4 Hilbert’s Basis Theorem and Gröbner basis We define Gröbner bases of ideals in multivariate polynomial rings and see how they work in tandem with the division algorithm. We look again at the standard questions: ideal generation, ideal membership and analysing the solutions of systems of polynomials. Since Gröbner basis makes profit from contemplating the leading terms of equations, we look for a moment at the univariate case. Imagine trying to solve f (t) = 0 over C using only the leading term of f . Of course, that throws almost all the information away. But you can still distinguish between three cases: f = 0, f is a nonzero constant and deg f > 0. That doesn’t sound much, but pronounced geometrically—the set of solutions is either one dimensional, empty or a nonempty finite set—it captures the three possible quite different types of behaviour. 4.1 The leading terms in an ideal Here is the definition of Gröbner basis out of the blue. Definition 19 Let I ⊂ k[x1 , . . . , xn ] be a nonzero ideal. Then a subset G ⊂ I is a Gröbner basis of I if and only if G is finite and for all f ∈ I there is some g ∈ G whose leading monomial LM(g) divides LM(f ). This definition is about as unenlightening as it could possibly be. Since the definition wisely requires that G be finite, it is not even clear that Gröbner bases exist, so here is one class of examples. Lemma 20 Let I = (m1 , . . . , mr ) ⊂ k[x1 , . . . , xn ] be an ideal generated by monomials m1 , . . . , mr . Then G = {m1 , . . . , mr } is a Gröbner basis of I. Proof Any f ∈ I has an expression f = a1 m1 + · · · + ar mr for some polynomials ai ∈ k[x1 , . . . , xn ]. In other words, every term of f is divisible by some mi , so it is certainly true of the leading term of f . Q.E.D. The division algorithm with Gröbner basis In the last section, we said that we could characterise those sets of polynomials that work well with the division algorithm. First we see that the problem of nonuniqueness of the remainder in the Division Algorithm is solved. Theorem 21 Let G = {g1 , . . . , gs } be a Gröbner basis of I ⊂ k[x1 , . . . , xn ]. Then for any f ∈ k[x1 , . . . , xn ] there are polynomials q1 , . . . , qs , r ∈ k[x1 , . . . , xn ] such that f = q1 g1 + · · · + qs gs + r where r is reduced with respect to G and LT(f ) ≥ LT(qi ) LT(gi ) for every i = 1, . . . , s. Moreover r is uniquely determined by these conditions. 1 P In other words, however we compute f = qi gi + r, whether we use the division algorithm or not, as long as we obey the conditions on leading terms, the remainder r will always turn out to be the same polynomial. The qi , by contrast, are not well determined at all, and we compute the extent to which they can be varied in Part IV. Proof The existence of such an r and the conditions on leading terms follows immediately from the division algorithm. We only need to prove that r is unique. Suppose that we have two expressions f = g1 + r1 = g2 + r2 satisfying the conclusions of the theorem. We must prove that r1 = r2 . Certainly r1 − r2 = g2 − g1 ∈ I, so by the definition of Gröbner basis there is a polynomial g ∈ G so that LM(g) divides LM(r1 − r2 ). But both r1 − r2 is reduced with respect to G, since this is true for both r1 and r2 , so there is no monomial in r1 − r2 that is divisible by LM(g) for any g ∈ G. That is a contradiction unless r1 − r2 = 0, and that is what we had to prove. Q.E.D. This proof makes no condition on the order of elements in G, so the following notation does make sense. Definition 22 The unique polynomial r in the theorem is called the remainder of G f after division by G and is denoted either by f or f mod G. This solves many of our questions: part (b) of the corollary below is the ideal membership problem for ideals with a given Gröbner basis. Corollary 23 Let G = {g1 , . . . , gs } be