First: listed name, prereq=593, and first semi-assignment: (please write out: 1. name, 2. previous prep in (comm)alg, 3. topics you wish to cover, 4. your purpose in taking this course, 5. preferences regarding ofc hours) Structure of the course is fluid. Lecture and some discussion. Ask questions. I'll try to distribute lecture notes, but not timely. Using Hochster notes & Atiyah-MacDonald (on res), but also Mats I (on res.), Mats II, MacLane, and others. Coverage includes: -basic category theory -Localization & fractions -Exact sequences -Flatness -Tensor Products -Integrality and Normalization -Noetherian & Artinian rings & modules -finitely generated K-algebras (K a field) -Hilbert Basis Theorem -Spec -power series and completion -Primary Decomposition and Ass -some dimension theory -number theory flavor (UFDs, PIDs, DVRs, class group, Dedekind domains) -Artin-Rees lemma Commutative algebra is a cool thing. Its roots are in algebraic geometry and number theory, and it contains many tools for doing these subjects. You can prove, using commalg, that if f,g,h \in C[x,y], then fgh \in (f^2, g^2, h^2). (It, and many statements like it, is a consequence of the "Briancon-Skoda Theorem", originally from analysis..) Unsolved problem: The Jacobian Conjecture. The simplest case of it states that for f,g \in C[x,y], f and g generate C[x,y] iff Jac(f,g) \in C. Ring = comm assoc ring w/ unity = (R,+,.,1) with . associative, distrib over addition, commutative, with identity 1 = 1_R (1x = x all x\in R) Ring Homomorphism = map f:R->S that preserves +, ., and 1. Module = abelian group M together with ring homo p:R -> End(M). Write r.x = p(r)(x). R has the R-module structure defined by addition. An (R)-submodule N <= M is a subgroup such that for every r in R, rN <= N. An ideal I is an R-submodule of R. The quotient ring is the group R/I, which inherits the obvious ring structure. The surjection R ->> R/I is a ring homo, and induces an order-preserving bijection between the ideals of R that contain I and the ideals of R/I. kernels, images, and cokernels of module homos are modules. kernels of ring homos are ideals, while images are rings. (im f = R/ker(f)) r \in R is a zero divisor on M if there is nonzero x \in M with rx= 0. A ring with no zero divisors is called an integral domain. An ideal p is PRIME if R/p is an integral domain. An ideal m is called maximal if m is maximal among _proper_ ideals of R. R/m is a field [exercise], hence an integral domain, hence maximal ideals are prime. Zorn's Lemma: Let P be a nonempty poset. Suppose every totally ordered subset of P has an upper bound in P. Then P has a maximal element. Prop: Any nonzero ring R has maximal ideals. pf: Let S = {proper ideals of R}. S is nonempty since 0 \in S. Let C be a totally ordered subset of S. Then J = U (C) is a proper ideal of R. J is proper, since if 1 \in J, then 1 \in some element I of C, and then I = R, contradiction. J is an ideal, since if x, y \in J, there is some I \in C such that x,y \in I, so that rx, x+y \in I <= J. Hence J is an upper bound in S for C. Thus by Zorn's lemma, S has a maximal element, i.e. a maximal ideal. Spec(R) := {prime ideals of R}, Omega(R) := {maximal ideals of R} Cor: R = 0 iff Spec R = \empty iff Omega(R) = \empty. x \in R is nilpotent if x^n = 0 for some n \geq 1. \N(R) = "nilradical of R" = {nilpotent elements of R} Prop: \N = \N(R) is an ideal. In fact, N = \bigcap \Spec(R). pf: x,y \in \N, x^n=0, y^m = 0 => (rx)^n = 0 and (x+y)^{n+m-1} = 0, so \N is an ideal. Note first that N <= \p \forall \p \in \Spec(R). (x \in \N => x^n = 0 \in \p => x \in \p) Let x \in R \setminus \N. Let S = {I : no power of x is in I}. The same argument as before shows that S has maximal elements. Let p be such a maximal element. Let y,z \in R such that neither is in \p. By maximality of \p in S, neither \p+(y) nor \p+(z) is in S. So x^n \in \p+(y) and x^m \in \p+(z). So x^{n+m} \in \p+(yz), hence \p+(yz) \notin S, hence yz \notin \p. Thus, \p is a prime ideal that does not contain x. Next time: Categories!