Math 4370 Exam 1 Handed out March 9th 2010 Due March 18th 2010 Problem 1. Recall from problem 1.4.6.e in the book, that a generating set {f 1,..., f s } of I is minimal if I is not the ideal generated by any proper subset of {f 1,..., f s }. a) Show that a generating set {f 1,..., f s } of I is minimal if and only if for each i, one has f i f 1,..., f i 1, f i+1,..., f s. b) Use the above problem, and Dickson s Lemma to prove that every monomial ideal I in k[x 1,..., x n ] has a minimal monomial generating set (i.e. a minimal generating set that also consists of monomials), and that such a generating set is unique. c) Use b) to give a different proof of the fact that if G and G are minimal Gröbner bases, then lt(g) = lt( G). d) Let f k[x 1,..., x n ]. Show that if f x 1,..., x n, then x 1,..., x n, f = k[x 1,..., x n ]. Solution. a) If {f 1,..., f s } is not minimal, then there is some proper subset that generates I, and hence this proper subset is contained in {f 1,..., f i 1, f i+1,..., f s } for some i. Since this proper subset generates, and f i I, one must have f i f 1,..., f i 1, f i+1,..., f s. For the converse, note that if f i f 1,..., f i 1, f i+1,..., f s, then by definition {f 1,..., f s } is not minimal, since the set without f i generates I. b) By Dickson s Lemma, a monomial ideal I has a finite monomial generating set {m 1,..., m s }. Using part a), we can remove the elements of this list one by one until we can t remove any more. This gives us a minimal monomial generating set {m 1,..., m s }. We now need to show that such a set is unique. Suppose M and M are minimal monomial generating sets for I, and let m M. Then since m I = M, there exists a monomial n 1 such that m = n 1 m for some m M. Similarly, there exists a monomial n 2 such that m = n 2 m for m M. Therefore m = n 1 n 2 m for some m M. However, since M is minimal, this forces m = m and hence n 1 = n 2 = 1. Therefore, we have an inclusion M M. By symmetry, we also have M M, and thus M = M. c) The condition that G is a minimal Gröbner basis is exactly the same that lt(g) is a minimal monomial generating set for lt(i). Therefore if G and G are minimal Gröbner bases, then their set of lead terms are both minimal monomial generating sets for lt(i) and thus are the same. d) Let f x 1,..., x n, and set I = f, x 1,..., x n. Then we can write f = a 1 x 1 + +a n x n +b with b k nonzero (otherwise, f would be in the ideal). Therefore, b = f a 1 x 1 a n x n I, and hence 1 = 1 b b I, and thus I = k[x 1,..., x n ]. Problem 2. Let a = (a 1,..., a n ) be a vector in R n. For a monomial x α k[x 1,..., x n ], define the weight of x α (denoted wt(x α )) by wt(x α ) := a α, where is the usual dot product. Define an order > a on monomials by x α > a x β if wt(x α ) > wt(x β ). If the weights are equal, then we break ties using the lex ordering.
a) Does every vector a give a monomial ordering? If not, what conditions must one assume to guarantee that a gives a monomial ordering? b) Find vectors that give the same monomial ordering as grlex, lex, and revlex, or prove that they do not exist. c) What is one condition on the vector a that ensures that no ties are possible? Solution. a) Not every vector gives a monomial ordering. The first two conditions do not impose any restrictions upon the vector a. Indeed, since R is totally ordered, there are only three outcomes of a comparison of weights, >, < or =. In case the weights are equal, we fall back on the lex ordering, which we know is a total ordering. The multiplication condition also holds by the linearity of the dot product. For > a to be a well-ordering, one needs a R n 0. Indeed, suppose that a has a coordinate a i that is negative. Then x i < a 1 (as monomials), which violates Corollary 2.4.6. Since this is a characterization of well-orderings, this condition is also sufficient to guarantee that < a is a monomial ordering. b) It is quite clear that the vectors a = (1, 1,..., 1) and a = (0, 0,..., 0) give the grlex and lex orderings, respectively. For revlex, the question was a bit ambiguous. If one uses revlex as in the definition from class, this is not a monomial order (as was mentioned by some of you). If this was mentioned, full marks were given. Indeed, for the ordering to behave like revlex, the a i must be negative, which we showed was impossible for a monomial ordering in part a). For grevlex (or if you thought revlex was grevlex), note that since x k i < x l j for any i j and k < l, we have that ka i < la j and hence a i a j < l k. Since k