a Gröbner basis of I ⊂ k[x1 , . . . , xn ] and f ∈ k[x1 , . . . , xn ]. Then (a) there is a unique r that is reduced with respect to G and so that f − r ∈ I. G (b) f ∈ I if and only if f = 0. Part (b) follows from (a) because if f ∈ I then r = 0 satisfies the conditions of (a) and so it must be the unique remainder on applying the division algorithm. Notice finally that the Gröbner basis really does generate I as an ideal—it is a natural slip of the brain to assume this without checking, but it is certainly not part of the definition, and fortunately we have not used this fact at all yet. Equivalent characterisations of Gröbner basis There are lots of different ways of saying what a Gröbner basis is. These two propositions give two of them; both are famous. Proposition 24 Let I ⊂ k[x1 , . . . , xn ] be an ideal and G = {g1 , . . . , gs } ⊂ I. Then G is a Gröbner basis of I if and only if for every f ∈ I, there is an expression f = q1 g1 + · · · + qs gs with LT(f ) ≥ LT(qi ) LT(gi ) for every i = 1, . . . , s. 2 Proof Given such an expression, LM(f ) = LM(qi gi ) for at least one i. Then LT(gi ) divides LT(f ). Q.E.D. Given a monomial order > on k[x1 , . . . , xn ], every polynomial f ∈ k[x1 , . . . , xn ] has a leading term LT(f ) with respect to >. If I is an ideal, we can consider the leading terms of every polynomial in I. Definition 25 Let I ⊂ k[x1 , . . . , xn ] be an ideal with monomial order >. The initial ideal of I, denoted LT(I), is the ideal generated by the leading terms of every element of I: in other words, LT(I) = h{LT(f ) : f ∈ I}i ⊂ k[x1 , . . . , xn ]. We can rephrase the definition of Gröbner basis in terms of LT(I). Proposition 26 Let I ⊂ k[x1 , . . . , xn ] be an ideal and G = {g1 , . . . , gs } ⊂ I. Then G is a Gröbner basis of I if and only if {LT(g1 ), . . . , LT(gs )} generates the ideal LT(I). By Lemma 20, if the leading terms LT(gi ) do generate the ideal LT(I) then they are a Gröbner basis for it. This is an exercise in the division algorithm, and we leave the proof to the exercises. 4.2 S-polynomials and the Buchberger criterion We make our routine technique of cancelling leading terms more formal by defining the S-polynomial of two polynomials. Definition 27 Let f, g ∈ k[x1 , . . . , xn ] be two nonzero polynomials. Then the Spolynomial of f and g is S(f, g) = L L f− g LT(f ) LT(g) where L = LCM(LM(f ), LM(g)). For example, let f = 2yx − y and g = 3y 2 − x. Suppose > is the grlex monomial order with y > x. Then LT(f ) = 2yx, LT(g) = 3y 2 so L = LCM(yx, y 2 ) = y 2 x and 2 2 y x (2yx − y) − y3yx2 (3y 2 − x) S(f, g) = 2yx = (y/2)(2yx − y) − (x/3)(3y 2 − x) = −(1/2)y 2 + (1/3)x2 . Dividing this S-polynomial by f, g using the division algorithm gives S(f, g) = −(1/6)g + ((1/3)x2 − (1/6)x) so we see that f,g S(f, g) = (1/3)x2 − (1/6)x. So if I = (f, g), then LT(I) includes (1/3)x2 , and since this is not in the ideal generated by LT(f ), LT(g) we conclude that {f, g} is not a Gröbner basis of I. This is the central observation of Buchberger’s criterion. 3 Theorem 28 (Buchberger’s criterion) Let G = {g1 , . . . , gs } ⊂ k[x1 , . . . , xn ] be a set of nonzero polynomials, and let I = (g1 , . . . , gs ) ⊂ k[x1 , . . . , xn ] be the ideal they generate. Then G is Gröbner basis for I if and only if G S(f, g) = 0 for every 1 ≤ i < j ≤ s. This theorem is central to our theory. We prove it in the next section. I believe it comes to us in exactly this form from Buchberger’s 1965 PhD thesis—you can get Buchberger’s own account from his webpage—although many other people contributed ideas before that. The first point Example 1 Let f = y 2 − 1, g = xy + 1 and define the ideal I = (f, g). Using Buchberger’s criterion, show that the given basis f, g of I is not a Gröbner basis. Show that f, g, h, where h = x + y, is a Gröbner basis for I. (In particular, you