Math 502 (Rezk, 10:00 MWF, 141 Altgeld Hall).
Here is the syllabus.
Homework
- Friday, September 2. Matsumura 1.2, 1.3, 1.5, 2.1, 3.1,
3.2, 3.6. Here are solutions.
- Monday, September 19.
- Matsumura 4.4, 4,5, 4.6, 4.7, 4.10.
- Matsumura 9.1, 9.7.
- Let k be a prime field (Q or
Fp), and let K be an algebraically closed field having
infinite transcendence degree over k. Let A=K[X1, ...,
Xr]. The following exercise gives a quick proof of the
Nullstellensatz for such rings (e.g., the case K=the complex numbers).
- Let P be a prime ideal of A. Show that there exist a finite set
of elements
c1,...,cn of K such that, if
F is any subfield of K containing them, and Q is the contraction of P
to F[X1,...,Xr], then P=QA.
- Let be F is a subfield as in part 1 which has finite
transcendence degree over k. Show that the fraction field of
F[X1,...,Xr]/Q can be embedded in K, extending
the inclusion of $F$ in $K$. (This uses
the hypotheses on the field K.)
- Show that if f is an element of A-P, there is an
element a=(a1,...,ar) in Kr such that
a is in Z(P) but f(a)≠0. (Use a suitable embedding as in part 2 to
produce a candidate a.)
- Conclude that every prime ideal P of A is an intersection of
ideals of the form
(X1-a1,...,Xr-ar), with a
in Z(P).
- Wednesday, October 5.
- Let A be a ring which as an abelian group is finitely generated.
Let P be a prime ideal of A. Show: (a) if P is maximal, then A/P is a
finite field, and (b) if P is not maximal, then it is a minimal prime.
- Matsumura 6.3, 6.4, 6.5, 6.7.
- Show that if A is a Noetherian local ring with maximal ideal M,
the M-primary ideals I are exactly those such that I contains
Mn for some n. Show that if A is a Noetherian ring, the
P-primary ideals are exactly those which are contractions of
PAP-primary ideals of the localization AP.
- Wednesday, November 2.
- Matsumura 7.1, 7.3, 7.9.
- Matsumura 8.1, 8.2, 8.5.
- Friday, November 18.
- Matsumura 10.7.
- Matsumura 11.2, 11.3, 11.8, 11.9, 11.10.
- Matsumura 13.1, 13.5.
- Friday, December 9.
- Matsumura 14.1.
- Matsumura 15.1.
- Suppose a homomorphism from A to B is an integral extension of
Noetherian rings, and M a finite B-module. Show that
dimB (M) = dimA (M).
Last modified 30 November 2004 by Charles Rezk. Email: