5 The existence of Gröbner basis

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1 5 The existence of Gröbner basis We use Buchberger s criterion from the previous section to give an algorithm that constructs a Gröbner basis of an ideal from any given set of generators Hilbert s Basis Theorem is used to show that the algorithm terminates. Here is Buchberger s criterion again: Theorem. Let G = {g,..., g s } k[x,..., x n ] be a set of nonzero polynomials, and let I = (g,..., g s ) k[x,..., x n ] be the ideal they generate. Then G is Gröbner basis for I if and only if S(f, g) G = 0 for every i < j s. Remember that the notation S(f, g) G denotes the reduced remainder of the S- polynomial S(f, g) after division by the polynomials in G you get the same reduced remainder however you do the division because G is a Gröbner basis. 5. The proof of Buchberger s 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 f,..., f s k[x,..., x n ] have the same leading term, LT(f i ) = X 0. If f = c i f i with c i k satisfies LT(f) < X, then there are d i,j k so that f = i<j d i,js(f i, f j ). Proof Let LT(f i ) = a i X so that S(f i, f j ) = (/a i )f i (/a j )f j. The coefficient of X in f is zero by hypothesis, so c a + c s a s = 0. The proof is complete by rearranging the defining expression of f into S-polynomial terms: f = c f ( + c ) s f s ( ) = c a a f + + c s a s a s f s ( ) ( ) = c a a f a 2 f 2 + (c a + c 2 a 2 ) a 2 f 2 a 3 f 3 ( ) +(c a + c 2 a 2 + c 3 a 3 ) a 3 f 3 a 4 f 4 ( ) + + (c a + + c s a s ) a s f s a s f s ( ) +(c a + + c s a s ) a s f s since most terms cancel in the addition and the final c i a i factor is zero.

2 Proof of Buchberger s criterion LT(f) for some i. We can write Let f I. We must show that LT(g i ) divides f = h g + + h s g s () but of course we do not have any bound on the leading terms on the right-hand side of (). Define the largest such term to be X = max i {LM(h i g i ) : i =,..., s}. We can suppose that the expression () 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: S X = {i {,..., s} : LM(h i g i ) = X}, and for i S X we write h i = c i X i + h i where c i k and X > LT(h i). Define g = i S X c x (X i g i ). This has the same leading part as that of the expression (), and that completely cancels out. So LM(g) < X although LM(X i g i ) = X for every i S. So by Lemma 29, g is really a k-linear combination of S-polynomials: there are d i,j k for which g = d i,j S(X i g i, X j g j ). i,j S,i<j The proof is almost complete, bar the calculation: these S-polynomials are almost S(g i, g j ), 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 (), which contradicts the minimality of X. It is easy to see that ( S(X i g i, X j g j ) = X LCM(LM(g i ), LM(g j )) so that S(X i g i, X j g j ) = v h i,j,vg v for h i,j,v satisfying So we conclude that ) S(g i, g j ) max{lm(h i,j,v } = LM(S(X i g i, X j g j )) < X. v f = i S X h i g i + i/ S X h i g i = i S X (c i X i + h i)g i + = i S X c i X i g i + i S X h ig i + = i,j S X,i<j d i,j v S X h i,j,v g v = ( ) v S X i,j S X,i<j d i,jh i,j,v g v Now every expression p v g v on the right-hand side has leading term < X. So this is an expression f = q i g i with max i {LM(q i g i } < X, contradicting the minimality of (). 2

3 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 (Buchberger s algorithm). input I = (f,..., f s ) k[x] output Gröbner basis G I start H := {f,..., f s } repeat G := H P := {{f, g} : f, g H and f g} for {f, g} P do h := S(f, g) H if h 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 (f, f 2,... ) = (f i ) i= of distinct f i I. Consider the ideals I j = (f,..., f j ) which form a chain of inclusions I I 2 I 3. 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 I, I 2,... for which I I 2 I 3. An ascending chain is said to terminate if there is some K N for which I j = I K whenever j K. So an ascending chain that terminates looks like I I 2 I 3. I K I K = I K+ = I K+2 = and the inclusions I j I j+ for j < K may be strict or not. 3