must show that h ∈ I—you’ve already done this using the division algorithm.) In fact, f, h is also a Gröbner basis: it is easy to check that g is a combination of these two. ♥ Homework Q.1. Let I be the ideal with basis y 2 − 1, xy + 1 (with respect to the lex order with x > y). Work out why each of the two equivalent characterisations of Gröbner basis above fail for this basis. Show that this is not a Gröbner basis by using the Buchberger criterion. Q.2. Repeat the exercise with the ideal generated by xy 2 − 1 and x2 − y 3 . Q.3. Show that any non-zero ideal in any polynomial ring has infinitely many elements. Q.4. Let f be any polynomial and {m1 , . . . , ms } be a set of monomials. The division P algorithm (applied with some order chosen on the mi ) computes an expression f = qi mi + r, where r is the reduced remainder. Show that r is simply the sum of those terms of f that are not divisible by any of the mi . Conclude that the division algorithm will return the same reduced remainder, whatever order the mi are written in. 4 Example 2 We know that computing the kernel of a matrix M is the same as solving a system of linear equations. Let M be the 3 × 3 matrix of linear forms   x−y z y − 2z 2z x − 4z  . M = y x+y x−y y For different values of x, y, z, the dimension of the solution space will change, that is to say that rank of the matrix M will vary with x, y, z. It is easy to say when M has full rank: that is when det M is not zero, and this happens for all values of x, y, z which do not satisfy x3 − 2x2 y − 5x2 z + 9xyz + y 3 − y 2 z = 0. (1) At values that do satisfy this equation, the rank of M is at most 2. For what values of x, y, z does M have rank at most 1? From linear algebra, we know that this happens at any values for which all of the 2 × 2 minors of M vanish. So we consider the ideal defined by these minors. Let R = k[x, y, z] have the lex order with x > y > z and let I ⊂ R be the ideal generated by the 2 × 2 minors of M , so I has basis 2xz − 3yz, x2 − 2xy − xz + y 2 − yz, xy − 2xz − y 2 − 2yz −x2 + xy + 4xz + y 2 − 6yz, −2xz + 2y 2 − 2yz, x2 + xy − 4xz − y 2 − 4yz xz − 2yz, −xy + 2xz + y 2 − yz, −x2 + xy + 4xz − 2yz. These look complicated, and so the hope is that, as in the univariate case, considering leading terms of these polynomials will give us some information about the solutions. The leading monomials that we can see are x2 , xy, xz. However, it is easy to find elements of I whose leading term is not a combination of these. For example, the leading term of (2xz − 3yz) − 2(xz − 2yz) is yz, which does not have x as a factor so is not a combination. Finding another element with leading term y 2 is also easy. In fact, in this case the problem is now completely solved, since one can check that I = (x2 , xy, xz, y 2 , yz). In other words, the rank of M is at most 1 exactly along the z axis. If we think of points p = (x, y, z) ∈ A3k parametrising the matrices Mp = M , the picture is very pretty. For a general point p ∈ A3k , the matrix Mp has full rank. If p lies on the hypersurface (1) but not on the z-axis, then the rank of Mp is 2. Moreover, the z-axis is exactly the singular locus—the locus where all the derivatives of the equation (1) vanish, and so the locus where the tangent plane is not defined—of this hypersurface. For p on the z-axis but not at the origin, the rank of M is 1. And the origin is certainly a distinguished point of V (I), since the generators of I all vanish twice only there. And we see immediately that special case of rank zero is achieved only when x, y, z are all zero, that is, when p is the origin. 