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
MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 1 Contents 1. Integrality 1 1.1. Integral closure 3 1.2. Integral elements as lattice points 3 2. Transcendental extensions 6 2.0. Purely transcendental extensions 6 2.1. Transcendence basis 6 2.2. Noether Normalization Theorem 9 3. Outline of rest of Part B 12 3.1. Valuation rings 12 3.2. Noetherian rings 13 4. Algebraic spaces 14 4.0. Preliminaries 14 4.1. Hilbert s Nullstellensatz 15 4.2. Algebraic sets and varieties 18 5. Noetherian rings 21 5.1. Hilbert basis theorem 21 5.2. Noetherian modules 22 5.3. Associated primes 24 5.4. Primary decomposition 29 5.5. Spec(R) 33 6. Local rings 34 6.1. Basic definitions and examples 34 6.2. Nakayama s Lemma 35 6.3. Complete local rings 37 6.4. Discrete valuation rings 38
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 1 + + 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 2 + + 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 2 + + 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.
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.
MATH 131B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 3 1.1. 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. 1.2. 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 1.10. 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. 1.2.1. 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 1.11. 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.