Lecture 3
Proposition [Existence of primes from Lecture 2]:
A ring, S multiplicative set, I an ideal disjoint from S.
Then there exists an ideal P containing I that is maximal
w.r.t. condition P disjoint from S, and any such P is prime.
==
Radical: x in A is _nilpotent_ if x^n = 0 (for some n). For
nilpotent x,y (with x^n=0, y^m=0) the binomial theorem gives
(x+y)^{n+m} = 0. Hence the set of nilpotents of A is an
ideal, the _nilradical_ rad A of A.
In the same way, for an ideal I, the radical rad I is the
set of f in A such that some f^n in I. In this context
rad A equals rad 0 (radical of the 0 ideal.)
Cor. of Proposition
nilrad(A) = intersection of all primes P of A
rad(I) = intersect all primes P of A containing I
If x is nilpotent, it is certainly in every prime P.
Conversely, if x is not nilpotent, the multiplicative set S
= {1,x,..x^n ..} is disjoint from {0} and Proposition
provides a prime P not containing any x^n.
This is an important analog in abstract commutative algebra
of the NSS in algebraic geometry. However, in my experience,
not particularly memorable. Come back to this point after
introducing some more basic ideas.
==
Localisation S^-1A and A_P:
An integral domain A is contained in its field of fractions A in
K = Frac A, and there may be many different partial ring of
fractions S^-1A in Frac A, obtained simply by choosing the
multiplicative set S of denominators to allow. For example,
a popular choice is A[1/f] for some polynomial, or (say)
k[x1,x2,x3][1/(x1*x2)].
For A that is not an integral domain, we need to be more careful,
because we have to decide how S interacts with the zero divisors
of A. The right way to think of this is via the categorical
device of a UMP (universal mapping property).
Recall that a multiplicative set S in A contains 1 and is closed
under multiplication. (There is no sanity check on S here -- it
may contain zerodivisors.)
Definition. A ring, S a multiplicative set in A. Then the
homomorphism t: A -> S^-1A is the universal ring homomorphism
such that for each s in S, multiplication by the image t(s) is
bijective on the ring S^-1A. This means that either t(s) is a
multiplicative unit of a nonzero ring S^-1A , or S^-1A = 0.
(Here _universal_ simply means that any t1: A -> B with the
property is obtained by composing A -> S^-1A -> B. For example
ZZ -> ZZ[1/2] -> QQ.)
Trivial consequence: if a in A and s in S satisfy s*a = 0 in A,
then t(a) = 0. More drastically, if S contains a nilpotent, it
also contains 0, and then S^-1A = 0.
Construction-Proposition. Consider the set of possible
expressions {a/s | a in A and s in S} and introduce the
equivalence relation on it by
a1/s1 ~ a2/s2 <==> there exists t in S
such that t*(s2*a1 - s1*a2) = 0.
Then the quotient set
S^-1A = {a/s | a in A and s in S}
has a ring structure, given by the formulas for + and x that
you already know, making it a solution to the above UMP.
There are a few things to check (and no-one has the patience
to do them all). For example, if a1/s1 ~ a1'/s1' then
t*a1*s1' = t*a1'*sa, and the sum with a2/s2 has one of the
two shapes
(a1*s2+a2*s1)/(s1*s2) or (a1'*s2+a2*s1')/(s1'*s2).
To prove these are equal we multiply through by t*s1*s1' and
consider
t*s1'*(a1*s2+a2*s1) - t*s1*(a1'*s2+a2*s1').
Two terms equal t*a2*s1*s1' so cancel exactly, leaving
(t*a1*s1' - t*s1*a1')*s2
which cancel from a1/s1 ~ a1'/s1'.
====
Spec of a ring and Zariski topology
Spec A = set of prime ideals.
If phi: A -> B is a ring homomorphism, sending
P |-> phi^-1(P) defines a map of set Spec B -> Spec A
(note that is is contravariant).
The same does not work for maximal ideals, because phi^-1(m) is
prime but usually not maximal.
Proof: Let P in Spec B, and f,g in A with f,g notin phi^-1(P).
Then phi(f), phi(g) notin P, so the product
phi(f)*phi(g) = phi(f*g) notin P ==> f*g notin phi^-1(P).
As a formal and rather trivial example, let A be an integral
domain and A -> K = Frac A the inclusion to its field of
fractions. The ideal 0 in K is maximal, but 0 in A is only
a maximal ideal if A = K. Less trivially, a partial ring of
fractions such as
k[x,y] into k(x)[y],
the localisation S^-1 where S is the multiplicative set
(k[x]\0). Then (y) or (y-a) in k(x)[y] is maximal, but
(y) or (y-a) in k(x)[y] is obviously not
The Zariski topology on Spec A is the topology having
closed sets V(I) for I an ideal of A, where
V(I) = { P in Spec A | I in P }.
This is a topology.
Proof. The empty set is V(A) and the whole set V(0).
Arbitrary union: V(sum I_i) = intersect V(I_i)
Finite intersection: V(I1) union V(I2) = V(I1 intersect I2)
= V(I1*I2)
[UCA] Chapter 5 is mostly about the Zariski topology. I will
cover it in more detail later, together with some motivation
based on algebraic varieties in AA^n.