HILBERT’S NULLSTELLENSATZ
Edward S. Letzter
Math 4096 Spring 2009
Consider a system
f1 (x1 , . . . , xn )
f2 (x1 , . . . , xn )
..
.
=
=
..
.
0
0
..
.
fm (x1 , . . . , xn ) =
0
of polynomial equations, in the commuting variables x1 , . . . , xn , and with coefficients in
some field k; the solutions describe a set of points in k n . In these notes we learn how to view
such polynomial systems from a simultaneously algebraic and geometric perspective (thus
barely scratching the surface of the deep and profound subject of algebraic geometry).
Throughout these notes k will denote an arbitrary field.
1. Algebraic Sets
All of the polynomials mentioned in this section will have commuting variables and
coefficients in k.
We begin by defining a new topology on k n , a topology in which the closed subsets are
precisely the solution-sets to systems of polynomial equations. When k = R, we can recall
that the zero-set of a polynomial will be closed in, for example, the metric topology on Rn .
However, the metric topology does not distinguish between polynomial and non-polynomial
equations.
For the moment, our discussion uses very little “abstract algebra” – both to benefit
the reader unfamiliar with the subject, and to more clearly illustrate later where the
“abstraction” is needed.
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HILBERT’S NULLSTELLENSATZ
• As in high school algebra, we view the polynomials in (commmuting) variables x1 , . . . ,
xn in two ways: First, these polynomials are objects on which we perform various types
of algebraic manipulations. Second, each polynomial f (x1 , . . . , xn ) is a function on k n :
(λ1 ,... ,λn )7−→f (λ1 ,... ,λn )
k n −−−−−−−−−−−−−−−−→ k.
Now let f1 , . . . , fm be polynomials in the x1 , . . . , xn . We use V (f1 , . . . , fm ) to denote the
set of points (λ1 , . . . , λn ) ∈ k n such that
f1 (λ1 , . . . , λn ) = · · · = fm (λ1 , . . . , λn ) = 0.
Next, let S be any collection of polynomials in the x1 , . . . , xn . Then
V (S) = {(λ1 , . . . , λn ) ∈ k n | f (λ1 , . . . , λn ) = 0 for all f ∈ S}.
We refer to V (S) as an algebraic (sub)set (of k n ) or as an affine (sub)variety (of k n ).
If S 0 ⊆ S, then V (S) ⊆ V (S 0 ). However, distinct sets of polynomials can correspond
to the same algebraic subset of k n . For example, V (x + y) and V ((x + y)2 ) are the same
algebraic subset of k 2 .
Proposition. (i) Both k n and ∅ are algebraic subsets of k n .
T
(ii) If {Vα } is a collection of algebraic subsets of k n , then α Vα is also an algebraic
subset of k n .
(iii) If V1 , . . . , Vt are algebraic subsets of k n , then V1 ∪ · · · ∪ Vt is an algebraic subset of
kn .
Proof. (i) Observe that k n = V (0) and ∅ = V (1).
(ii) Let {Sα } be a collection of sets of polynomials in the x1 , . . . , xn . Then
!
\
α
V (Sα ) = V
[
Sα
.
α
(Verify!)
(iii) An easy induction argument shows that it suffices to consider the case when t = 2.
So suppose that S and T are sets of polynomials in the x1 , . . . xn ; we will show that
V (S) ∪ V (T ) is an algebraic set. To start, let
ST = {f g | f ∈ S and g ∈ T }.
It is straightforward to check that V (S) ∪ V (T ) ⊆ V (ST ).
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We claim that V (ST ) ⊆ V (S)∪V (T ). To prove the claim, choose (λ1 , . . . , λn ) ∈ V (ST );
we will show that (λ1 , . . . , λn ) ∈ V (S) ∪ V (T ). First, notice that (λ1 , . . . , λn ) is in both
V (S) and V (T ) if
f (λ1 , . . . , λn ) = g(λ1 , . . . , λn ) = 0
for all f ∈ S and g ∈ T . Therefore, we can assume that there exists some h ∈ S ∪ T
such that h(λ1 , . . . , λn ) 6= 0. Without loss of generality, assume that h ∈ T . Because
(λ1 , . . . , λn ) ∈ V (ST ), it follows that
f (λ1 , . . . , λn )h(λ1 , . . . , λn ) = 0
for all f ∈ S. But since h(λ1 , . . . , λn ) 6= 0, we see that f (λ1 , . . . , λn ) = 0 for all f ∈ S.
