Hilbert’s Nullstellensatz An Introduction to Algebraic Geometry Scott Sanderson Department of Mathematics Williams College April 6, 2013
Hilbert’s NullstellensatzAn Introduction to Algebraic Geometry
Scott Sanderson
Department of MathematicsWilliams College
April 6, 2013
Introduction
I My talk today is on Hilbert’s Nullstellensatz, a foundationalresult in the field of algebraic geometry.
I First proved by David Hilbert in 1900.
I Pronounced “nool-shtell-en-zatss”.
I The Nullstellensatz derives its name, like many other Germanwords, from a combination of smaller words: null (zero),stellen (to put/place), satz (theorem). It is generallytranslated as “theorem of zeros”, or more literally as “zeroplaces theorem”.
Introduction
I My talk today is on Hilbert’s Nullstellensatz, a foundationalresult in the field of algebraic geometry.
I First proved by David Hilbert in 1900.
I Pronounced “nool-shtell-en-zatss”.
I The Nullstellensatz derives its name, like many other Germanwords, from a combination of smaller words: null (zero),stellen (to put/place), satz (theorem). It is generallytranslated as “theorem of zeros”, or more literally as “zeroplaces theorem”.
Introduction
I My talk today is on Hilbert’s Nullstellensatz, a foundationalresult in the field of algebraic geometry.
I First proved by David Hilbert in 1900.
I Pronounced “nool-shtell-en-zatss”.
I The Nullstellensatz derives its name, like many other Germanwords, from a combination of smaller words: null (zero),stellen (to put/place), satz (theorem). It is generallytranslated as “theorem of zeros”, or more literally as “zeroplaces theorem”.
Introduction
I My talk today is on Hilbert’s Nullstellensatz, a foundationalresult in the field of algebraic geometry.
I First proved by David Hilbert in 1900.
I Pronounced “nool-shtell-en-zatss”.
I The Nullstellensatz derives its name, like many other Germanwords, from a combination of smaller words: null (zero),stellen (to put/place), satz (theorem). It is generallytranslated as “theorem of zeros”, or more literally as “zeroplaces theorem”.
Polynomial Rings
I Given a ring R, we can define an associated ring, R[x ], calledthe polynomial ring over R in the indeterminate x . Elementsof R[x ] are of the form:
cnxn + cn−1xn−1 + · · ·+ c1x1 + c0
where each ci is an element of R.
I Multiplication and addition are defined for polynomial rings tocorrespond with the usual rules for polynomials from highschool algebra.
“Evaluating” Polynomials
I For any r ∈ R, we can define a natural homomorphismφr : R[x ]→ R that corresponds to “evaluating” eachpolynomial in R[x ] at r in the usual way.
I These notions generalize naturally to polynomials defined overany number of indeterminates, which we denote byR[x1, x2, . . . , xn].
I They also generalize to fields of rational functions, which wedenote R(x1, x2, . . . , xn).
I Note: When no confusion will arise, we often denote φx(p)simply by p(x).
“Evaluating” Polynomials
I For any r ∈ R, we can define a natural homomorphismφr : R[x ]→ R that corresponds to “evaluating” eachpolynomial in R[x ] at r in the usual way.
I These notions generalize naturally to polynomials defined overany number of indeterminates, which we denote byR[x1, x2, . . . , xn].
I They also generalize to fields of rational functions, which wedenote R(x1, x2, . . . , xn).
I Note: When no confusion will arise, we often denote φx(p)simply by p(x).
Affine Varieties
Definition (Affine Variety)
Let S = {p1, p2, . . . , pn} be a subset of K [x1, x2, . . . , xn], with Kan algebraically closed field. We say that the intersection of thezero sets of the pi is the affine variety associated with S , and wedenote this set by V(S).
Ideals
Definition (Ideal)
An ideal of a ring R is a subset of R satisfying the followingconditions:
I For any a, b ∈ I , a + b ∈ I .
