LECTURE NOTES: A self-contained proof of the Hilbert s Nullstellensatz. Jesús A. De Loera Univ. of California, Davis

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1 LECTURE NOTES: A self-contained proof of the Hilbert s Nullstellensatz Jesús A De Loera Univ of California, Davis 1

2 1 Introduction Hilbert Nullstellensatz is a central result that tells us a necessary and sufficient condition for when a system of equations has a no solution This theorem was first proven by David Hilbert in 19 Here we present a self-contained proof Theorem 11 (Hilbert s Nullstellensatz (HNS)) Let f 1 =, f 2 =,, f s = be a system of polynomials in C[x 1,, x m ], then the system has no root, or solution, if and only if there exist polynomials α 1, α 2,, α s C[x 1,, x m ] such that 1 = n i=1 α i(x 1,, x m )f i (x 1,, x m ) The idea of the proof is by induction on the number of variables One settles it first for one variable and in the case one has more than one variable one needs to have a way to eliminate variables one at a time (this is done using resultants) The proof of the Hilbert Nullstellensatz in in one variable uses a familiar notion: Definition 12 Let f 1,, f s be univariate polynomials in C[x] (eg f 1 (x) = x 2x 3 + 3) We say a polynomial h is a Greatest Common Divisor (GCD) of f 1,, f s if 1) h f 1, h f 2,, h f s 2) p f i for i = 1,, s = p h The centuries old Euclidean division algorithm allow us to compute the GCD Recall Algorithm: (GCD computation) Input: f, g C[x] Output: gcd(f, g) Set i = 1 Set h = f, h 1 = g while h i do divide h i 1 by h i h i 1 = q i h i + r i where r i equals remainder(h i 1, h i ) Set h i = r i i = i + 1 Suppose f, g are univariable polynomials with complex coeffi- Lemma 13 cients 1) If we know q, r, f satisfy f = qg + r, then GCD(f, g) = GCD(g, f qg) = GCD(g, r) 2) GCD(f 1, f 2,, f s ) = GCD(f 1, GCD(f 2,, f s )) 3) If g 1, g 2 are both GCD(f 1,, f s ), then g 1 = cg 2 where c is a constant 2

3 Proof: : For part (1) GCD(f, g) g, and GCD(f, g) f qg = r this implies GCD(f, g) GCD(g, r) by definition: GCD(g, r) g, and GCD(g, r) r = f qg, which implies GCD(g, r) must divides f Also implies GCD(g, r) GCD(f, g) For the proof of part (3) GCD(f 1, f 2 f s ) f i ; for i = 1,, s This implies GCD(f 1, f 2,, f s ) GCD(f 2,, f s ) and it also divides f 1 Therefore GCD(f1, GCD(f 2, f 3,, f s )) is divided by GCD(f1, f 2, f 3,, f s ) Given these two quantities, if h f 1 and h GCD(f 2, f 3,, f s ) then h f i for i = 1 to s Thus GCD(f1, f 2, f 3,, f s ) must divide GCD(f1, GCD(f 2, f 3,, f s )) This proves they are equal Theorem 14 Euclidean Algorithm works: It returns the GCD(f, g) Proof: : We know h i+1 = remainder(h i 1, h i ) Which can means h i+1 = h i qh i (h i 1 ) By the lemma above GCD(h i 1, h i ) = GCD(h i+1, h i ), Since the degrees decrease and eventually h n =, for some n, h n 1 = GCD(h n 2, h n 3 ) = GCD(h n 3, h n 4 ) = = GCD(h, h 1 ) Now, we clearly obtain then an algorithm to compute GCD (f 1, f 2, f s ) using Lemma part (3) But remember our main goal, to find solutions of the system of polynomial equations How can we determine whether system g 1 (x) =,, g r (x) = has a common root? The key point is that if a is a common root, (x a) is a common factor Then when GCD(g 1 (x),, g r (x)) 1, then you have a common solution, any of the roots of this GCD will be a common root Lemma 15 if h = GCD(f 1, f 2, f s ), there exists polynomials a 1, a 2, a s such that h = a 1 f a s f s Proof: This is proved by induction on the number of polynomials Say for n = 2, apply Euclidean algorithm to f 1, f 2 while keeping track of divisors in each iteration h i+1 = h i g i+1 h 1 1 We can start by observing GCD = h n 1 = h n 2 q n 1 h n 3 ( ) In each iteration: h n 2 = h n 3 q n 2 h n 4 ( ) Substitute (**) into (*) and we obtain: GCD(f 1, f 2 ) = (1 q n 1 )h n 3 q n 2 h n 4 Repeat such back substitution using the identities h i+1 = h i g i+1 h 1 1 Eventually, a 1 f 1 + a 2 f 2 = GCD(f 1, f 2 ) Assume lemma is true for S 1 polynomials or less We know GCD(f 1, f 2, f s ) = GCD(f 1, GCD(f 2, f s )) Thus by induction: GCD(f 2, f s ) can be written as h = GCD(f 2, f s ) = b 2 f b s f s Which, by the case of 2 polynomials, gives, as desired: GCD(f 1, f 2, f s ) = GCD(f 1, h) = r 1 f 1 + sh = r 1 f 1 + s(b 2 f b s f s ) = r 1 f 1 + sb 2 f sb s f s ; thenr 1 = a 1, sb 2 = a 2,, sb s = a s 3