Therefore, (λ1 , . . . , λn ) ∈ V (S), and the claim follows.
Thus V (ST ) = V (S) ∪ V (T ), and (iii) follows. •
The Zariski topology on k n is formed by taking the closed subsets to be the algebraic
sets; it follows from the preceding that the appropriate axioms are satisfied. It is not hard
to see that the Zariski topology is the weakest topology on k n in which the algebraic sets
V (f ) are closed, for each polynomial f in the x1 , . . . , xn . However, the information we
have collected so far does not guarantee that this topology is very useful.
But the first major task of these notes is to show that each algebraic set in k n can be
written in terms of a suitably chosen finite system of polynomials, a result upon which the
effectiveness of the Zariski topology is based.
We will take the “scenic” route to this goal, encountering some interesting and useful
algebra along the way.
•
If ϕ: U → V is a k-linear transformation, then ϕ−1 (0) is the kernel of ϕ; recall that ϕ
is injective if and only if its kernel is zero.
(i) A k-vector space A, equipped with a bilinear multiplication map
(a,b)7−→ab
A×A −−−−−−→ A,
is an (associative) k-algebra (with identity) provided:
Multiplication is associative. a(bc) = (ab)c for all a, b, c ∈ A.
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HILBERT’S NULLSTELLENSATZ
There exists a multiplicative identity. There exists an element 1 ∈ A such that 1a =
a1 = a for all a ∈ A.
(ii) Let ϕ: A → A0 be a k-linear transformation of k-algebras. Then ϕ is a k-algebra
homomorphism provided ϕ(1) = 1 and ϕ(ab) = ϕ(a)ϕ(b) for all a, b ∈ A.
(iii) If A is a k-subspace of the k-algebra A0 , then A is a k-subalgebra of A0 if the inclusion map is a k-algebra homomorphism. In other words, k-subalgebras are k-subspaces,
containing 1, that are closed under multiplication.
(iv) A k-algebra A is commutative provided ab = ba for all a, b ∈ A.
(v) Note that we are not excluding the possibility that A is the trivial k-vector space,
in which case 0 = 1. Note however, that the trivial k-algebra is not a subalgebra of any
nontrivial k-algebra.
• For now, the most important example of a k-algebra is the commutative k-algebra comprised of all polynomials in the variables x1 , . . . , xn . This algebra is denoted k[x1 , . . . , xn ].
The k-vector space and multiplicative structures are the obvious ones.
Of course, k is itself a k-algebra.
An example of noncommutative k-algebra is the set of n×n matrices with entries in
k, denoted Mn (k). The k-vector space and multiplicative structures are those produced
by matrix addition, scalar multiplication, and matrix multiplication. The n×n identity
matrix is the 1 element in this algebra.
• Let A be a nonzero k-algebra. There is an embedding of k into A provided by the map
α 7→ α.1, for α ∈ k. We identify k with its image under this map.
Exercise. Let ϕ: A → A0 be a k-algebra homomorphism. Prove that the restriction of ϕ
to k induces the identity map k → k.
• Let A be a k-algebra, containing a subset S, such that A is k-linearly spanned by all
products of the form
s1 · · · st
for t = 0, 1, 2, . . . and s1 , . . . , st ∈ S. (We follow the standard convention that “s1 · · · st ”
is equal to 1 when t = 0, and we say that t is the length of the preceding product.) We
say that A is generated as a k-algebra by the elements of S, and we write A = khSi. If S
is finite (i.e., when A is finitely generated as a k-algebra), then A is k-affine.
• Let A be a commutative k-algebra containing a k-subspace I, and suppose that I is
closed under multiplication by elements in A (i.e., ai ∈ I for all a ∈ A and i ∈ I). Then I
is an ideal of A. We say that I is generated by a subset S of A if
I = {a1 s1 + · · · + at st | s1 , . . . , st ∈ S, a1 , . . . , at ∈ A},
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and we say that I is finitely generated when S is finite.
•
For the remainder of this section, A will denote an affine commutative k-algebra.
• Suppose that A = kha1 , . . . , an i. Observe that there is a surjective k-algebra homomorphism
k[x1 , . . . , xn ] −→
A
x1 , . . . , xn
7−→
a1 , . . . , an .