I For any a ∈ I and r ∈ R, ar ∈ I .
Examples - Ideals
I {0} is an ideal of every ring. It is called the trivial ideal.
I nZ, the subset of Z containing all multiples of n is an ideal ofZ.
Generating Ideals
For any set S ⊆ R, we can associate an ideal with S as follows:
Definition (〈S〉)For any subset S ⊆ R, we define the ideal generated by S ,denoted 〈S〉, to be the intersection of all ideals containing S . It isnot too hard to verify that 〈S〉 is given by:
〈S〉 = {r | r = r1s1 + r2s2 + · · ·+ rksk}
where si ∈ S and ri ∈ R.
Generating Ideals
For any set S ⊆ R, we can associate an ideal with S as follows:
Definition (〈S〉)For any subset S ⊆ R, we define the ideal generated by S ,denoted 〈S〉, to be the intersection of all ideals containing S . It isnot too hard to verify that 〈S〉 is given by:
〈S〉 = {r | r = r1s1 + r2s2 + · · ·+ rksk}
where si ∈ S and ri ∈ R.
Generating Ideals cont’d
Given a set of points V ⊆ Kn, we can construct an ideal ofK [x1, x2, . . . , xn] whose variety is precisely V .
Definition (I(V ))
Let V ⊂ Kn. I(V ) ⊆ K [x1, x2, . . . , xn] is the set of all polynomialsin K [x1, x2, . . . , xn] that vanish on every point in V .
We’ll leave it as an exercise to show that I(V ) is, in fact, an ideal.
Generating Ideals cont’d
Given a set of points V ⊆ Kn, we can construct an ideal ofK [x1, x2, . . . , xn] whose variety is precisely V .
Definition (I(V ))
Let V ⊂ Kn. I(V ) ⊆ K [x1, x2, . . . , xn] is the set of all polynomialsin K [x1, x2, . . . , xn] that vanish on every point in V .
We’ll leave it as an exercise to show that I(V ) is, in fact, an ideal.
Radical Ideals
Given an ideal I ⊆ R, we can construct a new ideal√
I , called theradical ideal of I , which is given by:√
I = {r | rn ∈ I}, r ∈ R, n ∈ N
More Special Rings
Definition (Noetherian Ring)
A ring R is said to be Noetherian if every ideal of R is of the form〈S〉, where S is a finite subset of R. (Such ideals are said to befinitely generated.
Hilbert’s Basis Theorem
Theorem (Hilbert’s Basis Theorem)
For any ring R, if R is Noetherian, then so is R[x1, x2, . . . , xn].
Corollary
K [x1, x2, . . . , xn] is Noetherian for any field K.
We will use this fact in our proof in a moment.
Hilbert’s Basis Theorem
Theorem (Hilbert’s Basis Theorem)
For any ring R, if R is Noetherian, then so is R[x1, x2, . . . , xn].
Corollary
K [x1, x2, . . . , xn] is Noetherian for any field K.
We will use this fact in our proof in a moment.
Hilbert’s Nullstellensatz
We now have all the vocabulary we need to state Hilbert’sNullstellensatz in both its strong and weak forms.
Theorem (Hilbert Nullstellensatz (Weak Form))
Let K be an algebraically closed field, and let I ⊆ K [x1, x2, . . . , xn]be an ideal such that V(I ) = ∅. Then I = K [x1, x2, . . . , xn].
Theorem (Hilbert Nullstellensatz (Strong Form))
Let K be an algebraically closed field, and let I ⊆ k[x1, x2, . . . , xn].Then I(V(I )) =
√I .
Hilbert’s Nullstellensatz
We now have all the vocabulary we need to state Hilbert’sNullstellensatz in both its strong and weak forms.
Theorem (Hilbert Nullstellensatz (Weak Form))
Let K be an algebraically closed field, and let I ⊆ K [x1, x2, . . . , xn]be an ideal such that V(I ) = ∅. Then I = K [x1, x2, . . . , xn].