4 Corollary 1 In the case of one variable, a system of equations f 1 (x) =,, f s (x) = has no common root if and only if 1 = a 1 f a s f s ; for some polynomialsa 1, a 2, a s Now that we have the Nullstellensatz for systems of polynomials in one single variable we will use Resultants to reduce any other system to this simpler case The resultant will be a special determinant obtained from f, g polynomials in R[x], where R is any integral domain (Think R = K[y 1,, y n ], or R = Z) The Sylvester matrix syl(f, g) will have the following crucial property: Theorem 1 Let R be an integral domain The polynomials f, g have a common root factor in R[X] if and only if det(syl(f, g)) = Lemma 18 Let f and g be polynomials in R[x] such that f, g are nonzero, where R is an integral domain, and deg(f) = n, deg(g) = m, then f, g have a common factor if and only if A, B polynomials exist in R[x] such that 1 Af + Bg = Bg = Af 2 deg (A) < m, deg (B) < n 3 A and B are nonzeros Proof: ( ) Suppose f = f 1 h, g = g 1 h, for some common h Let A = g 1 and B = f 1, then g 1 f + f 1 g = g 1 f 1 h g 1 f 1 h = ( ) By contradiction Suppose f, g have no common factor, that implies Ã, B polynomials in k[x] such that Ãf + Bg = 1 Then B = BÃf + B Bg = BÃf BAf = (BÃ BA)f implies deg(b) > n, thus this contradicts our assumption that deg(b) < n Example: f = 2x 3 + 4x 2 g = 2x 2 + 3x + 9 Solution: A = a 1 x + a B = b 2 x 2 + b 1 x + b = Af + Bg = (a 1 x + a )(2x 3 + 4x 2) + (b 2 x 2 + b 1 x + b )(2x 2 + 3x + 9) = (2a 1 + 2b 2 )x 4 + (2a + 3b 2 + 3b 1 )x 3 + (4a 1 + 9b 2 + 3b 1 + 2b )x 2 + ( 2a 1 + 4a + 9b 1 + 3b )x + ( 2a + 9b ) 4

