MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA

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Nora Bishop
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1 MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA I want to cover Chapters VIII,IX,X,XII. But it is a lot of material. Here is a list of some of the particular topics that I will try to cover. Maybe I won t get to all of it. (1) integrality (VII.1) (2) transcendental field extensions (VIII.1) (3) Noether normalization (VIII.2) (4) Nullstellensatz (IX.1) (5) ideal-variety correspondence (IX.2) (6) primary decomposition (X.3) [if we have time] (7) completion (XII.2) (8) valuations (VII.3, XII.4) There are some basic facts that I will assume because they are much earlier in the book. You may want to review the definitions and theorems: Localization (II.4): Invert a multiplicative subset, form the quotient field of an integral domain (=entire ring), localize at a prime ideal. PIDs (III.7): k[x] is a PID. All f.g. modules over PID s are direct sums of cyclic modules. And we proved in class that all submodules of free modules are free over a PID. Hilbert basis theorem (IV.4): If A is Noetherian then so is A[X]. Algebraic field extensions (V). Every field has an algebraic closure. If you adjoin all the roots of an equation you get a normal (Galois) extension. An excellent book in this area is Atiyah-Macdonald Introduction to Commutative Algebra. 0

2 MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 1 Contents 1. Integrality Integral closure Integral elements as lattice points 3 2. Transcendental extensions Purely transcendental extensions Transcendence basis Noether Normalization Theorem 9 3. Outline of rest of Part B Valuation rings Noetherian rings Algebraic spaces Preliminaries Hilbert s Nullstellensatz Algebraic sets and varieties Noetherian rings Hilbert basis theorem Noetherian modules Associated primes Primary decomposition Spec(R) Local rings Basic definitions and examples Nakayama s Lemma Complete local rings Discrete valuation rings 38

3 MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 1 1. Integrality Basic properties of integral extensions. All rings are commutative with 1. Definition 1.1. Suppose that R is a subring of S and 2 S. Then is integral over R if any of the following equivalent conditions is satisfied. (Equivalence proved later.) (1) is the root of a monic polynomial with coe cients in R, i.e., f( ) = n + r 1 n r n =0 for some r i 2 R. (2) The subring R[ ] S is a finitely generated (f.g.) R-module. (3) There exists a faithful R[ ]-module which is a f.g. R-module. For example any element r 2 R is integral since it it a root of the polynomial x r. The name integral comes from the following. Theorem 1.2. The only rational numbers which are integral over Z are the integers. Proof. Suppose that x = a/b 2 Q is integral over Z where a, b 2 Z are relatively prime. Then there are integers n, c 1,,c n so that x n + c 1 x n 1 + c 2 x n c n =0 multiplying by b n we get the integer equation a n + c 1 a n 1 b + c 2 a n 2 b b n =0 This implies that b divides a n. Since a, b are relatively prime this means b = ±1 and x = a/b 2 Z. Each condition in Definition 1.1 makes some aspect of integrality most apparent. The first condition immediately implies: Lemma 1.3. If is integral over R then is integral over any subring of S which contains R. For the last condition, I explained the definitions. Given any R-module M, the annihilator ann R (M) ofm in R is the set of all r 2 R so that Mr =0. Thisiseasilyseen to be an ideal. A faithful module M is one whose annihilator is 0: ann R[ ] (M) =0. Lemma 1.4. If R is a subring of S and S is f.g. as R module, then all elements of S are integral over R. Proof. For all 2 S, (1) S is a f.g. R-modules by assumption. (2) R[ ] actsfaithfullyons since 1 2 S cannot be annihilated. So, is integral over R by (3). Lemma 1.5. If R is a subring of T and T is f.g. as an R-module then any f.g. T -module is also f.g. as an R-module.

