MATH 603: Commutative Algebra (Spring 2011)
Title: Commutative Algebra
Course web site: http://www.math.umd.edu/~jmr/603/
Meeting times: MWF, 9:00am-9:50am (MTH 0411)
Instructor: Professor
Jonathan
Rosenberg. His office is room 2114 of the Math Building,
phone extension 55166, or you can contact him by
email.
Office hours are Mondays and Wednesdays 10-11, or by appointment.
Professor Rosenberg will be away in Japan from February 16 to
February 24. Professor Washington will teach the class on
February 16 and 18, and there will be no class on
February 21 or 23. Class will resume as usual on February 25.
Also, there will be no class on Monday,
April 25.
Text:
M. Atiyah and I. MacDonald, Introduction to Commutative Algebra,
Westview Press,
ISBN 978-0-201-40751-8. Other good references for certain topics are:
- N. Bourbaki, Commutative Algebra, various editions in French
and English. Reprinted, Springer, 1998. The bible on the subject,
but probably much more than you really want to know.
- D. Eisenbud, Commutative Algebra, with a view toward
algebraic geometry, Springer, 1995. A slightly different
spin than A-M, slightly more advanced also, and covering
more topics.
- H. Matsumura, Commutative Algebra, 2nd ed.,
Benjamin, 1980. A very good book, somewhat more advanced than A-M.
- M. Reid, Undergraduate Commutative Algebra,
Cambridge University Press, Cambridge, 1995. Slightly lower level
than A-M.
- R. Y. Sharp, Steps in commutative algebra,
London Mathematical Society Student Texts, 19.
Cambridge University Press, Cambridge, 1990. 2nd edition, 2000.
Less terse than A-M, and at a slightly lower level.
- O. Zariski and P. Samuel, Commutative Algebra, 2 vols., Springer, 1958.
The original classic text on the subject. Still valuable for some things.
Prerequisite: Math 600 or equivalent, or in other words,
one semester of graduate-level abstract algebra, getting up to basics
of rings, modules, and ideals. Math 601 can be taken concurrently.
Catalog description:
Ideal theory of Noetherian rings, valuations, localizations, complete local rings, Dedekind domains.
Course Description:
Commutative algebra means the study of commutative rings and
modules over them. The subject is motivated by
applications in
algebraic geometry, number theory, and algebraic topology.
This course will be a basic introduction to commutative algebra,
starting from the basics, and covering what you need to know
for basic
applications to algebraic geometry, number theory, and algebraic topology.
We will cover all of Atiyah-MacDonald, and then depending on time,
we will cover some more modern topics, especially those involving
homological and category-theoretic methods.
Course Requirements:
Homework will be assigned, collected and graded regularly,
but there will be no exams. If you want to get an "A" in the course, you
need to do most of the homework in a satisfactory way.
Course evaluations. Please
fill out the on-line course
evaluation questionnaire at CourseEvalUM before May 11.
Homework Assignments
Conventions for these problems: "ring" means commutative ring
with unit, unless stated otherwise. Integral domains are also
assumed commutative with unit. Z denotes the ordinary integers,
N denotes the non-negative integers,
Q denotes the rational numbers, and
Fq denotes the finite field of
q elements.
- Exercises on ideals (based on A-M, Ch. 1) (due Monday, January 31)
- Show that an integral domain with only finitely many elements
is a field. Deduce that if R is a ring with only finitely many
elements, then every prime ideal of R is maximal, and the nilradical
and Jacobson radical of R coincide.
- Show that if R is a ring, the set of zero-divisors in R
is a union of prime ideals of R. (This is A-M, Ch. 1, #14.)
- If R is a ring and J is an ideal in R, the
nilradical of J, denoted √J, is the set of
x ∈ R with xn ∈ J
for some n. Show that √J is the intersection
of the prime ideals containing J. In particular, it's an ideal
containing J.
- Let R = F3[x,
y]/(xy, x2). Give as explicit a list
as possible of all the prime ideals in R. Which ones are maximal?
