Hilbert s Nullstellensatz

Transcription

1 Hilbert s Nullstellensatz An Introduction to Algebraic Geometry Scott Sanderson Department of Mathematics Williams College April 6, 2013

2 Introduction My talk today is on Hilbert s Nullstellensatz, a foundational result in the field of algebraic geometry.

3 Introduction My talk today is on Hilbert s Nullstellensatz, a foundational result in the field of algebraic geometry. First proved by David Hilbert in 1900.

4 Introduction My talk today is on Hilbert s Nullstellensatz, a foundational result in the field of algebraic geometry. First proved by David Hilbert in Pronounced nool-shtell-en-zatss.

5 Introduction My talk today is on Hilbert s Nullstellensatz, a foundational result in the field of algebraic geometry. First proved by David Hilbert in Pronounced nool-shtell-en-zatss. The Nullstellensatz derives its name, like many other German words, from a combination of smaller words: null (zero), stellen (to put/place), satz (theorem). It is generally translated as theorem of zeros, or more literally as zero places theorem.

6 Polynomial Rings Given a ring R, we can define an associated ring, R[x], called the polynomial ring over R in the indeterminate x. Elements of R[x] are of the form: c n x n + c n 1 x n c 1 x 1 + c 0 where each c i is an element of R. Multiplication and addition are defined for polynomial rings to correspond with the usual rules for polynomials from high school algebra.

7 Evaluating Polynomials For any r R, we can define a natural homomorphism φ r : R[x] R that corresponds to evaluating each polynomial in R[x] at r in the usual way.

8 Evaluating Polynomials For any r R, we can define a natural homomorphism φ r : R[x] R that corresponds to evaluating each polynomial in R[x] at r in the usual way. These notions generalize naturally to polynomials defined over any number of indeterminates, which we denote by R[x 1, x 2,..., x n ]. They also generalize to fields of rational functions, which we denote R(x 1, x 2,..., x n ). Note: When no confusion will arise, we often denote φ x (p) simply by p(x).

9 Affine Varieties Definition (Affine Variety) Let S = {p 1, p 2,..., p n } be a subset of K[x 1, x 2,..., x n ], with K an algebraically closed field. We say that the intersection of the zero sets of the p i is the affine variety associated with S, and we denote this set by V(S).

10 Ideals Definition (Ideal) An ideal of a ring R is a subset of R satisfying the following conditions: For any a, b I, a + b I. For any a I and r R, ar I.

11 Examples - Ideals {0} is an ideal of every ring. It is called the trivial ideal. nz, the subset of Z containing all multiples of n is an ideal of Z.

12 Generating Ideals For any set S R, we can associate an ideal with S as follows:

13 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 is not too hard to verify that S is given by: where s i S and r i R. S = {r r = r 1 s 1 + r 2 s r k s k }

14 Generating Ideals cont d Given a set of points V K n, we can construct an ideal of K[x 1, x 2,..., x n ] whose variety is precisely V. Definition (I(V )) Let V K n. I(V ) K[x 1, x 2,..., x n ] is the set of all polynomials in K[x 1, x 2,..., x n ] that vanish on every point in V.

15 Generating Ideals cont d Given a set of points V K n, we can construct an ideal of K[x 1, x 2,..., x n ] whose variety is precisely V. Definition (I(V )) Let V K n. I(V ) K[x 1, x 2,..., x n ] is the set of all polynomials in K[x 1, x 2,..., x n ] that vanish on every point in V. We ll leave it as an exercise to show that I(V ) is, in fact, an ideal.

16 Radical Ideals Given an ideal I R, we can construct a new ideal I, called the radical ideal of I, which is given by: I = {r r n I }, r R, n N

17 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 be finitely generated.

18 Hilbert s Basis Theorem Theorem (Hilbert s Basis Theorem) For any ring R, if R is Noetherian, then so is R[x 1, x 2,..., x n ].

19 Hilbert s Basis Theorem Theorem (Hilbert s Basis Theorem) For any ring R, if R is Noetherian, then so is R[x 1, x 2,..., x n ]. Corollary K[x 1, x 2,..., x n ] is Noetherian for any field K. We will use this fact in our proof in a moment.

20 Hilbert s Nullstellensatz We now have all the vocabulary we need to state Hilbert s Nullstellensatz in both its strong and weak forms.

21 Hilbert s Nullstellensatz We now have all the vocabulary we need to state Hilbert s Nullstellensatz in both its strong and weak forms. Theorem (Hilbert Nullstellensatz (Weak Form)) Let K be an algebraically closed field, and let I K[x 1, x 2,..., x n ] be an ideal such that V(I ) =. Then I = K[x 1, x 2,..., x n ].

22 Hilbert s Nullstellensatz We now have all the vocabulary we need to state Hilbert s Nullstellensatz in both its strong and weak forms. Theorem (Hilbert Nullstellensatz (Weak Form)) Let K be an algebraically closed field, and let I K[x 1, x 2,..., x n ] be an ideal such that V(I ) =. Then I = K[x 1, x 2,..., x n ]. Theorem (Hilbert Nullstellensatz (Strong Form)) Let K be an algebraically closed field, and let I k[x 1, x 2,..., x n ]. Then I(V(I )) = I.

23 Hilbert s Nullstellensatz Despite the unusual nomenclature, the strong and weak Nullstellensatze are actually equivalent!

24 Hilbert s Nullstellensatz Despite the unusual nomenclature, the strong and weak Nullstellensatze are actually equivalent! We won t have time to prove the full Nullstellensatz today. The usual way of doing so is to first prove the weak Nullstellensatz, then prove that the strong is equivalent to the weak.

25 Proof: Strong Weak Let K be an algebraically closed field and I an ideal in K[x 1, x 2,..., x n ] with V(I ) =.

26 Proof: Strong Weak Let K be an algebraically closed field and I an ideal in K[x 1, x 2,..., x n ] with V(I ) =. Since V(I ) =, I(V(I )) is the set of polynomials that vanish on every point in the empty set.

27 Proof: Strong Weak Let K be an algebraically closed field and I an ideal in K[x 1, x 2,..., x n ] with V(I ) =. Since V(I ) =, I(V(I )) is the set of polynomials that vanish on every point in the empty set. But this statement is (vacuously) true for all polynomials in K[x 1, x 2,..., x n ], so I(V(I )) = K[x 1, x 2,..., x n ].

28 Proof: Strong Weak Assuming now the strong Nullstellensatz, we have I = I(V(I ))

29 Proof: Strong Weak Assuming now the strong Nullstellensatz, we have I = I(V(I )) and we just showed that I(V(I )) = K[x 1, x 2,..., x n ]

30 Proof: Strong Weak Assuming now the strong Nullstellensatz, we have I = I(V(I )) and we just showed that I(V(I )) = K[x 1, x 2,..., x n ] so it follows that I = K[x1, x 2,..., x n ] which implies that 1 I. By the definition of a I there exists an n N such that 1 n I so 1 I, which implies that I = K[x 1, x 2,..., x n ].

31 Proof: Weak Strong We now move on to the proof that the weak Nullstellensatz is actually equivalent to the strong.

32 Proof: Weak Strong We now move on to the proof that the weak Nullstellensatz is actually equivalent to the strong. The proof we looking at today is due to J.L. Rabinowitsch. Though remarkably short, it is, as mentioned earlier, a bit tricky.

33 Proof: Weak Strong We now move on to the proof that the weak Nullstellensatz is actually equivalent to the strong. The proof we looking at today is due to J.L. Rabinowitsch. Though remarkably short, it is, as mentioned earlier, a bit tricky. Tricky enough, in fact, that the technique used in this proof has become known as the Trick of Rabinowitsch

34 Weak Strong Let K be an algebraically closed field, and let I be an ideal in K[x 1, x 2,..., x n ]. Assuming the weak Nullstellensatz, we want to show that I(V(I )) = I.

35 Weak Strong: I(V(I )) I We first show that I(V(I )) I. This direction is easy.

36 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 p vanishes 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 p d I, which means that: 0 = φ x (p d ) = p d (x) = d p(x) i=1

37 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 p vanishes 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 p d I, which means that: 0 = φ x (p d ) = p d (x) = d p(x) Since p d I, p d (x) = 0, and since K is a field, it has no zero divisors, so it immediately follows that p(x) = 0. Thus p vanishes on every point in V(I ), so it is in I(V(I )). i=1

38 Weak Strong: I(V(I )) I Proof Idea Given an ideal I in K[x 1, x 2,..., x n ] and a polynomial f that vanishes everywhere in V(I ), we want to show that there exists some natural number d such that f d I. Our general strategy for doing so is a common one in mathematics: we consider a related object in a higher dimensional space and project back down to the space we actually care about. Specifically, we construct a related ideal, I in a larger polynomial ring by adding another indeterminate, which we call y. We do so in a way that guarantees that V(I ) =, which lets us use the weak Nullstellensatz to conclude that I = K[x 1, x 2,..., x n, y]. This allows us to construct an expression for f d in terms of products of the generators of I, which is sufficient to show that f d I.

39 Weak Strong: I(V(I )) I Proof. Let K be an algebraically closed field, let I be an ideal in K[x 1, x 2,..., x n ], and let f be a polynomial in I(V(I )), i.e. a polynomial that vanishes everywhere on V(I ).