Theorem (Hilbert Nullstellensatz (Strong Form))
Let K be an algebraically closed field, and let I ⊆ k[x1, x2, . . . , xn].Then I(V(I )) =
√I .
Hilbert’s Nullstellensatz
We now have all the vocabulary we need to state Hilbert’sNullstellensatz in both its strong and weak forms.
Theorem (Hilbert Nullstellensatz (Weak Form))
Let K be an algebraically closed field, and let I ⊆ K [x1, x2, . . . , xn]be an ideal such that V(I ) = ∅. Then I = K [x1, x2, . . . , xn].
Theorem (Hilbert Nullstellensatz (Strong Form))
Let K be an algebraically closed field, and let I ⊆ k[x1, x2, . . . , xn].Then I(V(I )) =
√I .
Hilbert’s Nullstellensatz
I Despite the unusual nomenclature, the strong and weakNullstellensatze are actually equivalent!
I We won’t have time to prove the full Nullstellensatz today.The usual way of doing so is to first prove the weakNullstellensatz, then prove that the strong is equivalent to theweak.
Hilbert’s Nullstellensatz
I Despite the unusual nomenclature, the strong and weakNullstellensatze are actually equivalent!
I We won’t have time to prove the full Nullstellensatz today.The usual way of doing so is to first prove the weakNullstellensatz, then prove that the strong is equivalent to theweak.
Proof: Strong ⇒ Weak
Let K be an algebraically closed field and I an ideal inK [x1, x2, . . . , xn] with V(I ) = ∅.
Since V(I ) = ∅, I(V(I )) is the set of polynomials that vanish onevery point in the empty set.
But this statement is (vacuously) true for all polynomials inK [x1, x2, . . . , xn], so I(V(I )) = K [x1, x2, . . . , xn].
Proof: Strong ⇒ Weak
Let K be an algebraically closed field and I an ideal inK [x1, x2, . . . , xn] with V(I ) = ∅.
Since V(I ) = ∅, I(V(I )) is the set of polynomials that vanish onevery point in the empty set.
But this statement is (vacuously) true for all polynomials inK [x1, x2, . . . , xn], so I(V(I )) = K [x1, x2, . . . , xn].
Proof: Strong ⇒ Weak
Let K be an algebraically closed field and I an ideal inK [x1, x2, . . . , xn] with V(I ) = ∅.
Since V(I ) = ∅, I(V(I )) is the set of polynomials that vanish onevery point in the empty set.
But this statement is (vacuously) true for all polynomials inK [x1, x2, . . . , xn], so I(V(I )) = K [x1, x2, . . . , xn].
Proof: Strong ⇒ Weak
Assuming now the strong Nullstellensatz, we have
√I = I(V(I ))
and we just showed that
I(V(I )) = K [x1, x2, . . . , xn]
so it follows that √I = K [x1, x2, . . . , xn]
which implies that 1 ∈√
I . By the definition of a√
I there existsan n ∈ N such that
1n ∈ I
so 1 ∈ I , which implies that I = K [x1, x2, . . . , xn].
Proof: Strong ⇒ Weak
Assuming now the strong Nullstellensatz, we have
√I = I(V(I ))
and we just showed that
I(V(I )) = K [x1, x2, . . . , xn]
so it follows that √I = K [x1, x2, . . . , xn]
which implies that 1 ∈√
I . By the definition of a√
I there existsan n ∈ N such that
1n ∈ I
so 1 ∈ I , which implies that I = K [x1, x2, . . . , xn].
Proof: Strong ⇒ Weak
Assuming now the strong Nullstellensatz, we have
√I = I(V(I ))
and we just showed that
I(V(I )) = K [x1, x2, . . . , xn]
so it follows that √I = K [x1, x2, . . . , xn]
which implies that 1 ∈√
I . By the definition of a√
I there existsan n ∈ N such that
1n ∈ I
so 1 ∈ I , which implies that I = K [x1, x2, . . . , xn].