What are the radical and nilradical of R?
- Exercises on modules (based on A-M, Ch. 2) (due Friday, February 11)
- Show that the finite generation hypothesis in Nakayama's Lemma cannot
be removed, even over a local ring. (Hint: Let R be the local ring
of germs at 0 of continuous (real-valued) functions on small intervals
(-ε, -ε) of
the real line. Show that the maximal ideal M of R is equal to
M2. Why would this contradict Nakayama's Lemma if it weren't
for the finite generation hypothesis?)
- Show that if R is a ring and M is an R-module
with the property that every finitely generated submodule of M is
flat, then M is flat.
- Let R be a ring. An R-module is called torsion-free
if x in M, a in R, and ax = 0
implies x = 0 or a = 0. Show that if R is a PID
and an R-module M is torsion-free, then M is flat.
Use the previous problem plus the fact (which you can assume) that every
finitely generated torsion-free module over a PID is a free module.
- Let R be a ring. Show that an R-module M is flat
if and only if it has the property that every linear dependence
relation in M comes from relations in R, i.e., that
if Σi
ai xi = 0 in
M, with ai in R and
xi in M,
then there exist finitely many bij in
R and yj in M such that
xi = Σj
bij yj for each i
and Σi bij
ai = 0 for each j.
- Exercises on localization (based on A-M, Ch. 3) (due Friday, February 25)
- Let R be a local ring with maximal ideal p. Show
that the localization map R → Rp
is an isomorphism.
- Use Nakayama's Lemma to show that a finitely generated projective module
P over a local ring R
is free (of finite rank). (Hint: reduce P
modulo mP, m the maximal ideal, to get a vector space over the field
R/m, choose
a basis and pull back. You want to show you get a free basis for
P. Use the projectivity by doing this simultaneously for
P and for a complement Q, whose direct sum is a free module.)
Then show that if R
is any ring and P is a finitely generated projective module, then
for any prime ideal p of R, Pp is a finitely
generated projective module over Rp, hence a free
module of finite rank, and
p → rank Pp is a well-defined continuous
function Spec R → N.
Solution to the hardest part:
If P is a finitely generated projective R-module,
and if p is a prime ideal of R,
then Pp is a finitely generated projective module
over the local ring Rp, hence it is free by the
previous part, and so its rank is well defined. We need to show that
the rank is a continuous function of p ∈ Spec R.
If P is a direct summand in Rm, then
P = RmA (as a left R-module) for some
idempotent m × m matrix A with entries in
R.
Then rank Pp is the rank of A after
localizing the entries at p.
So rank Pp = n means there is an
n × n submatrix of A
with determinant not in p, and that any larger square
submatrix of A does have determinant in p. But the
minor determinants of A are elements of R, and for one
of them to lie in p is a closed condition in the Zariski topology.
(The prime ideals containing a given ideal are a closed set in Spec
R.) Thus the condition of Pp
having rank less than a specified value is a closed condition. But
the condition of Pp
having rank greater than a specified value is equivalent to the
complement of P in Rn having rank
less than a specified value, which is a closed
condition on minor determinants of 1 − A, so rank
Pp is both upper and lower semicontinuous and is
thus continuous.
- Let R = Z[x]/(x2 − 1).
Show that J = (x − 1) is a prime ideal in
R and show that RJ is isomorphic to
Q. What is the kernel of the map R →
RJ ?
- Suppose (a) is an ideal in R, R a PID, and
suppose the cyclic R-module R/(a) vanishes when
localized at (p), p irreducible. What are the
possibilities for a?
- Exercises on primary decomposition (based on A-M, Ch. 4) (due Monday,
March 7)
- Show that if R is the ring of continuous real-valued
functions on [0, 1], then the zero ideal in R is not decomposable.
