier now the following element containe in I; n x x 1 x 2... x j 1 x j+1 x j+2 x n. j1 If i j the summan x 1 x 2... x j 1 x j+1 x j+2 x n is containe in p i, since x i appears as a factor. So x p i if an only if x 1 x 2... x i 1 x i+1 x i+2 x n p i. If so, then at least one factor x j (j i) is containe in p i (since p i is a prime ieal). Contraiction. So x is not containe in the union of the p i s, hence not containe in I. Contraiction. So there is an i such that I is containe in p i.
13 33 Département e mathématiques et e statistique, Université e Montréal, C.P. 6128, succursale Centre-ville, Montréal (Québec), Canaa H3C 3J7 aress: broera@dms.umontreal.ca
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