5 Conclusion At heart, this example is just about the leading terms of an ideal that happens to be generated by the nine 2 × 2 minors of a matrix with polynomial entries. Let B be the basis of 2 × 2 minors used to define I. Then the ideal generated by the leading terms of polynomials in B is only (x2 , xy, xz). After calculating (presumably by cancelling leading terms, if you tried it yourself) the basis for I was x2 , xy, xz, y 2 , yz. So certainly the leading terms of different bases for the same ideal can generate different monomial ideals. What we want to look at is the ideal consisting of the leading terms of every f ∈ I. This is what Gröbner basis computes. ♥ 6 5 5.1 The existence of Gröbner bases Back to S-polynomials and the Buchberger criterion Let’s think again about the Buchberger criterion. One implication in this theorem is clear: S-polynomials lie in the ideal I and so reduce to zero on division by a Gröbner basis G. The converse is the point. By construction, S-polynomials are expressions that kill a pair of leading terms. The next lemma says that any k-linear expression that kills leading terms is a combination of S-polynomials. Lemma 29 Suppose P f1 , . . . , fs ∈ k[x1 , . . . , xn ] have the same leading term, LT(fi ) = X 6= 0. If ci fi with ci ∈ k satisfies LT(f ) < X, then there are di,j ∈ k so Pf = that f = i<j di,j S(fi , fj ). Proof Let LT(fi ) = ai X so that S(fi , fj ) = (1/ai )fi − (1/aj )fj . The coefficient of X in f is zero by hypothesis, so c1 a1 + · · · cs as = 0. The proof is complete by rearranging the defining expression of f into S-polynomial terms: f = c1 f1 + · · · cs fs + · · · + cs as a1s fs = c1 a1 a11 f1 − a12 f2 + (c1 a1 + c2 a2 ) a12 f2 − a13 f3 +(c1 a1 + c2 a2 + c3 a3 ) a13 f3 − a14 f4 1 1 + · · · + (c1 a1 + · · · + cs−1 as−1 ) as−1 fs−1 − as fs +(c1 a1 + · · · + cs as ) a1s fs = c1 a1 1 f a1 1 since most terms cancel in the addition and the final P ci ai factor is zero. Q.E.D. Proof of Buchberger’s criterion Let f ∈ I. We must show that LT(gi ) divides LT(f ) for some i. We can write f = h1 g1 + · · · + hs gs (2) but of course we do not have any bound on the leading terms on the right-hand side of (2). Define the largest such term to be X = max {LM(hi gi ) : i = 1, . . . , s} . i We can suppose that the expression (2) for f has been chosen so that this X is minimal (with respect to the monomial order) among leading terms appearing in such expressions. If X = LM(f ) then we are done. So supposing not, we identify the worst offending terms: SX = {i ∈ {1, . . . , s} : LM(hi gi ) = X} , and for i ∈ SX we write hi = ci Xi + h0i where ci ∈ k and X > LT(h0i ). 7 P Define g = i∈SX cx (Xi gi ). This has the same leading part as that of the expression (2), and that completely cancels out. So LM(g) < X although LM(Xi gi ) = X for every i ∈ S. So by Lemma 29, g is really a k-linear combination of S-polynomials: there are di,j ∈ k for which X g= di,j S(Xi gi , Xj gj ). i,j∈S,i<j The proof is almost complete, bar the calculation: these S-polynomials are almost S(gi , gj ), and indeed we check that they do reduce to zero under G; but then we can make an expression for f with smaller leading term X than (2), which contradicts the minimality of X. It is easy to see that X S(Xi gi , Xj gj ) = S(gi , gj ) LCM(LM(gi ), LM(gj )) P so that S(Xi gi , Xj gj ) = v hi,j,v gv for hi,j,v satisfying max{LM(hi,j,v } = LM(S(Xi gi , Xj gj )) < X. v So we conclude that P P f = Pi∈SX hi gi + i∈S gi / X hiP 0 = Pi∈SX (ci Xi + hP )g + i i 1 P = Pi∈SX ci Xi gi +P i∈SX h0i gi + P 1 P = d h g + + i,j i,j,v v v∈SX i,j∈SX ,i<j 2 1 P P P P = d h g + + v v∈SX i,j∈SX ,i<j i,j i,j,v 2 1. Now every expression P pv gv on the right-hand side has leading term < X. So this is an expression f = qi gi with maxi {LM(qi gi } < X, contradicting the minimality of (2). Q.E.D. 5.2 Existence of Gröbner bases and Buchberger’s algorithm Buchberger’s criterion allows one to check whether a given finite set is a Gröbner basis, but it does not explain why an ideal should have a Gröbner basis. It is easy to see that