Solution to this problem: Suppose
we had primary ideals Q1, ...,
Qn, with intersection equal to the zero
ideal. The associated prime ideals P1, ...,
Pn would have to arise as radicals of annihilators
of elements fj of R. But the annihilator
I of
a function f ∈ R consists of all functions
g ∈ R that vanish on the closure of the set E
of all x ∈ [0, 1] where f(x) ≠ 0. This
ideal I is determined by E, which can be any closed
subset of [0, 1] with
dense interior. Furthermore, I is always equal to its
radical, since R/I ≅ C(E) has no
nilpotents. But I can never be a prime ideal, since if
I ≠ R, f(x) ≠ 0 on some open interval
(a, b), and C(E) has zero-divisors
(take a function vanishing on [a, (a + b)/2]
and another one vanishing on [(a + b)/2,
b]). So there can't be any associated prime ideals.
- Let R = C[x,
y]/(x − y2). Find a
minimal primary decomposition of the ideal I in R
generated by [the image of] x − 1, resp., (x
− 1)2.
- (A-M, #16, p. 56) Assume that every ideal in R is
decomposable. Show that the same property holds for any localization
S−1R of R.
- Exercises on integral closure (based on A-M, Ch. 5) (due Monday,
March 14)
- Show that if k is a field and R = k[x,
y]/(x2 − y3), then
R is an integral domain but is not integrally closed
(in its field of fractions). (From the algebraic geometry point of view,
that means the curve x2 = y3
is not a normal variety.) What is the integral closure S?
Show explicitly that S is finitely generated as an
R-module.
- Do A-M problem 1 from Ch. 5, that an integral homomorphism of
rings induces a closed mapping of spectra.
- Let d be a square-free integer (so d is not a
perfect square and is not divisible by any perfect square > 1),
and let R =
Z[√d]. Show that R is integrally closed
in its field of fractions unless d ≡ 1 mod 4, in
which case it is not integrally closed.
- Do A-M problem 2 from Ch. 5, that given a homomorphism A
→ Ω with A a ring and Ω an algebraically
closed field, and given B ⊇ A integral over A, the
homomorphism extends to one B → Ω. Exhibit a case
where B is an extension of A which is not integral
and no such extension exists. Is integrality necessary for
the extension to exist? Why or why not?
- Exercises on going down, valuations, Noether
normalization, and the Nullstellensatz (based
on A-M, Ch. 5) (due Monday, March 28)
- Show that the hypothesis that B be an integral domain is
not superfluous in the Going Down Theorem (A-M Theorem 5.16), as
follows. Let A = Z, B = Z[x] /
(2x, x2 − x). Check that
A is integrally closed in its field of fractions and that
B is integral over A, but that B is not an
integral domain. Then check that ``going down''
fails for (0) ⊂ (2) in Z, (2, x − 1) over
(2) in B. (Here we are using the slight abuse of notation of
letting x also stand for its image in B.)
- One version of the Nullstellensatz asserts that if K is
an algebraically closed field and A is a finitely generated
K-algebra ("finitely generated" here means as an algebra, not
as a K-vector space), then the inclusion K →
A splits (i.e., there is a homomorphism of K-algebras
in the other direction, such that the composite K →
A → K is the identity). Prove this assertion as
follows. Choose a maximal ideal I in A. Then show that
A / I has to be algebraic over K, hence equal
to K. (Hint: Show that a transcendental extension of
K cannot be finitely generated as a K-algebra.) Show
also that the theorem is false if K =
R, or if A is only assumed countably generated, but
not finitely generated, over K.
- An ordered abelian group means an abelian group
(written additively) with a translation-invariant transitive and
reflexive relation ≥, such that for every
x, either x ≥ 0 or −x ≥ 0, and if
both x ≥ 0 and
−x ≥ 0, then x = 0. Let K be a
field. A valuation on K is
defined to be a homomorphism ν: K× →
G, where G is an ordered abelian group (caution: we write
G additively but K× multiplicatively)
with the property that ν(x + y) ≥
min(ν(x), ν(y)) whenever this makes sense (i.e.,
none of x, y, or x + y are zero).