and l are arbitrary numbers with k < l, this shows that a i = a j. Therefore, the vector a must be (a, a,..., a). Note also that lex is the same as grevlex when there are only two variables, and is indeed different when there are more than two variables (indeed, xz > lex y 2 and xz < grevlex y 2 ), so there is such a vector a when there are two variables, but no such vector when there are more than two variables. c) For ties to not occur, there should not exist α and β in N n so that α a = β a. In other words, there should not exist γ so that γ a = 0, or there do not exist γ 1,..., γ n N so that i γ ia i = 0. A necessary and sufficient condition for this to hold is that the a i be linearly independent real numbers over Q. Problem 3. Let I be an ideal of k[x 1,..., x n ], and > any term order. a) Show that if lt(i) is a radical ideal, then I is a radical ideal. b) Show by example that the converse does not hold. Solution. a) Let f m I. Then lt(f m ) lt(i), and thus lt(f) m lt(i) (this follows from the properties of monomial ordering and need not be proven since we used it so much in class). Since lt(i) is radical, it follows that lt(f) lt(i), from which it also follows that lt(f) lt(i). Thus, there exists g I such that lt(f) = lt(g). Consider f = f g. One has lt(f ) < lt(f), and also f m = (f g) m I since f m I and g I. Therefore, using the argument above we again have that lt(f ) lt(i), so we can find a g I so that lt(f ) = lt(g ). Setting f = f g, we can continue this process. This gives us a descending chain of monomials lt(f) > lt(f ) > lt(f ) >
that since > is a monomial ordering must terminate. Therefore, f is a finite sum of elements of I, and hence is in I. b) Consider I = x 2 + x. This is clearly a radical ideal since the generator is a reduced polynomial. Its ideal of lead terms is lt(i) = x 2 which is not radical. Problem 4. Let r(t) = (t, t 2 + t, t 3 + t) be a parameterization of a curve in R 3. Use a computer algebra package (or by hand, but it is long) to compute the defining equation of the tangent surface of the curve r(t). Recall that we worked through the case of the twisted cubic (t, t 2, t 3 ) in class. Prove carefully (using both the elimination and the extension theorems, or their corollaries) that your answer defines the correct variety over the real numbers. (If using Sage Notebook, the output may be too wide to read properly. To fix this, click on the Text tab to see the correct output). See the sample file onl the blog to see how to perform this computation for the twisted cubic. Solution. The tangent surface S to the curve r(t) is parameterized by (x, y, z) = r(t) + ur (t), or x = t + u y = t 2 + t + 2ut + u z = t 3 + t + 3ut 2 + u So, we let I = x t u, y t 2 t 2ut u, z t 3 t 3ut 2 u, and compute a Gröbner basis of I using the lex term order, most certainly doing so using some computer algebra system such as Macaulay2. The pertinent elements of the Gröbner basis are listed below: g 1 : 7x 4 6x 3 y 4x 3 z + 10x 3 + 3x 2 y 2 18x 2 y 6x 2 z x 2 + 12xy 2 + 6xyz + 2xz 4y 3 z 2 g 2 : 2ux 2 + 2ux 2uy 2x 3 3x 2 + 3xy + x z g 3 : u 2 x 2 x + y g 4 : t + u x The first is the only element of the Gröbner basis that only involves x, y, z, so by the Polynomial Implicitization Theorem, we know that V (g 1 ) = V (I 2 ) is the smallest variety that contains the tangent surface. The first question is whether or not each of the solutions to g 1 extend to full solutions on the tangent surface. Here, we first look at g 3. Note that the leading coefficient of the highest power of u is 1, and hence every partial solution (x, y, z) extends to a solution (u, x, y, z). Similarly, the leading coefficient of t in g 4 is 1, and thus every partial solution (u, x, y, z) extends to a solution (t, u, x, y, z). Note that all these points are over C, which we must work over since the Extension theorem only holds over algebraically closed fields. If we wish to check that the solutions to g 1 = 0 over the real numbers is precisely the parameterized surface with real parameters u and v, then we must check that if (x, y, z) R 3, that (t, u, x, y, z) R 5. Indeed, then equation g 2 shows that one may solve for u in terms of x, and that u is real, provided that x 2 + x y is nonzero. In this case, we can also look at g 3 to see that in this case, u = 0 R as well. Also, if u and x are real, then t is real by g 4. Therefore, we have that (x, y, z) R 3 implies that (t, u, x, y, z) R 5, and so the equation g 1 = 0 does indeed describe exactly the tangent surface over R as well.