Indeed, a commutative k-algebra is affine if and only if it is a homomorphic image of a
polynomial algebra (with coefficients in k).
• Let S be a subset of k[x1 , . . . , xn ], and let I be the ideal generated by S. Since S ⊆ I,
we immediately conclude that V (I) ⊆ V (S). On the other hand, every polynomial f in I
has the form
a1 s1 + · · · + at st ,
for a1 , . . . , at ∈ k[x1 , . . . , xn ] and s1 , . . . , st ∈ S. Now if (λ1 , . . . , λn ) ∈ V (S), then
s1 (λ1 , . . . , λn ) = · · · = st (λ1 , . . . , λn ) = 0,
and so f (λ1 , . . . , λn ) = 0. Hence, V (S) ⊆ V (I), and so V (S) = V (I).
Consequently, the algebraic subsets of k n are precisely the algebraic sets V (I), for ideals
I of k[x1 , . . . , xn ]. In the next section we will see that every ideal of k[x1 , . . . , xn ] is finitely
generated, from which we can subsequently deduce that every algebraic set is determined
by a finite system of polynomial equations.
The Hilbert Basis Theorem
We now move to a slightly more general setting.
•
A nonempty set L is additive when there exists an addition map
(l,m)7−→l+m
L×L −−−−−−−−→ L,
and a zero element 0 ∈ L such that for all a, b, c ∈ L: a + (b + c) = (a + b) + c, a + b = b + a,
0 + a = a, and there an additive inverse −a ∈ L such that a + (−a) = 0. We use a − b to
denote a + (−b).
Let L and M be additive sets equipped with a function ϕ: L → M . We say that ϕ is
additive provided ϕ(a + b) = ϕ(a) + ϕ(b) for all a, b ∈ L; the kernel of ϕ is ϕ−1 (0). An
additive map is injective if and only if its kernel is {0}.
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HILBERT’S NULLSTELLENSATZ
If L, M , and N are additive sets, then a function
(a,b)7−→ϕ(a,b)
L×M −−−−−−−−−→ N
is bi-additive provided ϕ(a + b, c) = ϕ(a, c) + ϕ(b, c) and ϕ(a0 , b0 + c0 ) = ϕ(a0 , b0 ) + ϕ(a0 , c0 ),
for all a, b, c ∈ L and a0 , b0 , c0 ∈ M .
•
An additive set R, equipped with a bi-additive multiplication map
(a,b)7−→ab
R × R −−−−−−→ R,
and a multiplicative identity 1 ∈ R, is a ring provided: a(bc) = a(bc), and 1a = a1 = a,
for all a, b, c ∈ R.
If ab = ba, for all a, b ∈ R, then R is commutative.
If R and S are rings, then an additive map ϕ: R → S is a homomorphism provided
ϕ(1) = 1 and ϕ(ab) = ϕ(a)ϕ(b) for all a, b ∈ R.
If R is an additive subset of a ring S, then R is a subring of S when R is closed under
the multiplication in S, and when the mutliplicative identity of S is contained in R.
• Basic examples of commutative rings include k and k[x1 , . . . , xn ]. Indeed, any k-algebra
is automatically a ring, and Mn (k) is a noncommutative ring.
The integers Z form a commutative ring, as do the integers modulo m, Z/mZ.
For any ring R (commutative or not), one can form the n×n matrix ring Mn (R), using
the usual rules for addition and multiplication of matrices.
For any commutative ring R, we can form the polynomial ring R[x] with coefficients in
R.
• Let R be a commutative ring, and let I be an additive subset of R closed under
multiplication by R (i.e., r.i ∈ I for all r ∈ R and i ∈ I). As before, we say that I is an
ideal of R. When R is a k-algebra, the two definitions of ideal are equivalent. (Verify.)
As before, if S is a subset of R then the ideal generated by S is
{r1 s1 + · · · + rt st | r1 , . . . , rt ∈ R, s1 , . . . , st ∈ S, t = 1, 2, . . . },
and is denoted hSi.
An ideal I of R is maximal provided there exist no ideals J of R such that I ( J ( R.
Continuing to assume that R is a commutative ring, we say that R is noetherian when
every ideal is finitely generated.
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Exercise. (i) Let ϕ: R → S be a surjective homomorphism of commutative rings. Prove
(a) that every ideal in S is the ϕ-image of an ideal in R, (b) if R is noetherian, then so is
S.