Proof: Weak ⇒ Strong
I We now move on to the proof that the weak Nullstellensatz isactually equivalent to the strong.
I The proof we looking at today is due to J.L. Rabinowitsch.Though remarkably short, it is, as mentioned earlier, a bittricky.
I Tricky enough, in fact, that the technique used in this proofhas become known as the “Trick of Rabinowitsch”
Proof: Weak ⇒ Strong
I We now move on to the proof that the weak Nullstellensatz isactually equivalent to the strong.
I The proof we looking at today is due to J.L. Rabinowitsch.Though remarkably short, it is, as mentioned earlier, a bittricky.
I Tricky enough, in fact, that the technique used in this proofhas become known as the “Trick of Rabinowitsch”
Proof: Weak ⇒ Strong
I We now move on to the proof that the weak Nullstellensatz isactually equivalent to the strong.
I The proof we looking at today is due to J.L. Rabinowitsch.Though remarkably short, it is, as mentioned earlier, a bittricky.
I Tricky enough, in fact, that the technique used in this proofhas become known as the “Trick of Rabinowitsch”
Weak ⇒ Strong
Let K be an algebraically closed field, and let I be an ideal inK [x1, x2, . . . , xn]. Assuming the weak Nullstellensatz, we want toshow that I(V(I )) =
√I .
Weak ⇒ Strong: I(V(I )) ⊇√I
We first show that I(V(I )) ⊇√
I . This direction is easy.
Let p ∈√
I . To show that p ∈ I(V(I )) we must show that pvanishes on every point in the variety of I . Let x be in V(I ).We know that there exists a natural number d such that pd ∈ I ,which means that:
0 = φx(pd) = pd(x) =d∏
i=1
p(x)
Since pd ∈ I , pd(x) = 0, and since K is a field, it has no zerodivisors, so it immediately follows that p(x) = 0. Thus p vanisheson every point in V(I ), so it is in I(V(I )).
Weak ⇒ Strong: I(V(I )) ⊇√I
We first show that I(V(I )) ⊇√
I . This direction is easy.
Let p ∈√
I . To show that p ∈ I(V(I )) we must show that pvanishes on every point in the variety of I . Let x be in V(I ).We know that there exists a natural number d such that pd ∈ I ,which means that:
0 = φx(pd) = pd(x) =d∏
i=1
p(x)
Since pd ∈ I , pd(x) = 0, and since K is a field, it has no zerodivisors, so it immediately follows that p(x) = 0. Thus p vanisheson every point in V(I ), so it is in I(V(I )).
Weak ⇒ Strong: I(V(I )) ⊇√I
We first show that I(V(I )) ⊇√
I . This direction is easy.
Let p ∈√
I . To show that p ∈ I(V(I )) we must show that pvanishes on every point in the variety of I . Let x be in V(I ).We know that there exists a natural number d such that pd ∈ I ,which means that:
0 = φx(pd) = pd(x) =d∏
i=1
p(x)
Since pd ∈ I , pd(x) = 0, and since K is a field, it has no zerodivisors, so it immediately follows that p(x) = 0. Thus p vanisheson every point in V(I ), so it is in I(V(I )).
Weak ⇒ Strong: I(V(I )) ⊆√I
Proof IdeaGiven an ideal I in K [x1, x2, . . . , xn] and a polynomial f thatvanishes everywhere in V(I ), we want to show that there existssome natural number d such that f d ∈ I . Our general strategy fordoing so is a common one in mathematics: we consider a relatedobject in a higher dimensional space and project back down to thespace we actually care about.
Specifically, we construct a related ideal, I ∗ in a larger polynomialring by adding another indeterminate, which we call y . We do so ina way that guarantees that V(I ∗) = ∅, which lets us use the weakNullstellensatz to conclude that I ∗ = K [x1, x2, . . . , xn, y ]. Thisallows us to construct an expression for f d in terms of products ofthe generators of I , which is sufficient to show that f d ∈ I .