how to use it to make an algorithm that computes a Gröbner basis from a given (finite) basis. Algorithm 1 (Buchberger’s algorithm) I = (f1 , . . . , fs ) ⊂ k[x] input output Gröbner basis G ⊂ I H := {f1 , . . . , fs } start repeat G := H P := {{f, g} : f, g ∈ H and f 6= g} 8 for {f, g} ∈ P do H h := S(f, g) if h 6= 0 then H := H ∪ {h} end if end for until H = G return G. The key point in the proof of this algorithm is its termination. As you must expect, this follows from Hilbert’s basis theorem, and so we discuss that now. Finite generation and the ascending chain condition Let R be any ring and I ⊂ R a non-zero ideal. Certainly I contains infinitely many non-zero elements, so we can make an infinite sequence (f1 , f2 , . . . ) = (fi )∞ i=1 of distinct fi ∈ I. Consider the ideals Ij = (f1 , . . . , fj ) which form a chain of inclusions I1 ⊂ I2 ⊂ I3 ⊂ · · · . Can one ever choose a sequence so that every inclusion in this chain is strict? Definition 30 Let R be a commutative ring. An ascending chain of ideals in R is a sequence of ideals I1 , I2 , . . . for which I1 ⊂ I2 ⊂ I3 ⊂ · · · . An ascending chain is said to terminate if there is some K ∈ N for which Ij = IK whenever j ≥ K. So an ascending chain that terminates looks like I1 ⊂ I2 ⊂ I3 ⊂ · · · . ⊂ IK−1 ⊂ IK = IK+1 = IK+2 = · · · and the inclusions Ij ⊂ Ij+1 for j < K may be strict or not. Example Just for the purposes of this exercise, imagine that the only ideals you work with are monomial ideals. We can try to make an ascending chain of monomial ideals I1 ⊂ I2 ⊂ I3 ⊂ · · · in k[x, y] where each inclusion is strict. Suppose we start with I1 = (x4 , x3 y, x2 y 3 , xy 4 , y 6 ). Only 14 monomials are not in I1 : namely 1, x, x2 , x3 , y, xy, x2 y, y 2 , xy 2 , x2 y 2 , y 3 , xy 3 , y4, y5. (Drawing a picture of the exponents (i, j) ∈ N2 of monomials xi y j makes this clear.) To make I2 strictly bigger than I1 , we must include at least one of these missing monomials. Suppose we include xy 2 , so that I2 = (x4 , x3 y, xy 2 x2 y 3 , xy 4 , y 6 ) = (x4 , x3 y, xy 2 , y 6 ). 9 Only 11 monomials are not in I2 : namely 1, x, x2 , x3 , y, xy, x2 y, y2, y3, y4, y5. To make I3 strictly bigger than I2 , we must include at least one of these missing monomials. Suppose we include x2 , so that I3 = (x4 , x3 y, xy 2 , y 6 , x2 ) = (x2 , xy 2 , y 6 ). Only 8 monomials are not in I3 : namely 1, x, y2, y3, y4, y5. y, xy, And so we can continue. But, whatever choices we make for monomial to add, eventually we will make an ideal that contains all the monomials (and so is equal to k[x, y]) and cannot be enlarged at all. So we cannot make an infinite and strictly increasing chain of ideals: our chain of ideals terminates. Of course, we could have allowed ourselves to make steps Ik = Ik+1 along the way—that is allowed in the definition of ascending chain, after all. But that won’t help: there has to come a point in the increasing chain I1 ⊂ I2 ⊂ I3 ⊂ · · · where the ideals stop increasing and are all equal—any ascending chain of monomial ideals we build will terminate. ♦ Lemma 31 Let R be a ring. Statements (a) and (b) below are equivalent. (a) Every ideal in R has a finite basis. (b) Every ascending chain of ideals in R terminates. Proof Suppose that (a) holds and let I1 ⊂ I2 ⊂ ·S · · be an ascending chain of ideals in R. It is easy to check that the subset I = ∞ j=1 Ij is an ideal of R. Let f1 , . . . , fs ∈ I be a basis of I, which exists by (a). Each fi must lie in some ideal of the chain, say fi ∈ Ij(i) . Of course, then fi ∈ Ik whenever k ≥ j(i). In particular, if N = max{j(1), . . . , j(s)} then every fi ∈ IN . So IN ⊂ IN +1 ⊂ · · · ⊂ I ⊂ IN , and therefore every inclusion in this chain is in fact equality. The converse is easy: if an ideal does not have a finite basis, then the ascending chain constructed as in the example above cannot terminate. Q.E.D. Theorem 32 (Hilbert’s Basis Theorem) If I ⊂ k[x1 , . . . , xn ] is an ideal, then there are finitely many polynomials f1 , . . . , fs ∈ I so that I = (f1 , . . . , fs ). 