(ii) Prove that Z and k[x] are noetherian.
(iii) Prove that a commutative ring R is a field if and only if the only ideals in R are
(0) and R itself.
Theorem. If R is noetherian then R[x] is noetherian.
Proof. (Due to H. Sarges, following [Kunz].) Suppose that R is noetherian but that I is an
ideal of R[x] that is not finitely generated. Choose f1 ∈ I of minimal degree n1 (viewing
f1 as a polynomial in the variable x) and with (necessarily nonzero) leading coefficient a1 .
Having chosen f1 , . . . , ft ∈ I with respective degrees and leading coefficients n1 , . . . , nt
and a1 , . . . , at , choose
ft+1 ∈ I \ hf1 , . . . , ft i
of minimal degree. Note that n1 ≤ n2 ≤ · · · .
Next, consider the infinite series of ideals,
ha1 i ⊆ ha1 , a2 i ⊆ ha1 , a2 , a3 i ⊆ · · · ,
and suppose that
ha1 , . . . , at i = ha1 , . . . , at , at+1 i
for some positive integer t. Then
at+1 = r1 a1 + · · · rt at
for some r1 , . . . , rt ∈ R.
Next, for 1 ≤ i ≤ t, the polynomial
ri xnt+1 −ni fi
is contained in I, has nt+1 -degree coefficient ai ri , and has degree no greater than nt+1 .
(Check.) Therefore, the nonzero polynomial
g = ft+1 −
t
X
ri xnt+1 −ni fi ∈ I \ hf1 , . . . , ft i
i=1
has degree no greater than nt+1 , and the nt+1 -degree coefficient is
at+1 − (r1 a1 + · · · + rt at ) = 0.
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HILBERT’S NULLSTELLENSATZ
Consequently, the degree of g is strictly less than nt+1 , a contradiction to the choice of
ft+1 . We conclude that
ha1 , . . . , at i ( ha1 , . . . , at+1 i
for all positive integers t.
Now let J denote the ideal of R generated by the set {ai | i = 1, 2, . . . }, and suppose
that J is generated by a finite set S. For some t, it must be the case that S ⊆ {a1 , . . . , at }.
But then
J = ha1 , . . . , at i = ha1 , . . . , at+1 i,
a contradiction to the preceding conclusion.
Thus every ideal of R[x] is finitely generated, and R[x] is noetherian.
Corollary. (Hilbert Basis Theorem) Every affine commutative k-algebra is noetherian.
Proof. Exercise
• We now see that every every algebraic set in k n is of the form V (f1 , . . . , fm ), for a
suitable choice of finitely many polynomials f1 , . . . , fm ∈ k[x1 , . . . , xn ].
Our next major goal is to more precisely tie the affine geometry of k n (i.e., k n equipped
with the Zariski topology) to the algebraic structure of k[x1 , . . . , xn ]. Once again we will
begin with a more abstract discussion.
Noetherian Modules
• Let R be a ring, and let M be an additive set equipped with a bi-additive map (termed
an R-action),
(r,m)7−→r.m
R×M −−−−−−−−→ M,
such that 1.m = m and r(s.m) = (rs).m, for all r, s ∈ R and m ∈ M . Then M is a (left)
R-module. Right R-modules can be similarly defined. Unless the designation “right” is
used, we will assume that “module” means “left module.”
Additive subsets of M closed under the R-action are R-submodules, and an additive
map ϕ: M → M 0 of R-modules that is R-equivariant (i.e., ϕ(r.m) = r.ϕ(m) for all r ∈ R
and m ∈ M ) is an R-module homomorphism. Note that R can itself be regarded as either
a left or right R-module, using the multiplication map.
When R = k, the R-modules are simply k-vector spaces. More generally, when R is a
k-algebra, the R-modules are also k-vector spaces (check). Furthermore, if R is a k-algebra
and L is a submodule of the R-module M , then N is also a k-subspace of M .
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• Let R be a ring, and let M be a left R-module. It is useful at this point to start doing
arithmetic with additive subsets, rather than element by element:
To start, if X and Y are additive subsets of M , then X + Y refers to the additive subset
{x + y | x ∈ X, y ∈ Y }.