4 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 I I 2 I 3 in k[x, y] where each inclusion is strict. Suppose we start with I = (x 4, x 3 y, x 2 y 3, xy 4, y 6 ). Only 4 monomials are not in I : namely, x, x 2, x 3, y, xy, x 2 y, y 2, xy 2, x 2 y 2, y 3, xy 3, y 4, y 5. (Drawing a picture of the exponents (i, j) N 2 of monomials x i y j makes this clear.) To make I 2 strictly bigger than I, we must include at least one of these missing monomials. Suppose we include xy 2, so that I 2 = (x 4, x 3 y, xy 2 x 2 y 3, xy 4, y 6 ) = (x 4, x 3 y, xy 2, y 6 ). Only monomials are not in I 2 : namely, x, x 2, x 3, y, xy, x 2 y, y 2, y 3, y 4, y 5. To make I 3 strictly bigger than I 2, we must include at least one of these missing monomials. Suppose we include x 2, so that Only 8 monomials are not in I 3 : namely I 3 = (x 4, x 3 y, xy 2, y 6, x 2 ) = (x 2, xy 2, y 6 )., x, y, xy, y 2, y 3, y 4, y 5. 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 I k = I k+ 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 I I 2 I 3 where the ideals stop increasing and are all equal any ascending chain of monomial ideals we build will terminate. Lemma 3. Let R be a ring. Statements (a) and (b) below are equivalent. (a) (b) Every ideal in R has a finite basis. Every ascending chain of ideals in R terminates. Proof Suppose that (a) holds and let I I 2 be an ascending chain of ideals in R. It is easy to check that the subset I = j= I j is an ideal of R. Let f,..., f s I be a basis of I, which exists by (a). Each f i must lie in some ideal of the chain, say f i I j(i). Of course, then f i I k whenever k j(i). In particular, if N = max{j(),..., j(s)} then every f i I N. So I N I N+ I I N, 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. Theorem 32 (Hilbert s Basis Theorem). If I k[x,..., x n ] is an ideal, then there are finitely many polynomials f,..., f s I so that I = (f,..., f s ). 4

5 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[x,..., x n ] be an ideal generated by a (possibly infinite) set of monomials A I, then there is a finite subset {m,..., m s } A so that I = (m,..., m s ). 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(g ),..., LT(g s ) for g,..., g s I. By construction, {g,..., g s } 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 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 f,..., f s 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 H to be the set H after one pass through this loop, H 2 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 H H 2 H 3. Let I j be the ideal generated by {LT(h) : h H j }, and consider the ascending chain I I 2 I 3. I claim that I j I j+ is a strict inclusion for every j. Certainly there is some h H j+ \H j. Moreover, it arises from the algorithm as h = S(f, g) H j for some pair f, g H j. Certainly LT(h) I j+. But equally clearly LT(h) / I j for h is reduced with respect to H j by construction. We have constructed an ascending chain in I that does not terminate. By Lemma 3, 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. 5

6 Homework Q.. Follow the Buchberger algorithm step by step to compute Gröbner basis of the ideal I = (x 2 y, xy ) 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. For example, LT(x 2 y) = x 2 and LT(xy ) = xy, and the ideal generated by these two leading terms is simply (x 2, xy). Now the first (reduced) S-polynomial you compute will turn out to be x y 2. This has leading term x. Including that in the ideal basis leading terms gives (x 2, xy, x) = (x) the ideal has grown a bit, as is clear by drawing the Newton polygon. Now compute another S-polynomial and see whether including its leading term makes the ideal still bigger. Q.2. Follow the Buchberger algorithm step by step to compute Gröbner basis of the ideal I = (x 3, xy ) 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. Q.3. Follow the Buchberger algorithm step by step to compute Gröbner basis of the ideal I = (x 3 y, xy 2 ) 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. Q.4. Let I = (x 2, xy 2, y 3 ), an ideal in k[x, y]. Find an ascending chain I I 2 for which the first three inclusions are strict (that is, I I 2 I 3 I 4 ). Sketch your ideals in N 2, by plotting their exponents (i, j) for x i y j. Can you find an example for which I 4 k[x, y]? Q.5. Explain why a sequence of monomial ideals that starts with the ideal I = (x 2 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.) 6