- Show that any ordered abelian group is torsion-free. (Hint:
first show that x ≥ 0 implies nx ≥ x
for n positive.)
- Let ν: K× → G be a
valuation. Show that it must send any root of unity to 0. (Use #1.)
- Give an example of a valuation ν on Q, with G =
Z with the usual ordering and with ν(2) = 1. (By #2, you
know that ν(−1) = 0.)
- Show that if ν is a valuation on a field K, then
{x : ν(x) ≥ 0} ∪ {0} is a valuation ring
in K (in the sense of A-M, Ch. 5).
- What is the valuation ring in your example from #3?
- Exercises on chain conditions and Noetherian
rings (based on A-M, Chs. 6-7) (due Wednesday, April 6)
- Do A-M exercises 5, 6, 8, 12 on page 79. These are all about
Noetherian topological spaces and applications to Spec of a ring.
Terminology: "quasi-compact" means "compact but not necessarily
Hausdorff", i.e., each open covering has a finite subcovering, but
distinct points need not have disjoint neighborhoods.
- Let R be a Noetherian ring and let S be the set
of non-zero divisors in R. Show that
S−1R
has only finitely many maximal ideals. Also show by example that this
is not necessarily the case if R is not Noetherian.
Solution to this
problem
- Exercises on Artinian
rings and discrete valuation rings
(based on A-M, Chs. 8-9) (due Wednesday, April 13)
- Let G be a finite abelian group, let K be a field,
and let R = KG be the group ring of G over
K. Show that R is Artinian. What can the local factors of
R look like? Show that R is semisimple (has trivial radical)
exactly when all the local factors of R are fields, which happens
if and only if the characteristic of K is relatively prime to
the order of G. (Hint: if x ∈ G has order
q and q is a power of the characteristic of K,
then (x − 1)q = 0.)
- (A-M Ch. 9, exercise 3) Show that any Noetherian valuation
ring is automatically a discrete valuation
ring.
- Deduce from exercise 2 above that if R is a PID (for example,
Z or a polynomial ring in one variable over a field) and if
P is a non-zero prime ideal of R, then the localization of
R at P is a discrete valuation
ring.
- Exercises on Dedekind domains
and completions
(based on A-M, Chs. 9-10) (due Friday, April 29)
- Let R=R[x,
y]/(x2+y2−1) be
the ring of polynomial functions on the unit circle.
- Show that R is a Dedekind domain.
- If P is a non-zero prime ideal in R, show that
R/P is isomorphic to either R or C.
Show that both possibilities can occur, and that in the first
case, P = (x−α, y−β),
where α, β ∈ R and α2
+ β2 = 1. Show that in the second case,
P is a principal ideal generated by a linear polynomial.
- Deduce from #2 that the class group of R is generated
by the ideals of the form (x−α, y−β).
Show that all these ideals define the same element of the class group,
which is of order 2, by showing that if P is a prime ideal
of this form, then P is not principal but
P2 is principal, and that
if P1 and P2 are distinct
prime ideals of this form, then P1P2
is also principal.
- For each of the following Z-modules M, compute the
P-adic completion M^, where P = (2).
Does M^ coincide with Z^
⊗Z M?
- M = Z/(3).
- M = Z/(2).
- M = Q/Z.
- M = Q.
Solutions to
this assignment
- Exercises on dimension theory
(based on A-M, Ch. 11) (due Monday, May 9)
- Is there a purely topological proof of the fact that
Noetherian rings have the descending chain condition on prime ideals?
In other words, answer the following questions:
- What property of a Noetherian topological space X
corresponds to the descending chain condition on prime ideals
of R when X = Spec R?
- Does every Noetherian topological space X have the
above property? Give a proof or a counterexample.
- Compute the Hilbert function of the graded ring
R = K[x, y, z]/(x2z
− y3 − yz2),
where K is a field and
x, y, z each have degree 1.
- Show that a regular Noetherian local ring of dimension d is
integrally closed (see comments in A-M, bottom of p. 123).
Solutions to
this assignment