Weak ⇒ Strong: I(V(I )) ⊆√I
Proof.Let K be an algebraically closed field, let I be an ideal inK [x1, x2, . . . , xn], and let f be a polynomial in I(V(I )), i.e. apolynomial that vanishes everywhere on V(I ).
Since K is a field, it is Noetherian, so Hilbert’s Basis Theorem tellsus that K [x1, x2, . . . , xn] is also Noetherian, which means that Ican be written as 〈p1, . . . , pk〉, where each pi ∈ I .
Thus we need to show that there exists a d ∈ N andq1, . . . , qk ∈ K [x1, x2, . . . , xn] such that
f d =k∑
i=1
piqi
.
Weak ⇒ Strong: I(V(I )) ⊆√I
Proof.Let K be an algebraically closed field, let I be an ideal inK [x1, x2, . . . , xn], and let f be a polynomial in I(V(I )), i.e. apolynomial that vanishes everywhere on V(I ).
Since K is a field, it is Noetherian, so Hilbert’s Basis Theorem tellsus that K [x1, x2, . . . , xn] is also Noetherian, which means that Ican be written as 〈p1, . . . , pk〉, where each pi ∈ I .
Thus we need to show that there exists a d ∈ N andq1, . . . , qk ∈ K [x1, x2, . . . , xn] such that
f d =k∑
i=1
piqi
.
Weak ⇒ Strong: I(V(I )) ⊆√I
Proof.Let K be an algebraically closed field, let I be an ideal inK [x1, x2, . . . , xn], and let f be a polynomial in I(V(I )), i.e. apolynomial that vanishes everywhere on V(I ).
Since K is a field, it is Noetherian, so Hilbert’s Basis Theorem tellsus that K [x1, x2, . . . , xn] is also Noetherian, which means that Ican be written as 〈p1, . . . , pk〉, where each pi ∈ I .
Thus we need to show that there exists a d ∈ N andq1, . . . , qk ∈ K [x1, x2, . . . , xn] such that
f d =k∑
i=1
piqi
.
Weak ⇒ Strong: I(V(I )) ⊆√I
Consider the ideal I ∗ ⊆ K [x1, x2, . . . , xn, y ] given by〈p1, . . . , pk , 1− yf 〉, where f and the pi are now considered aselements of K [x1, x2, . . . , xn, y ] that happen not to contain theindeterminate y .
Suppose x ∈ V(I ∗). x must be a zero of each pi , which meansthat it must also be a zero of f . But this implies that x cannot bea zero of 1− yf , a contradiction, since 1− yf ∈ I ∗. ThusV(I ∗) = ∅, which by the weak Nullstellensatz implies thatI ∗ = K [x1, x2, . . . , xn, y ].
Weak ⇒ Strong: I(V(I )) ⊆√I
Consider the ideal I ∗ ⊆ K [x1, x2, . . . , xn, y ] given by〈p1, . . . , pk , 1− yf 〉, where f and the pi are now considered aselements of K [x1, x2, . . . , xn, y ] that happen not to contain theindeterminate y .
Suppose x ∈ V(I ∗). x must be a zero of each pi , which meansthat it must also be a zero of f . But this implies that x cannot bea zero of 1− yf , a contradiction, since 1− yf ∈ I ∗. ThusV(I ∗) = ∅, which by the weak Nullstellensatz implies thatI ∗ = K [x1, x2, . . . , xn, y ].