Next, let X be an additive subset of R, and let Y be an additive subset of M . We use
X.Y to denote all of the sums of all of the x.y’s, for x ∈ X and y ∈ Y . When X and
Y are additive subsets of R, and Z is an additive subset of M , it is not hard to check
that (XY ).Z = X.(Y Z). (The notation at first is a bit misleading, and some caution is
required.) When M = R, we use XY instead of X.Y .
•
We now describe the Noether isomorphism theorems for rings and their modules.
Let R be a ring, and let M be an R-module.
(i) Let L be an R-submodule of M . For each m ∈ M , set
m + L = {m + l | l ∈ L}.
Set
M/L = {m + L | m ∈ M },
an R-module under the operations
(m + L) + (m0 + L) = (m + m0 ) + L,
r.(m + L) = r.m + L,
for r ∈ R and m, m0 ∈ M . We now have the so-called “canonical” surjective R-module
homomorphism,
m7−→m+L
π: M −−−−−−−→ M/L.
(i) If ϕ: M → N is an R-module homomorphism whose kernel contains L, then
m+L7−→ϕ(m)
ϕ: M/L −−−−−−−−−→ W
is a well-defined injective R-module homomorphism, and ϕ(m) = (ϕ◦π)(m) for all m ∈ M .
Now let R be a commutative ring, and let I be an ideal of R. For each r ∈ R we set
r + I = {r + i | i ∈ I},
and
A/I = {r + I | r ∈ R}.
Note that R/I is a ring under the operations
(a + I) + (b + I) = (a + b) + I,
(a + I)(b + I) = ab + I.
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HILBERT’S NULLSTELLENSATZ
We have the “canonical” surjective ring homomorphism,
a7−→r+R
R −−−−−−→ R/I,
and if ϕ: R → S is a ring homomorphism whose kernel contains I, then
r+I7−→ϕ(r)
ϕ: R/I −−−−−−−→ S
is a well-defined injective ring homomorphism, and ϕ(r) = (ϕ ◦ π)(r) for all r ∈ R.
(ii) If K ⊆ L ⊆ M is a chain of R-modules (i.e., K is a submodule of L, and L is a
submodule of M ), then there is a surjective R-module homomorphism,
m+K7−→m+L
M/K −−−−−−−−−→ M/L,
whose kernel is L/K. We obtain an isomorphism of R-modules
(M/K)/(L/K) ∼
= M/L.
(iii) Suppose that M contains the R-submodules K and L. Then K + L is also an
R-submodule of M , and there is surjective R-module homomorphism
l7−→l+K
L −−−−−→ L + K/K
with kernel L ∩ K. We can now write the R-module isomorphism
L/L ∩ K ∼
= L + K/K.
(iv) All of (i), (ii), and (iii) remain true if the word “ring” is replaced with “k-algebra”
throughout.
Exercise. Prove (i), (ii), (iii), and (iv) in the preceding, taking special care to verify that
all of the homomorphisms and operations described are well defined.
• Let R be a ring. Given a family of R-modules Mα , for α in some index set X, the
direct sum
M
Mα
α
consists of the X-tuples (mα ) = (mα )α∈X , equipped with the following additive structure
and R-action:
(mα ) + (m0α ) = (mα + m0α ), r.(mα ) = (r.mα ).
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Exercise. Let R be a ring, and let N be an R module containing submodules L and M .
Suppose further that N = L + M and that L ∩ M is the trivial submodule (0). Prove that
(l,m)7−→l+m
L ⊕ M −−−−−−−−→ N
is an R-module isomorphism.
Proposition. Let R be a ring and let M be an R-module. The following conditions are
equivalent:
(1) Every R-submodule of M is finitely generated (i.e., if M 0 is an R-submoudle of M
then M 0 = R.m1 + · · · + R.mt for some m1 , . . . , mt ∈ M .)
(2) Every ascending chain of R-submodules of M stabilizes (i.e., if M1 ⊆ M2 ⊆ · · · are
R-submodules of M , then for some t, Mt = Mt+1 = Mt+2 = · · · .)
(3) Every non-empty set of submodules of M , ordered by inclusion, contains a maximal
member.
S
Proof. “(1) ⇒ (2)”: Set M 0 = i Mi , for an ascending chain
M1 ⊆ M2 ⊆ · · ·
of submodules of M . Then M 0 = R.m1 + · · · + R.ms for some m1 , . . . , ms ∈ M 0 . But we
can then choose a positive integer t such that m1 , . . . , ms ∈ Mt . Notice, now, that
M 0 = Mt = Mt+1 = Mt+2 = · · · ,
and so (1) ⇒ (2).
