MATH 250B: COMMUTATIVE ALGEBRA
39
14. lecture 14
Example 14.1. With a little more effort, Zariski’s lemma can be used to find
the maximal ideals when k is not algebraically closed because maximal ideals are
related to finite extensions of the field. More precisely, if m is a maximal ideal of
k[x1 , . . . , xn ] then the field k[x1 , . . . , xn ]/m is a finite extension of k by Zariski so is
isomorphic to a subfield of the algebraic closure k, so the maximal ideal corresponds
n
to a point in k . However 2 points can give the same maximal ideal. Galois theory
shows this happens exactly when the 2 points are conjugate under the absolute
Galois group. For example, if k = R then the maximal ideals of R[x1 , . . . , xn ]
n
correspond to orbits of C under complex conjugation.
Knowing the maximal ideals of k[x1 , . . . , xn ] allows one to find the maximal
ideals of any finitely generated extension of a field, because they are just images of
some of the ideals of k[x1 , . . . , xn ]. So the weak nullstellensatz can be used to find
all maximal ideals of any finitely generated algebra over a field.
Theorem 14.2. (The strong Nullstellensatz) If I is an ideal of k[x1 , . . . and √
Z(I)
is the set of its zeros, then the polynomials vanishing on Z(I) form the ideal I.
Proof. We use the Rabinowitsch trick to reduce the strong Nullstelllensatz
√ to the
weak one by adding an extra variable. It is obvious that elements of I vanish
on Z(I), so we have to show that if f vanishes on Z(I) then some power of Z
is in I. The ideal (I, 1 − f x0 ) of k[x0 , x1 , . . .] is not contained in any maximal
ideals (using the weak Nullstellensatz and the fact that these polynomials
have no
P
common roots) so is the unit ideal. Therefore we can write 1 =
ai bi + a(1 − f x0 )
where the elements a, ai are in k[x0 , x1 , . . .] and the elements bi are in I. Setting
x0 = 1/f and multiplying by a high power of f to clear denominators, we find that
this high power of f is in I.
Remark. The strong Nullstellensatz is equivalent to saying that any radical ideal
of k[x1 , . . .] is an intersection of maximal ideals, as the maximal ideals correspond
to points of k n . In any ring, all radical ideals are intersections of the prime ideals
containing them. (Proof: If I is an ideal and b is not in its radical, then the
powers of b are disjoint from I so there is a prime containing I but not b, such as
any maximal ideal among those containing I and disjoint from b.) So the strong
Nullstellensatz can be rephrased as saying that k[x1 , . . .] is a Jacobson ring: this
means all prime ideals are intersections of maximal ideals. In particular the result in
the textbook saying that a polynomial ring over a Jacobson ring is Jacobson is a sort
of generalization of the strong Nullstellensatz. Roughly speaking, Jacobson rings
are those where the maximal spectrum can be recovered from the spectrum (the
points of the spectrum are the irreducible closed subsets of the maximal spectrum).
Most rings in the early days of commutative algebra are Jacobson (fg over a field or
Z) which is roughly why some early work used the maximal spectrum rather than
the spectrum. Typical examples of non-Jacobson rings are local rings of positive
dimension.
We can also ask for computational nullstellensatz: can we find the radical of
an ideal explicitly, or given polynomials f generating the unit ideal can we write
1 explicitly in terms of them. This seems quite hard: For example if the fi are
polynomials of degree at most d ≥ 3 in n variables then the best known bound of the
degrees of polynomials gi such that σfi gi = 1 is about dn , so we end up doing linear
40
RICHARD BORCHERDS
2
algebra in spaces of dimension about dn , which is prohibitively expensive unless
d and n are both rather small. Unfortunately n can be quite large in applications:
for example, resolving singularities of varieties tends to introduce a large number
of extra variables. Making the algorithms for this more efficient is an active area
of computational commutative algebra.
14.1. Integrality. An R-algebra S i called finite over R if it is a finitely generated
R-module. (This is much stronger than saying it is finitely generated.) An element
of an R-algebra S is called integral over R if it satisfies a polynomial with leading
coefficeint 1. If b ∈ S is integral over R then R[b] is finite over R.
