MATH 250B: COMMUTATIVE ALGEBRA
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2. Lecture 2 Hilbert’s theorems
Hilbert’s theorem follows (using induction) from the more general theorem:
Theorem 2.1. If a ring R is Noetherian, so is the ring R[x].
Theorem 2.2. The following are equivalent for a ring R:
Every ideal is finitely generated
Every non-empty set of ideals has a maximal element
Every strictly increasing chain of ideals is finite.
Proof. (1) implies (3) because the union of the chain is finitely generated, so a set
of generators can be found after a finite number of steps of the chain. (3) implies
(1) If an ideal I is not finitely generated, we can find an infinite increasing chain
(a1 ) ⊂ (a1 , a2 ) ⊂ · · · of ideals, where at each step we add some element of I not
generated by the elements we already have.
(2) and (3) are equivalent: this has little to do with ring theory, but is a general
property of partially ordered sets. If there is a strictly increasing chain of ideals,
this gives a nonempty set of ideals with no maximal element. Conversely if we have
a nonempty set of ideals with no maximal element, then for every ideal in the set
we can find a larger one, so (using the axiom of choice) we can find an infinite
strictly increasing sequence. (2) implies (1)
Example 2.3. In the ring k[X1 , x2 , . . .], the ideals (x1 ), (x1 , x2 ), ... form a strictly
increasing chain, and there is no maximal element of this set.
Example 2.4. In the ring of integers the ideals (2), (4), (8), ... form a strictly
decreasing chain and this set of ideals has no minimal element.
Example 2.5. Informally, a ring of functions being Noetherian has something to
do with zeros of the functions being under control. We have rings R[x], holomorphic
functions= power series of infinite radius of convergence, analytic functions on R,
germs of analytic functions at 0 = power series of finite radius of convergence,
smooth functions, germs of smooth functions, formal power series. The rings of
holomorphic functions or analytic functions or smooth functions are not Noetherian
(take the ideal of functions vanishing at all but a finite number of integers) and the
rings of germs of smooth functions is not (The ideal of functions vanishing at an
infinite number of points 1/n is not finitely generated) but the ring of polynomials,
germs of analytic functions, and formal power series are; in fact they are all PIDs.
In particular any ring that is a finitely generated algebra over a Noetherian
ring is also Noetherian. This includes polynomial rings over the integers and over
fields, the cases in Hilbert’s papers. This is perhaps the most fundamental theorem
in commutative algebra as many of the deeper theorems only work for Noetherian
rings. Hilbert’s theorem assures us that many of the rings we come across in number
theory or algebraic geometry are indeed Noetherian.
Example. The ring of smooth functions on the real line is not Noetherian. The
ideal of all functions vanishing to infinite order at 0 is not finitely generated.
Example. The ring of holomorphic functions is not Noetherian. (Look at the
ideal of functions vanishing at all but a finite number of integers.) Although the
ring of polynomials is in some sense an approximation to the ring of holomorphic
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RICHARD BORCHERDS
functions, as it is dense in the compact open topology, one is Noetherian and the
other is not. The ideal above is not closed.
Now we can prove Hilbert’s theorem that if R is Noetherian then so is R[x].
Suppose I is an ideal of R[x]. Look at the ideals I0 , I1 , ... of leading coefficients of
polynomials in I of given degree. Obviously I0 ⊆ I1 ⊆ · · · as we can multiply by
x, so as R is Noetherian this sequence is eventually constant, say at In . Then In
generates I, so if take a set of polynomials for each i ≤ n whose leading coefficients
generated the R-ideal Ii , these generate the ideal I. End of proof.
Exercise. Show that if R is Noetherian then so is the power series ring R[[x]]
of all formal power series with coefficients in R. (The proof is similar to that for
polynomials, but upside down: In is now the ideal of coefficients of xn of all power
series in I with no terms of degree less than n.)
Example: While any ideal of R[x, y, ...] is finitely generated, this is certainly not
true for subrings. For example, the ring generated by 1 and the ideal (x) is not
finitely generated.
Example. Any ideal of k[x] is generated by 1 element, so it is natural to guess
that any ideal of k[x1 , · · · , xn ] is generated by n elements. This is false. For
example, the ideal (xn , xn−1 y, . . . , y n ) of k[x, y] cannot be generated by less than
n + 1 elements. However a weaker version of this is true, related to the fact that
k[x1 , · · · , xn ] has depth n. This mean that any ideal I has at most n elements
such that if we quotient out by these, then the image of I in the quotient consists
entirely of zero divisors.
Now we want to prove finiteness of the number of basic invariants of a finite group
acting on a field of characteristic 0, a special case of Hilbert’s finiteness theorem.
In general as we have seen an arbitrary subalgebra of a polynomial ring need not
be finitely generated, so we need to use some special property of invariant rings.
We will use the existence of a Reynolds operator ρ: this takes any polynomial to
its average under the group. (It is well defined because the group is FINITE and
we are working in characteristic 0 so can divide by the order of the group.) The
Reynold operator satisfies ρ(1) = 1 and has the following key properties:
(2.1)
ρ(ab) = aρ(b)
whenever a is invariant. So ρ is a homomorphism of modules (over the invariants)
from polynomials to invariants. In other words, the injection of modules from
invariants to polynomials SPLITS.
Now we can prove finiteness of the algebra of invariants. Let I be the graded
algebra of invariants inside the graded polynomial ring. We let J be the graded ideal
generated by invariants of nonzero degree. It is a finitely generated ideal, and we
can find a finite set of generators a1 , . . . , am that are invariants. We will show that
the ai generate the ring of invariants. So let b be any invariant of positive degree.
It is in the ideal J so we can write b = a1 c1 + · · · where the ci are polynomials.
Now applying the Reynolds operator we find
X
(2.2)
b = ρ(b) =
ai ρ(ci )
The elements ci are invariants of degree strictly less than that of b (as all the ai
have strictly positive degree), so by induction on the degree they are polynomials
in the ai . Therefore b is also a polynomial in the ai . This proves that the algebra
of invariants is finitely generated, at least for finite groups in characteristic 0.
MATH 250B: COMMUTATIVE ALGEBRA
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Reynolds is the same guy as in the ”Reynolds number” in fluid mechanics. He
introduced the idea of studying a fluid flow by taking the average fluid flow at
a given point. This is called the Reynolds operator: it corresponds to taking an
average under the group of time translations, and the term ”Reynolds operator”
was later extended to taking the average under any group.
We can easily extend the proof to cover the case of infinite groups such as the
special linear group. First of all, the proof works whenever we can find a Reynolds
operator. In particular it works for all COMPACT groups (acting continuously
on a vector space) because we can take averages of any continuous function over
a compact group by integrating using Haar measure on the group. The special
linear group is not compact, but it behaves like one at least for finite dimensional
representations. The point is that the Lie algebras of SLn (R) and SUn are not the
same, but become the same when complexified. This means that finite dimensional
complex representations of SLn (R) and SU (n) are essentially the same, so SLn (R)
has a Reynolds operator. (Infinite dimensional representations are totally different:
the problem is that the exponential map no longer converges, so the correspondence
between representations of Lie algebras and Lie groups breaks down.) In particular
this proves the finite generation of rings of invariants for binary quantics and so on.