“(2) ⇒ (3)”: Let X be a collection of submodules of M . Choose M1 ∈ X. If M1 is not
maximal, choose M2 ∈ X such that M2 ) M1 . If M2 is not maximal, choose M3 ∈ X such
that M3 ) M2 . If X does not contain a maximal member, the preceding process continues
indefinitely, producing an ascending chain that does not stablize.
“(3) ⇒ (1)”: Let M 0 be a submodule of M , and let X be the set of finitely generated
submodules of M 0 . Note that X is not empty, because the trivial module (0) is finitely
generated. Let M 00 be a maximal member of X. If M 00 6= M 0 , choose m ∈ M 0 \ M 00 . But
M 00 +R.m is a finitely generated submodule of M 0 properly containing M 00 , a contradiction.
Therefore, M 00 = M 0 , and M 0 is finitely generated. • Let R be a ring. A (left) module satisfying any of the three equivalent conditions in
the preceding proposition is said to be (left) noetherian. Right noetherianity is similarly
defined for right modules.
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HILBERT’S NULLSTELLENSATZ
An additive subset I of R closed under left multiplication by R (i.e., ri ∈ I for all i ∈ I
and r ∈ R) is called a left ideal. Right ideals are similarly defined. An additive subset
that is both a left and right ideal of R is called an ideal of R; sometimes, for emphasis,
ideals are referred to as two-sided ideals.
Of course, when R is commutative the definitions of left, right, and two-sided ideals are
equivalent.
If R is noetherian as a left R-module it is said to be left noetherian, if R is noetherian as
a right R-module it is said to be right noetherian, and if R is noetherian on both the right
and left it is said to be noetherian. This notation coincides with the usage we established
for commutative rings. (Check.)
Exercise. The free k-algebra in X and Y , denoted k{X, Y }, is the k-algebra (finitely)
generated by {X, Y }, such that the noncommuative monomials
W1i1 W2i2 · · · Wtit ,
for non-negative integers t, for positive integers i1 , . . . , it , and for W1 , . . . , Wt ∈ {X, Y },
comprise a k-basis for k{X, Y }. Prove that k{X, Y } is not noetherian.
Proposition. Let R be a ring, and let M be a left R-module containing an R-submodule
L. Then M is noetherian if and only if both L and M/L are noetherian.
Proof. “⇒”: Exercise.
“⇐”: (Following [Lang].) To each submodule K of M , assign the associated pair
(K ∩ L, (K + L)/L).
Let E ⊆ F be submodules of M . Suppose, for the moment, that the associated pairs of
E and F are equal. Choose x ∈ F . Because (E + L)/L = (F + L)/L, there exist a, b ∈ L
and y ∈ E such that y + a = x + b. Consequently, x − y = a − b ∈ F ∩ L = E ∩ L. Thus
x ∈ E, since y ∈ E. We now see that E = F if and only if their associated pairs are the
same.
Hence, if E1 ⊆ E2 ⊆ · · · is an ascending chain in M , the associated pairs (Ei ∩ L, Ei +
L/L) form a pair of ascending chains in L and M/L.
Therefore, if L and M/L are noetherian, so is M . Corollary. If R is left noetherian and M is a finitely generated left R-module, then M is
noetherian.
Proof. Exercise.
Exercise. Let R be a ring, and let M be a noetherian R-module. Suppose that S is a
generating set for M . Prove that M is generated by a finite subset of S.
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Hilbert’s Nullstellensatz
•
Let R be a ring. Then
Z(R) = {z ∈ R | zr = rz for all r ∈ R}
is the center of R. Note that Z(R) is a commutative subring of R.
Let K be a subring of Z(R). We say in this case that R is a K-algebra. Let S be a
subset of R, and suppose that the products of elements (of arbitrary finite length) from S
span R as a K-module. We now say that S generates R as a K-algebra, and write KhSi.
When S is finite we say that R is K-affine.
Theorem. (Artin-Tate Lemma) Let R be a commutative noetherian ring, and suppose
that T = Rhy1 , . . . , ym i is an affine R-algebra. Let S be a subring of Z(T ) such that
R ⊆ S, and such that T is finitely generated as an S-module. Then S is R-affine.