4 2 MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA Proof. Let x 1,,x n be generators of T as an R-module. Then any t 2 T can be written as t = P x j r j.ifm is a f.g. T module with generators y 1,,y m then any element of M can be written as X yi t i = X y i x j r ij So, the products y i x j generate M as an R-module. This implies: Lemma 1.6. If R T are subrings of S, 2 S is integral over T and T is f.g. as an R-module then is integral over R. Proof. T [ ] isaf.g.t-module. By Lemma 1.5, T [ ] isaf.g.r-module. By Lemma 1.4 all elements of T [ ] areintegraloverr. Proof of equivalence of three definitions. (We cannot use any of the lemmas since they all assume equivalence of the three definitions.) (1) ) (2) since 1,,, n 1 generate R[ ] asanr-module. (2) ) (3) M = R[ ] isafaithfulf.g.r[ ]-module. (3) ) (1). Suppose that M is a faithful R[ ]-module which is generated by w 1,w 2,,w n as an R-module. Then, for each w j, (1.1) w j = nx a ij w i i=1 for some a ij 2 R. ThenIclaimthat is a root of the characteristic polynomial of the n n matrix A =(a ij ) f(t) =det(ti n A)=t n Tr At n 1 + +( 1) n det A The reason is that Equation (1.1) can be written in matrix form as: or If we multiply by the adjoint matrix ( I n we get: (w 1,w 2,,w n ) I n =(w 1,w 2,,w n )A (w 1,w 2,,w n )( I n A)=(0, 0,, 0) A) ad and use the fact that ( I n A)( I n A) ad =det( I n A)I n = f( )I n (w 1,,w n )f( )I n =(f( )w 1,f( )w 2,,f( )w n )=(0, 0,, 0) Since f( )w j =0forallgeneratorsw j of M we get f( )M =0. Thisimpliesthat f( ) =0sinceM is a a faithful R[ ]-module.

5 MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA Integral closure. Proposition 1.7. If R is a subring of S then the set of all 2 S which are integral over R forms a ring (which contains R). This is called the integral closure of R in S. Proof. Suppose that, 2 S are integral over R. Then is integral over R[ ]bylemma 1.3. Any element of R[, ]isintegraloverr[ ]bylemma1.4. Soeveryelementof R[, ](e.g., +, )isalsointegraloverr by Lemma 1.6. Therefore, + and are integral over R and the integral elements form a ring. By Theorem 1.2 we have: Theorem 1.8. Z is the integral closure of Z in Q. Definition 1.9. Adomain(=entirering:nozerodivisors)iscalledintegrally closed if it is integrally closed in its fraction field. The last theorem shows that Z is integrally closed Integral elements as lattice points. Suppose V is a vector space over a field k of characteristic 0 and B = {b 1,,b n } is a basis for V. Then the additive subgroup ZB generated by B forms a lattice L in V.(Alattice in V is defined to be an additive free subgroup whose free basis elements form a basis for V as a vector space over k.) Theorem Suppose that K is an algebraic number field, i.e., a finite extension of Q. Let O K be the integral closure of Z in K. Then (1) O K is a lattice in K as a vector space over Q. (So, O K is the free additive group generated by some Q-basis for K.) (2) O K contains any other subring of K which is finitely generated as an additive group. (So, O K contains any subring of K which is a lattice.) To prove this theorem we need to review the properties of the trace example. Take K = Q(i) wherei = p 1. Then Q(i) hasanautomorphism given by complex conjugation (a + bi) =a bi Q(i) isagaloisextensionofq with Galois group Gal(Q(i)/Q) ={1, } Proposition The ring of integers in Q(i) (= the integral closure of Z in Q(i)) is O Q(i) = Z[i] ={a + bi a, b 2 Z} Proof. Certainly, Z[i] O Q(i) since 1 and i are integral elements of Q(i). Conversely, suppose that = a + bi is integral. Then ( ) =a bi is also integral. So, the sum and product of, ( ) whicharecalledthetrace and norm of are also elements of the ring O Q(i).SinceO Q(i) \ Q = Z (Theorem 1.8), these are rational integers: Tr K/Q ( ) := + ( ) =2a 2 Z N K/Q ( ) := ( ) =a 2 + b 2 2 Z Also, Tr(i ) = 2b 2 Z. Theseimplythata, b 2 Z as claimed.

Extended Index. 89f depth (of a prime ideal) 121f Artin-Rees Lemma. 107f descending chain condition 74f Artinian module

Extended Index. 89f depth (of a prime ideal) 121f Artin-Rees Lemma. 107f descending chain condition 74f Artinian module Extended Index cokernel 19f for Atiyah and MacDonald's Introduction to Commutative Algebra colon operator 8f Key: comaximal ideals 7f - listings ending in f give the page where the term is defined commutative