Weak ⇒ Strong: I(V(I )) ⊆√I
In particular, we now have that 1 ∈ I ∗, so there existq1, . . . qk+1 ∈ K [x1, x2, . . . , xn, y ] such that
1 = p1q1 + · · · pnqn + (1− yf )qn+1
Consider now the homomorphism
ψ : K [x1, x2, . . . , xn, y ]→ K (x1, x2, . . . , xn) which takes
y i → (1/f )i
We have:
ψ(1) = ψ(p1q1 + · · · pnqn + (1− yf )qn+1)
1 = ψ(p1)ψ(q1) + · · ·+ ψ(pn)ψ(qn) + ψ(1− yf )ψ(qn+1)
Weak ⇒ Strong: I(V(I )) ⊆√I
In particular, we now have that 1 ∈ I ∗, so there existq1, . . . qk+1 ∈ K [x1, x2, . . . , xn, y ] such that
1 = p1q1 + · · · pnqn + (1− yf )qn+1
Consider now the homomorphism
ψ : K [x1, x2, . . . , xn, y ]→ K (x1, x2, . . . , xn) which takes
y i → (1/f )i
We have:
ψ(1) = ψ(p1q1 + · · · pnqn + (1− yf )qn+1)
1 = ψ(p1)ψ(q1) + · · ·+ ψ(pn)ψ(qn) + ψ(1− yf )ψ(qn+1)
Weak ⇒ Strong: I(V(I )) ⊆√I
Since y does not appear in the p1, they remain unchanged (exceptthat we now interpret them as rational functions with denominator1). The expression 1− yf gets taken to 1− 1 = 0, and each qi
becomes a sum of rational functions which contain only powers off in their denominators. Thus we have:
1 = p1ψ(q1) + · · ·+ pnψ(qn) + (1− 1)ψ(qn+1)
1 = p1ψ(q1) + · · ·+ pnψ(qn)
Weak ⇒ Strong: I(V(I )) ⊆√I
Let d be the maximum degree of f that appears in the denominatorof any of the ψ(qi ). Multiplying on both sides by f d , we have:
f d = p1q∗1 + · · ·+ pkq∗
k
where the q∗i are now all rational functions with denominator 1.
The subring of K (x1, x2, . . . , xn) containing entries of denominator1 is trivially isomorphic to K [x1, x2, . . . , xn], so we can map back toour original ring to obtain the same equality:
f d = p1q∗1 + · · ·+ pkq∗
k
where now all the terms are understood as elements ofK [x1, x2, . . . , xn]. But this shows that for some d ∈ N, f d can bewritten as a sum of products of the pi , which implies that f ∈
√I ,
which is what we wanted to show.
Weak ⇒ Strong: I(V(I )) ⊆√I
Let d be the maximum degree of f that appears in the denominatorof any of the ψ(qi ). Multiplying on both sides by f d , we have:
f d = p1q∗1 + · · ·+ pkq∗
k
where the q∗i are now all rational functions with denominator 1.
The subring of K (x1, x2, . . . , xn) containing entries of denominator1 is trivially isomorphic to K [x1, x2, . . . , xn], so we can map back toour original ring to obtain the same equality:
f d = p1q∗1 + · · ·+ pkq∗
k
where now all the terms are understood as elements ofK [x1, x2, . . . , xn].
But this shows that for some d ∈ N, f d can bewritten as a sum of products of the pi , which implies that f ∈
√I ,
which is what we wanted to show.
Weak ⇒ Strong: I(V(I )) ⊆√I
Let d be the maximum degree of f that appears in the denominatorof any of the ψ(qi ). Multiplying on both sides by f d , we have:
f d = p1q∗1 + · · ·+ pkq∗
k
where the q∗i are now all rational functions with denominator 1.
The subring of K (x1, x2, . . . , xn) containing entries of denominator1 is trivially isomorphic to K [x1, x2, . . . , xn], so we can map back toour original ring to obtain the same equality:
f d = p1q∗1 + · · ·+ pkq∗
k
where now all the terms are understood as elements ofK [x1, x2, . . . , xn]. But this shows that for some d ∈ N, f d can bewritten as a sum of products of the pi , which implies that f ∈
√I ,
which is what we wanted to show.
Conclusion/Questions
HOORAY! Questions?