Proof. Choose {w1 , . . . , wl } ⊇ {y1 , . . . , ym } such that T = S.w1 + · · · + S.wl . Hence,
wi wj =
l
X
aij
t wt ,
t=1
for some aij
t ∈ S and 1 ≤ i, j, t ≤ l.
0
0
Set S 0 = Rhaij
t | 1 ≤ i, j, t ≤ li ⊆ S ⊆ Z(T ). Then T = S .w1 + · · · + S .wl , since
S 0 .w1 + · · · + S 0 .wl contains y1 , . . . , ym and all of their products. By the Hilbert Basis
Theorem, S 0 is a noetherian ring, and so T is a noetherian S 0 -module, since T is finitely
generated as an S 0 -module. Consequently, S is a notherian S 0 -module, and so S is S 0 -affine.
But then S is R-affine, since S 0 is R-affine. We now present the main result of these notes; the proof will be temporarily deferred.
Theorem. (Hilbert’s Nullstellensatz) A k-affine field must be finite dimensional over k.
• Recall that a field K is algebraically closed whenevery every polynomial f (x) in K[x]
has a factorization
(x − α1 )(x − α2 ) · · · (x − αt )
for a suitable choice of α1 , . . . , αt ∈ K. For example, C is algebraically closed, but R is
not.
Exercise. If K is algebraically closed, and if K 0 is a finite field extension of K (i.e., K 0
is finite dimensional as a K-vector space), then K 0 = K.
We thus obtain:
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HILBERT’S NULLSTELLENSATZ
Corollary. If k is algebraically closed, the maximal ideals of k[x1 , . . . , xn ] are precisely
the ideals of the form
hx1 − λ1 , x1 − λ2 , . . . , xn − λn i,
for λ1 , . . . , λn ∈ k.
Proof. First, if M = hx1 − λ1 , x1 − λ2 , . . . , xn − λn i, then it is not hard to see that
k[x1 , . . . , xn ]/M = k, from which it follows that M is a maximal ideal.
Now let M be an arbitrary maximal ideal of k[x1 , . . . , xn ]. Then k[x1 , . . . , xn ]/M is
a k-affine field, and so k[x1 , . . . , xn ]/M = k. Hence x1 + M, . . . , xn + M ∈ k, and so
M ⊇ M 0 = hx1 − λ1 , . . . , xn − λn i, for suitable choices of λ1 , . . . , λn ∈ k. From the first
paragraph we know that M 0 is maximal, and so M = M 0 . The corollary follows. • The remainder of the section will be devoted to a proof of the Nullstellensatz. Some
familiarity with field theory and unique factorization will be helpful for this portion of our
discussion. (This material will not be used later.)
• Let K be a field extension of k, and let S be a subset of K such that every finite set of
products of elements (of arbitrary finite length) from S is k-linearly independent. We say
that S is algebraically independent over k. (Note that {1} is an example of such a set.) By
Zorn’s Lemma, we can assume that S has been chosen to be maximal among those subsets
of K that are algebraically independent over k. Let k(S) denote the field generated, over
k, by S. Exercise: K is algebraic over k(S). (The Noether Normalization Theorem is an
elaboration on this theme.)
• Next, we will require the following facts, presented here without proofs (which can be
found in any first year graduate algebra text): Let p be a polynomial, in k[x1 , . . . , xn ],
such that
p = f g ⇒ f or g is scalar,
where f and g are polynomials in k[x1 , . . . , xn ]. We say that p is irreducible. Every
polynomial in k[x1 , . . . , xn ] has a factorization into finitely many irreducible polynomials;
this factorization is unique up to scalar multiples of the factors. When two polynomials
are scalar multiples of each other, they are said to be associates. Finally, if the irreducible
polynomial p divides a product of polynomials in k[x1 , . . . , xn ], then p divides one of the
factors.
We use k(y1 , . . . , yt ) to denote the field of rational functions f /g, where f ∈ k[y1 , . . . , yt ]
and g ∈ k[x1 , . . . , xn ] \ {0}. We view k[y1 , . . . , yt ] as a k-subalgebra of k(y1 , . . . , yt ), by
identifying the polynomial f with f /1.
