MA4J8 Commutative Algebra II

Worksheet 0

These issues are intended as easy reminders of the course prerequisites.
Please think about them and let me or the TA Marc Truter know at once if
you have insuperable difficulties with them.

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(I) Exercises on B\'ezout's identity that illustrate the
Dedekind-Weber parallelism.

(1) Let a,b be coprime integers. Prove that a*x + b*y modulo a*b
takes all possible values for x in [0..b-1] and y in [0..a-1].
Deduce in particular that
   a*x + b*y = 1 + a*b
holds for some positive integers x,y.

(2) Let a,b in k[x] be coprime polynomials of degree d = deg a and
e = deg b. Prove that the polynomials
   a*f + b*g with degree f <= e-1 and deg b <= d-1
provide d-e linearly independent polynomials in k[x]_{<= d+e-1},
so base that vector space. In particular, there exist
a*f + b*g = 1 with (deg f, deg g) in that range

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(II) Reminders on localisation: Let A be a ring, M an A-module and
S in A a multiplicative set. Does f/s \twiddles g/t define an
equivalence relation on pairs f/s? Look at the definition of \twiddles
in the textbooks. [Explanation for those seeing this for the first time:
S is allowed to have zerodivisors. What happens if s*m = 0 and we try to
make s invertible.]

Define the ring of fractions S^-1A and the module of fractions S^-1M. If
j: M -> S^-1M is the standard map taking m to m/s, determine the kernel
of j.

An ideal I in A is prime if and only if the complement S = A - I
is multiplicative. Explain why we include 1 in S as part of the
definition of multiplicative set.

Prove that M -> S^-1M has the property that S^-1M is an A-module
on which S acts bijectively, and has the universal mapping property
(UMP) for that condition. (You need to figure out what that means.)

Recall the result of MA3G6 that localisation commutes with taking
quotient modules. Deduce that if M1 -a-> M2 -b-> M3 is an exact sequence
of A-modules then S^-1M1 -> S^-1M2 -> S^-1M3 is also an exact sequence.
(Hint: write out ker a and im a and the same for b, then join their
localisations together.) This can be restated as the slogan that S^-1
preserves exact sequences, or S^-1 is an exact functor.

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(III) Basic exercises on local rings. Prove that the different
definitions of _local ring_ are equivalent:
 (i) A ring A in which the set of non-units form an ideal.
 (ii) A ring A with a unique maximal ideal m in A.
 (iii) A ring with an ideal m such that 1+x is a unit
for every x notin m.

Prove that the ring k[[x]] of formal power series over a field is a
local ring. This is a precursor to the definition of completion (or
formal completion) later in the course.

If A is a ring, P a prime ideal with complement S = A - P, prove that
the localisation A_P = S^-1A is a local ring with maximal ideal P*A_P.

Prove Nakayama's lemma: For A,m a local ring and M a finite A-module,
m*M = M implies that M = 0. [Hint: let u1,.. un be a set of generators
of M. Use the given relation M = m*M to show that M is generated by only
n-1 of the ui.]

This has the effect of reducing many issues in linear algebra over a
local ring A into linear algebra over the residue field k = A/m.

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(IV) For ideals I,J in A, it is trivial that the product I*J is
contained in I intersect J. Give several counterexample to the converse.

Now let P be a prime ideal. Prove that P contains I*J implies that
it also contains I intersect J. This seems like a highbrow mystery until
you get the point, when it is trivial.

Let m1 and m2 be distinct maximal ideals of a ring A. Prove that
 (i) m1 + m2 = A,
 (ii) m1*m2 = m1 intersect m2, and
 (iii) A/(m1 intersect m2) \iso A/m1 \directsum A/m2.

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(V) As you know, a maximal ideal is prime. Many arguments use the idea
that an ideal that is maximal w.r.t. some property is prime.

(a) A basic result, "sufficiently many primes". If A is a ring,
I a nontrivial ideal of A and S a multiplicative set, consider an ideal
J that is maximal among ideals containing I and disjoint from S. Prove
that J is prime.

(b) Let M be an A-module. Define the annihilator of m in M by
  ann(m) = {f in A | fm = 0}.
Prove that an ideal that is maximal among the ann(m) for nonzero m in M
is prime. [Hint: if fg in ann(m), argue on gm and ann(gm).]

Deduce that if A is Noetherian and M <> 0 then M contains a nonzero
element m with prime annihilator, that is, a submodule A*m isomorphic to
A/P for some prime P in Spec A. This is called an _associated prime_ of
M. It is an key ingredient in many parts of the course because A/P is an
integral domain, so a solid plank for arguments.

(c) As a pretty illustration of the same principle, see the online
MA4J8_Exa1.pdf, Exc 12 for Cohen's theorem: an ideal of A that is not
finitely generated and maximal among all such is prime. It follows that
if every prime ideal is finitely generated, the same goes for every
ideal A, so that A is Noetherian.