Problem 5. Let S be the parametric surface given by x = uv y = uv 2 z = u 2 a) Find the equation of the smallest variety V that contains S (You may use a computer algebra system to perform this computation.) b) Over C, show that V contains points which are not on S, and determine exactly which points of V are not on S. (Hint: Use the elimination and extension theorems.) Solution. a) The equation of the smallest variety V that contains S is again found by computer. We compute a Gröbner basis of I = x uv, y uv 2, z u 2 with respect to the lex ordering to get {x 4 y 2 z, vyz x 3, vx y, v 2 z x 2, uy x 2, ux vz, uv x, u 2 z}. So, we see that V = V (x 4 y 2 z) is the smallest variety containing S. b) Let (x, y, z) V. We know that this point extends to a partial solution (v, x, y, z) provided (x, y, z) V V (yz, x, z) = V (x, z), so some points that we will have to check are X = {(0, y, 0) y C}. Any of these partial solutions (v, x, y, z) extend to full solution (u, v, x, y, z) since the coefficient in front of u 2 is 1. So what about the points of X? The extension theorem just tells us that these points do not necessarily extend, but they could. Indeed, the point (0, 0, 0) does in fact extend by choosing (u, v) = (0, 0) in the parameterization. However, if y 0, then we know that u and v are nonzero, and hence z is nonzero, thus the point (x, y, z) is not on the surface S. Therefore, the points that are on V but not on S are exactly {(0, y, 0) y 0}. Problem 6. Let I be an ideal of k[x 1,..., x n ], and let f be a polynomial. a) Show that (I : f) := {g k[x 1,..., x n ] gf I} is an ideal. b) Let (I : f ) := n 1 (I : f n ). Show that (I : f ) is an ideal and is equal to (I : f n ) for some n. (Hint: Hilbert Basis Theorem.) Solution. a) Since 0 I, one clearly has that 0 f = 0 I so that 0 I. Also, let g, h (I : f) and r k[x 1,..., x n ]. Then (g + h)f = gf + hf I since gf I and hf I since g, h (I : f), and also since I is closed under addition. Further, (rg)f = r(gf) I since I is closed under scaling from the polynomial ring. Therefore (I : f) is an ideal. b) First, note that (I : f n ) is an ideal for all n by part a). Also, (I : f n ) (I : f n+1 ), since for g (I : f n ), one has gf n+1 = (fg)f n = f(gf n ) I, since I is closed under scaling by polynomials. So, we have an infinite ascending chain of ideals (I : f) (I : f 2 ) (I : f 3 ) So, by the ascending chain condition on ideals in k[x 1,..., x n ], we have that there exists an n so that (I : f n ) = (I : f n+k ) for all k > 0. Therefore (I : f ) = (I : f i ) = (I : f n ), proving both the desired claims.
Problem 7. Let R = k[x 1,..., x n ] be the polynomial ring, let I be an ideal of R, G = {g 1,..., g t } be a Gröbner basis for I, and let R t be the set of t -tuples of polynomials. Let be the so-called first syzygy module of I. M = {(f 1,..., f t ) R t f 1 g 1 + + f s g s = 0} a) Show that the all zero vector in R t is in M. b) Show that if α, β M, then α + β M. c) Show that if α M, and f R, then fα M. d) Let S ij = S(g i, g j ) be the S -polynomial of g i and g j. Recall that by Buchberger s criterion S ij G = 0. Show that this fact produces an element of M, which we call Tij. e) Let T = {T ij i j} M be the collection of the T ij from part d). Let N = {r 1 α 1 + r k α k r i R and α i T } R t be the set of linear combinations of elements in T. Show using parts a), b) and c) that N is a subset of M. Remark. Note that a), b) and c) above say that M is a submodule of R t. One should think of submodules of R t in a very similar manner to ideals of R ; ideals are just submodules of R 1. Also, something else that is true (but a bit more difficult to prove) is that every element of M is an R -linear combination of elements of T. That is, N = M. Solution. a) One has 0 M since 0 g 1 + + 0 g s = 0. b) If α = (a 1,..., a s ) and β = (b 1,..., b s ) are in M, then one has (a 1 + b 1 )g 1 + + (a s + b s )g s = a 1 g 1 + + a s g s + b 1 g 1 + + b s g s Therefore, α + β M. = 0 + 0 = 0. c) In a similar manner as above, one can directly show that if r is a polynomial, then rα is in M by using the distributive property of polynomials. d) We know that S ij is of the form S ij = a i g i a j g j for some suitably chosen monomials a i and a j. Now since S ij G = 0, we know that we can use the division algorithm to write a i g i a j g j = b 1 g 1 + b s g s for suitable polynomials b i. Rearranging, we get an expression c 1 g 1 + c s g s = 0 where c k = b k for k {i, j}, c i = b i a i, and c j = b j + a j. This gives us an element γ = (c 1,..., c s ) in M. e) N is just the polynomial span of the elements of T, which we showed from the previous part is in M. Thus, parts a),b), and c) show that linear combinations of elements of T are again back in M, and hence N M.