10 Discussion in place of proof We will not prove this theorem—there is a standard proof that could appear in several courses. But there is an interesting point for us that is often omitted. Imagine that this theorem was proved for monomial ideals: that result is called Dickson’s Lemma, and a precise statement of it is if I ⊂ k[x1 , . . . , xn ] be an ideal generated by a (possibly infinite) set of monomials A ⊂ I, then there is a finite subset {m1 , . . . , ms } ⊂ A so that I = (m1 , . . . , ms ). Theorem 32 follows from this at once. Given I, consider its ideal of leading terms LT(I). By definition, LT(I) is generated by the set of all LT(g) for g ∈ I. By Dickson’s Lemma, it is generated by only finitely many of these monomials, LT(I) = (LT(g1 ), . . . , LT(gs ) for g1 , . . . , gs ∈ I. By construction, {g1 , . . . , gs } is a Gröbner basis for I, and so in particular it is a basis. One might expect a proof of Dickson’s Lemma (see [Cox, Little, O’Shea], for instance) to be easier that a proof of Hilbert’s Basis Theorem. But in fact, the proofs are an almost identical induction on the number of variables working with leading terms of polynomials. It is not usually emphasised, but the usual proof of Theorem 32 does prove the existence of Gröbner bases for every ideal. Proof of Algorithm 1 At every stage, the sets G, H, P are all finite, and so each operation can be carried out and the for-endfor loop is a finite loop. If at the end of the repeat-until loop, the two sets H and G are equal, then every S-polynomial between elements of H reduces to zero modulo H and so H (and therefore also G) is a Gröbner basis for the ideal it generates. Of course, that ideal is I because H contains the original basis f1 , . . . , fs of I, and every element h added to H was also in I. The only question is why does the repeat-until loop terminate. Define H1 to be the set H after one pass through this loop, H2 to be the set H after two passes through this loop, and so on. If the loop does not terminate, then it generates a sequence of strict inclusions H1 ⊂ H2 ⊂ H3 ⊂ · · · . Let Ij be the ideal generated by {LT(h) : h ∈ Hj }, and consider the ascending chain I1 ⊂ I2 ⊂ I3 ⊂ · · · . I claim that Ij ⊂ Ij+1 is a strict inclusion for every j ≥ 1. Certainly there is some Hj h ∈ Hj+1 \ Hj . Moreover, it arises from the algorithm as h = S(f, g) for some pair f, g ∈ Hj . Certainly LT(h) ∈ Ij+1 . But equally clearly LT(h) ∈ / Ij for h is reduced with respect to Hj by construction. We have constructed an ascending chain in I that does not terminate. By Lemma 31, this means that I does not have a finite basis. But that contradicts Hilbert’s basis theorem. So our initial hypothesis that the repeat-until loop did not terminate was wrong. Q.E.D. 11 Q.1. Let I1 = (x2 , xy 2 , y 3 ), an ideal in k[x, y]. Find an ascending chain I1 ⊂ I2 ⊂ · · · for which the first three inclusions are strict (that is, I1 6= I2 6= I3 6= I4 ). Sketch your ideals in N2 , by plotting their exponents (i, j) for xi y j . Can you find an example for which I4 6= k[x, y]? Q.2. Explain why a sequence of monomial ideals that starts with the ideal I1 = (x2 y, y 4 ) in k[x, y] must terminate. (Note that there are infinitely many monomials not lying in this ideal, so the basic finiteness argument that forces termination in the previous example needs something extra.) Q.3. Follow the Buchberger algorithm step by step to compute Gröbner basis of the ideal I = (x3 − y, xy 2 − 1) ⊂ k[x, y] with respect to the lex order with x > y. Compute the sequence of ideals generated by the leading terms of polynomials in the basis at each stage of the calculation. 12