hilbert's nullstellensatz - an introduction to algebraic ... · introduction i my talk today...

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

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 .

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.

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].

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].

√I = I(V(I ))

I(V(I )) = K [x1, x2, . . . , xn]

1n ∈ I

√I = I(V(I ))

I(V(I )) = K [x1, x2, . . . , xn]

1n ∈ I

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∏

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 )).

Let p ∈√

0 = φx(pd) = pd(x) =d∏

Let p ∈√

0 = φx(pd) = pd(x) =d∏

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 .

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∑

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 ].

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 ].

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)

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)

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)

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∗

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∗

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.

f d = p1q∗1 + · · ·+ pkq∗

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 ,

f d = p1q∗1 + · · ·+ pkq∗

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 ,

Conclusion/Questions

HOORAY! Questions?

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