5 setting the coefficients equal to zero: 2a 1 + 2b 2 = (1) 2a + 3b 2 + 2b 1 = (2) 4a 1 + 9b 2 + 3b 1 + 2b = (3) 2a 1 + 4a + 9b 1 + 3b = (4) 2a + 9b = (5) form the Sylvester matrix (S(f, g)): } {{ } S(f,g) Therefore, there is no common factor a 1 a b 2 b 1 b = det(s(f, g)) = 113 ALGORITHM: Input: f, g R[x] Output: Yes/No depending on whether they have a common factor Step 1: Compute the Sylvester matrix S(f, g) Step 2: Compute its determinant (= Resultant) If Resultant =, Return Yes there s a common factor Else Resultant Return No, there s no common factor Properties of Resultants : Proof: 1 Resultant GCD(f, g) = 1 2 Resultant(f, g) is a polynomial whose variables are the coefficients of f&g and it has integer coefficients 3 There exists polynomials C, D k[x], such that Cf+Dg = Resultant(f, g) 1 By lemma, resultant there is no common factor GCD(f, g) = 1 5

6 2 Resultant(f, g) = det a n b m a n 1 a n bm 1 b m a n 1 bm 1 a an b bm a b a b }{{} m }{{} n 3 Resultant(f, g) = Take C =, D =, then we are Done Resultant(f, g) det[s(f, g)] Because Resultant(f, g) GCD(f, g) = 1 Then there exist Cf + Dg = 1 Sylvester c n c n 1 c d m d m 1 d = 1 By Cramer s Rules: C i s and D j s can be written in the form: Resultant(f,g) Cf + Dg = 1 multiply by Resultant(f, g) The denominators cancel, hence Cf + Dg = Resultant(f, g) Lemma 19 B(x 1, x 2, x n )ɛc[x 1, x 2, x n ] polynomial that is not identically zero, then Z 1, Z 2, Z n ɛc such that B(Z 1, Z 2,, Z n ) Proof: By induction on n the number of variables If n = 1 we are done, because univariate polynomials have finitely many roots, thus B(x 1 ) in infinitely values Suppose the Lemma is true for n 1 variables and take

7 B(x 1, x 2, x n ) not identically zero Expand it in terms of x n, B(x 1, x 2, x n ) = P d (x 1, x 2, x n 1 )x d n + P d 1 (x 1, x 2,, x n 1 )zn d P 1 (x 1, x 2, x n 1 ) by hypothesis, P i (x 1, x 2, x n ) is not identically zero for some i Thus by induction z 1, z 2, z n 1 ɛc where P i (z 1, z 2, z n 1 ) Substitute x k = z k for k = 1, 2,, n 1 In B, B(z 1, z 2,, z n 1, z n ), compute its root β 1, β 2,, β n, let z n be any complex number β i Lemma 11 (Make me monic!) Let b(x 1, x 2,, x n ) be a polynomial all of whose monomials have total degree d There exist a change of coordinates x 1 = λ 1 y 1 + y n, x 2 = λ 2 y 2 + y n,, x i = λ i y i + y n x n = y n such that B(y 1 +λ 1 y n,, y n 1 +λ n 1 y n, y n ) = P (y 1, y n ) is a monic polynomial with respect to the variable y n This means with C d = 1 P (y 1, y n ) = C d (y 1, y n 1 )y d n + + C (y 1,, y n 1 ) Proof: Do substitution with parameter λ 1, λ 2,, λ n 1 ; say d is the highest degree for y n The leading coefficient, will be the polynomial in λ 1, λ 2, λ n 1 call leading coefficient C d From previous lemma we can find λ 1, λ 2, λ n 1 values in C that make C d Divide by constant C d (λ 1, λ 2, λ n 1 ), P 1 (y 1, y n ) = 1 yn+ d junk Theorem 111 (Hilbert s Nullstellensatz) Let f 1 =, f 2 =, f s = be a system of polynomial equation, f 1, f s ɛc[x 1, x n ] then there is no common solution over C n α 1, α 2, α s ɛc[x 1,, x n ] such that α 1 f 2 +α 2 f 2 ++α s f s = 1 Proof: By induction on the number of variables We already proved it for n = 1 using the GCD properties Suppose true for n 1 variables, you are now given a system with n variables (1) By Lemma about changes of variables you can assume f 1 is monic with respect to x n (2) Let y be an auxiliary variable Let Q(x 1, x 2,, x n, y) = f 2 + yf 3 + y 2 f y s 2 f s (3) Compute the resultant of f 1,Q with respect to x n namely Res(f 1, Q) = R d (x 1,, x n 1 )y d + R d 1 (x 1,, x n 1 )y d R (x 1,, x n 1 ) We saw there exist A, B C[x 1,, x n, y] Af + BQ = Res(f 1, Q, x n ) Because of induction, if I can prove R =, R 1 =,, R d = has no common root then done Thus we know now how, given a system of polynomials without a common root, we can construct a new polynomial whose resultant must also have no solution Here is all as a lemma Lemma 112 Given the system of polynomials f 1 = f 2 = = f s = without a common root, we let Q(x 1, x 2,, x n, y) = f 2 + yf 3 + y 2 f y s 2 f s