40 Weak Strong: I(V(I )) I Proof. Let K be an algebraically closed field, let I be an ideal in K[x 1, x 2,..., x n ], and let f be a polynomial in I(V(I )), i.e. a polynomial that vanishes everywhere on V(I ). Since K is a field, it is Noetherian, so Hilbert s Basis Theorem tells us that K[x 1, x 2,..., x n ] is also Noetherian, which means that I can be written as p 1,..., p k, where each p i I.

41 Weak Strong: I(V(I )) I Proof. Let K be an algebraically closed field, let I be an ideal in K[x 1, x 2,..., x n ], and let f be a polynomial in I(V(I )), i.e. a polynomial that vanishes everywhere on V(I ). Since K is a field, it is Noetherian, so Hilbert s Basis Theorem tells us that K[x 1, x 2,..., x n ] is also Noetherian, which means that I can be written as p 1,..., p k, where each p i I. Thus we need to show that there exists a d N and q 1,..., q k K[x 1, x 2,..., x n ] such that f d = k p i q i i=1.

42 Weak Strong: I(V(I )) I Consider the ideal I K[x 1, x 2,..., x n, y] given by p 1,..., p k, 1 yf, where f and the p i are now considered as elements of K[x 1, x 2,..., x n, y] that happen not to contain the indeterminate y.

43 Weak Strong: I(V(I )) I Consider the ideal I K[x 1, x 2,..., x n, y] given by p 1,..., p k, 1 yf, where f and the p i are now considered as elements of K[x 1, x 2,..., x n, y] that happen not to contain the indeterminate y. Suppose x V(I ). x must be a zero of each p i, which means that it must also be a zero of f. But this implies that x cannot be a zero of 1 yf, a contradiction, since 1 yf I. Thus V(I ) =, which by the weak Nullstellensatz implies that I = K[x 1, x 2,..., x n, y].

44 Weak Strong: I(V(I )) I In particular, we now have that 1 I, so there exist q 1,... q k+1 K[x 1, x 2,..., x n, y] such that 1 = p 1 q 1 + p n q n + (1 yf )q n+1

45 Weak Strong: I(V(I )) I In particular, we now have that 1 I, so there exist q 1,... q k+1 K[x 1, x 2,..., x n, y] such that 1 = p 1 q 1 + p n q n + (1 yf )q n+1 Consider now the homomorphism ψ : K[x 1, x 2,..., x n, y] K(x 1, x 2,..., x n ) which takes y i (1/f ) i We have: ψ(1) = ψ(p 1 q 1 + p n q n + (1 yf )q n+1 ) 1 = ψ(p 1 )ψ(q 1 ) + + ψ(p n )ψ(q n ) + ψ(1 yf )ψ(q n+1 )

46 Weak Strong: I(V(I )) I Since y does not appear in the p 1, they remain unchanged (except that we now interpret them as rational functions with denominator 1). The expression 1 yf gets taken to 1 1 = 0, and each q i becomes a sum of rational functions which contain only powers of f in their denominators. Thus we have: 1 = p 1 ψ(q 1 ) + + p n ψ(q n ) + (1 1)ψ(q n+1 ) 1 = p 1 ψ(q 1 ) + + p n ψ(q n )

47 Weak Strong: I(V(I )) I Let d be the maximum degree of f that appears in the denominator of any of the ψ(q i ). Multiplying on both sides by f d, we have: f d = p 1 q p k q k where the q i are now all rational functions with denominator 1.

48 Weak Strong: I(V(I )) I Let d be the maximum degree of f that appears in the denominator of any of the ψ(q i ). Multiplying on both sides by f d, we have: f d = p 1 q p k q k where the q i are now all rational functions with denominator 1. The subring of K(x 1, x 2,..., x n ) containing entries of denominator 1 is trivially isomorphic to K[x 1, x 2,..., x n ], so we can map back to our original ring to obtain the same equality: f d = p 1 q p k q k where now all the terms are understood as elements of K[x 1, x 2,..., x n ].

49 Weak Strong: I(V(I )) I Let d be the maximum degree of f that appears in the denominator of any of the ψ(q i ). Multiplying on both sides by f d, we have: f d = p 1 q p k q k where the q i are now all rational functions with denominator 1. The subring of K(x 1, x 2,..., x n ) containing entries of denominator 1 is trivially isomorphic to K[x 1, x 2,..., x n ], so we can map back to our original ring to obtain the same equality: f d = p 1 q p k q k where now all the terms are understood as elements of K[x 1, x 2,..., x n ]. But this shows that for some d N, f d can be written as a sum of products of the p i, which implies that f I, which is what we wanted to show.

50 Conclusion/Questions HOORAY! Questions?