SPRING 2006
15
Exercise. Let K be a field extension of k, and suppose that K = k(S) for some algebraically independent finite set S. Then K is isomorphic to the field of rational functions
k(y1 , . . . , yt ), for some t.
Lemma. The rational function field F = k(y1 , . . . , yt ) is not k-affine
Proof. Suppose that F is generated as a k-algebra by q1 , . . . , qm , with qi = fi /gi , for
1 ≤ i ≤ m and fi , gi ∈ k[y1 , . . . , yt ]. Let p ∈ k[y1 , . . . , yt ] be irreducible. Since 1/p ∈ F ,
we can write 1/p as a sum of products of the q1 , . . . , qm . It is not hard to see, therefore,
that p must divide some product of the g1 , . . . , gm . Hence, p is an irreducible factor of
some gi .
But there are only m choices for gi , and infinitely many pairwise-non-associate irreducible polynomials in k[x1 , . . . , xn ].
We have a reached a contradiction. Thus F is not k-affine.
Exercise. Prove that a k-affine algebraic field extension is necessarily a finite field extension.
Proof of Hilbert’s Nullstellensatz. Suppose that F is a k-affine field, not algebraic over k.
Let S be a subset of F that is maximal among subsets algebraically independent over k.
Set K = k(S). Notice that F is k-affine and algebraic over K, and so F is a finite field
extension of K. It now follows from the Artin-Tate Lemma that K is k-affine.
Since K is k-affine, there exists a finite subset S 0 of S such that K = k(S 0 ). (Exercise.)
It now follows that K is isomorphic to a field of rational functions k(y1 , . . . , yt ), for some
t. Therefore, K cannot be k-affine, by the preceding lemma.
We have arrived at a contradiction, and the theorem follows. Spectra of Commutative Rings
In this section, R denotes a commutative ring, and the set of maximal ideals of R is
denoted max R.
•
A nonzero element r of R is a zero divisor provided there exists a nonzero element s
of R such that rs = 0. (Note that zero is not a zero divisor.) We say that R is a domain
provided R is not the trivial ring and has no zero divisors. An ideal P of R is prime if
R/P is a domain.
•
Note that every maximal ideal M of R is prime, since R/M is a field. Also note that
Z and k[x] are domains.
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HILBERT’S NULLSTELLENSATZ
Exercise. Let P be an ideal of R. Prove that the following conditions are equivalent:
(a) P is prime
(b) For every pair of ideals I and J (of R) such that IJ ⊆ P , either I ⊆ P or J ⊆ P .
(c) For every pair of ideals I and J, both containing P , such that IJ ⊆ P , either I ⊆ P
or J ⊆ P .
•
If I is an ideal of R, set
V (I) = {M ∈ max R | M ⊇ I}.
Exercise. Prove the following:
(i) V ((0)) = max R, and V (R) = ∅.
(ii) V (I) ∪ V (J) = V (IJ).
(iii) V (I) ∩ V (J) = V (I + J).
• The Zariski topology on max R is obtained by taking the closed subsets to be those of
the form V (I) for ideals I of R.
Exercise. Use the previous exercise to prove that the Zariski topology on max R is indeed
a topology.
Exercise. Suppose that k is algebraically closed, and set R = k[x1 , . . . , xn ]. In the
corollary to Hilbert’s Nullstellensatz, we saw that
max R = {hx1 − λ1 , . . . , xn − λn i | λ1 , . . . , λn ∈ k}.
(i) Prove that the function
(λ1 ,... ,λn )7−→hx1 −λ1 ,... ,xn −λn i
k n −−−−−−−−−−−−−−−−−−−−−→ max R
is bijective.
(ii) Let f ∈ k[x1 , . . . , xn ], and let (λ1 , . . . , λn ) ∈ k n . Prove that
f (λ1 , . . . , λn ) = 0 ⇔ f ∈ hx1 − λ1 , . . . , xn − λn i.
(iii) Use (ii) to prove that the bijection between max R and k n is a homeomorphism
with respect to the Zariski topologies on each.
• In view of the last exercise (assuming that k is algebraically closed), we think of k n
and max R, each equipped with the Zariski topology, as the “same object.”
These notes were originally developed for an NSF VIGRE undergraduate research course
at Texas A&M University. Later drafts were used for graduate courses at Temple University and for the Temple University Mathematics Research Experience for Undergraduates,
supported by an NSF REU Site Grant.
c 2009 E. S. Letzter