8 Next, we compute the resultant: Res(f 1, Q, x 1 ) = R d (x 1,, x n 1 )y d +R d 1 (x 1,, x n 1 )y d 1 + +R (x 1,, x n ) And find the system R d = R d 1 = = R =, which also has no solution Proof: By contradiction, suppose not true, therefore, there must exist a solution to the system constructed Specifically, a 1, a 2,, a n 1 (let these be denoted by ā) such that R d (ā) = R d 1 (ā) = = R (ā) = Thus, Res(f 1, Q, x n )(ā) = Remember, since the resultant equals zero, the polynomials f 1 (ā, x n ) and Q(ā, x n, y) must have a common factor This means that they share a root Let this root be denoted as β So, f 1 (ā, β) = Q(ā, y, β) =, for any value of y Since y is unconstrained, the only way Q can be distinctly zero is if f 2 (ā, β) = f 3 (ā, β) = = Therefore, we have found a solution We will now use Hilbert s Nullstellensatz to prove the fundamental theorem that characterises which polynomials vanish in a variety! Theorem 113 (The Strong Hilbert Nullstellensatz) Let h i, g C[x 1,, x m ] We define the hypothesis variety as follows V = { x h 1 ( x) = h 2 ( x) = = h n ( x) = } Here V can be interpreted as the set of all positions or configurations where the hypotheses are satisfied, then g( x) = x V g r (x 1,, x m ) = n i=1 α i(x 1,, x m )h i (x 1,, x m ) where α i C[x 1,, x m ] Proof: (= ) Suppose g( x) = x V Let f i be defined as follows: f = 1 + g(x)t C[x 1, x 2,, x m, t] f 1 = h 1 f 2 = h 2 f i = h i There is no solution of this system (1 ) thus the Hilbert s Nullstellensatz implies 1 = α (1+tg(x))+α 1 h 1 +α 2 h 2 ++α n h n where α i C[x 1, x 2,, x m, t] If we set t = 1/g, then we get 1 = α 1 (x 1,, x m, 1/g)h 1 (x 1,, x m ) + + α n (x 1,, x m, 1/g)h n (x 1,, x m ) where the denominators look like g(x 1,, x m ) k Then pick the highest power of g that appears in the denominator, and multiply the expression by it we get the desired statement Now let us prove the converse, which is easy Given g n (x 1,, x m ) = n α i (x 1,, x m )h i (x 1,, x m ), i=1 when x is a root of the system of equations means = h i ( x) = = g r ( x) = = g( x) = References [1] Cox, D, Little, J, and O Shea, D Ideals, varieties, and Algorithms, Springer Verlag, Undergraduate Text, 2nd Edition, 199 8

9 [2] Cox, D, Little, J, and O Shea, D Using Algebraic Geometry, Springer Verlag, Undergraduate Text, 2nd Edition, 199 [3] Sturmfels, B Gröbner bases and convex polytopes, university lecture series, vol 8, AMS, Providence RI, (199) 9