Conversely if S is finite over R then every element of S is integral over R If R
is Noetherian this has a very easy proof: just look at the increasing sequence of
R-modules generated by the first k powers of b for k = 0, 1, 2, . . .. Two of these
must be the same, giving a monomial polynomial satisfied by b. For non-Noetherian
rings we need to work a bit harder. It follows from the following lemma (taking the
R-module to be S and taking the endomorphism φ to be multiplication by b).
Lemma 14.3. (Cayley-Hamilton for rings) Any endomorphism φ of a finitely generated R-module is integral.
Proof. Suppose the R-module M = Rn /N has generators m1 , · · · mn . Then φ is
given by some n by n matrix (not necessarily unique as the generators need not be
linearly independent). This means that (φ.1n − A)m = 0 (the left side is n by n
matrices with coefficients in R[φ] acting on M n and m on the right is the vector
(m1 , m2 , ...)T ). Since any matrix multiplied by its adjoint is a diagonal matrix
with entries given by the determinant, we see that det(φ.1n − A) ∈ R[φ] kills all
elements mi . On other words the endomorphism φ is a root of the monic polynomial
det(x.1n − A). (Over fields this is just the characteristic polynomial, but over more
general rings it depends on the choice of mi .)
So we see that S is finite over R if and only if it is generated by a finite number
of elements integral over R.
Geometric meaning of S finite over R: this implies that the fibers of the map
from Spec S to Spec R are finite. (This property is called quasifiniteness and is
weaker than finiteness: for example a localization of R is quasifinite over it but not
finite in general, and the projection of the hyperbola xy = 1 to the x-axis has finite
fibers but is not finite. Finite maps are also proper, which this example is not. One
form of Zariski’s main theorem in algebraic geometry says that any quasifinite map
satisfying some technical conditions is a composition of an open immersion and a
finite map.) Proof: suppose P is a prime of R corresponding to a point of Spec
R. We want to show there are only a finite number of primes of S with inverse P .
We may as well quotient out by P so can assume P = (0). Next we can localize at
P = 0 (in other words take the field of quotients of R) this will not affect the set
of primes of S with inverse image (0). So we can assume R is a field. As S is finite
over R it is a finite dimensional vector space, therefore Artinian, therefore has only
a finite number of primes (and the spectrum is finite and discrete).
In general if S is any R-algebra the elements integral over R form a subalgebra:
in particular they are closed under + and × (this follows because any 2 finite Rsubalgebras generate a finite R-subalgebra. FOr example, if we take R = Z and S
the algebraic closure of Q, then the elements integral over Z are just the algebraic
integers, so the algebraic integers form a ring. Notice that if α, β are algebraic
MATH 250B: COMMUTATIVE ALGEBRA
41
integers satisfying given polynomials ,it can be quite hard to find the polynomials
with α + β or αβ as roots. Try it for α + β with α, β roots of x2 − 2, x3 − 2.
If R is a domain and S its quotient field, then the normalization of R is defined
to be the integral closure of R in S.
Lemma 14.4. Any UFD is normal
Proof. Suppose b/a is an element of the quotient field of R that is integral over the
UFD R with a. b coprime. Looking at a monic polynomial with root b/a shows
that all prime factors of a divide b, so a is a unit and b ∈ R.
Example. In algebraic number theory, if K is an algebraic number field (finite
extension of Q we want to find rings inside it that are finite Z-modules and also
UFDs. There are plenty of different rings that are not UFDs such as Z[ni]. The
lemma shows that the only possibility for
of all algebraic
√ a UFD is to take the ring √
is
Z[
−3] which is
integers in K. In particular if K is Q[ −3] the obvious ring
√
not a UFD. However its normalization is the larger ring Z[( −3 + 1)/2] which is a
UFD (the ring of Eisenstein integers).
The ring of all algebraic integers in K need
√
not be a UFD; for example Z[ −5].