A brief introduction to computational algebraic geometry

Transcription

1 A brief introduction to computational algebraic geometry Evan M. Bullock This is a collection of lecture notes and problem sets from MATH 499: VIGRE Computational Algebraic Geometry at Rice University in Plane curve singularities A point p is a singular point of a plane curve f(x, y) = 0 if and only if f(p) = 0 so that p is on the curve, and ( f f p = x, f ) = 0. p y In single variable calculus, for studying points where the first derivative of a function g(x) is zero, it is often helpful to study the higher derivatives. For example, if g (a) = 0 and g (a) > 0, then g has a local minimum at a. More generally, if g is a sufficiently nice function 1, then g is represented near a by its Taylor series centered at a, g(x) = g(a) + g (a)(x a) + g (a) 2! p (x a) 2 + g(3) (a) (x a) ! and the behavior of g very close to a is completely determined (in a sense that we will make more precise in the future) by the first non-constant term of the Taylor series. In studying functions of several variables, when f p = 0, it also makes sense to look at higher derivatives of f at p. In fact, Taylor series work fine in several variables. The idea is the same as it is in the one variable case: we find a polynomial of degree n in several variables all of whose partial derivatives up to order n agree with those of the function f. Letting n, we get an infinite power series representation in several variables for f(x, y) centered at p = (x 0, y 0 ) that looks like: 1 The technical term is real analytic, which just means represented by its Taylor series near each point. Essentially all infinitely differentiable functions one encounters in practice are real analytic. Probably the simplest example of a smooth function that isn t real analytic is something like g(x) = { e 1/x2, if x 0; 0 if x = 0. You can check that this function is infinitely differentiable, but g(x) 0 so fast as x 0 that g (n) (0) = 0 for all n. Thus its Taylor series centered at x = 0 is identically zero, and is not equal to the function except at x = 0. Fortunately, we ll only be working with polynomials and power series, so we won t need to worry at all about pathological functions like this. In fact, we won t even really have to worry about convergence of the power series we deal with! 1

2 2 f(x, y) = f(p) + f x (x x 0 ) + f p y (y y 0 )] p + 1 2! 2 f x 2 (x x 0 ) p 2! 2 f x y (x x 0 )(y y 0 ) + 1 p 2! 2 f y 2 (y y 0 ) 2 p + 1 3! 3 f x 3 (x x 0 ) p 3! 3 f x 2 y (x x 0 ) 2 (y y 0 ) + 3 p 3! 3 f x y 2 (x x 0 )(y y 0 ) 2 p + 1 3! 3 f y 3 (y y 0 ) p 1 n ( ) n n f = n! k x k y n k (x x 0 ) k (y y 0 ) n k. p n=0 k=0 If f is a polynomial to start with, the resulting Taylor series will have only finitely many non-zero terms (Why?). For example, if we expand f(x, y) = y 2 x 3 +12x 16 around the singular point (2, 0), we get f(x, y) = y 2 6(x 2) 2 (x 2) 3. A computer algebra system can compute these Taylor series expansions for us. For example, the Sage command x,y = var("x y"); taylor(x^2*y + x*y^2, (x,3), (y,-1),10) produces the output 2 (y + 1) 2 (x 3) + (y + 1)(x 3) (y + 1) (y + 1)(x 3) (x 3) 2 5 x + 3 y + 12 where here we are asking Sage to write x 2 y + xy 2 as a series in x 3 and y + 1. The 10 in the command is telling Sage to only compute the terms of the series up to degree 10, though we know that really we re dealing with a polynomial of degree 3, so higher degree terms in x 3 and y + 1 couldn t possibly show up in the Taylor expansion (Why?). Exercises For these exercises, I would recommend using the Sage computer algebra system ( sagemath.org/), a free open-source alternative to Maple, Mathematica, etc.; for example, to graph the solution set to x 2 + y 2 = 2 in Sage, we might use the commands: x, y = var ( x y ) i m p l i c i t p l o t ( xˆ2+yˆ2 2, ( x, 3,3), ( y, 3,3)). show ( a s p e c t r a t i o =1) To get inline help on a command in Sage, use that command followed by a?, as in: i m p l i c i t p l o t? t a y l o r? 2 Really, I used the Sage command: to produce output I could copy directly into a.tex file. latex(taylor(x^2*y + x*y^2, (x,3), (y,-1),10))

3 2. MULTIPLICITY AND THE TANGENT CONE 3 (1) For each of the following curves, calculate the partial derivatives and find the singular points. Then calculate the Taylor expansion for the curve about each singular point. If possible, identify each singularity as a cusp or a node. Finally plot each curve. (a) x 2 = x 4 + y 4 (b) xy = x 6 + y 6 (c) x + y = x 3 y 3 (d) x 3 = y 2 + x 4 + y 4 (e) x 2 y + xy 2 = x 4 + y 4 (f) y 2 = x 3 + x 2 (g) (x 2 + y 2 ) 2 = x 2 y 2 (h) x 3 + xy = x + x 2 + y 2 (i) y 2 = x 5 2x 4 + x 3 (2) Prove that y 2 = x 3 + px + q has no singular points if and only if f(x) = x 3 + px + q has three distinct roots. [Hint: First show that a polynomial f(x) has a multiple root at a if and only if f(a) = f (a) = 0.] 2. Multiplicity and the tangent cone We ve now seen plane curves with various singularities and given a few types of them names, although we haven t give any precise mathematical definitions of these yet. Figure 1. Some singularities we ve seen (a) A node singularity (b) A cusp singularity (c) A tacnode singularity One simple question we should expect to be able to answer using calculus is what are the tangent lines to the branch or branches of the curve at the singular point? (In the node example above, we can see two distinct tangent lines in the picture, whereas in the cusp and tacnode cases, there should only be one tangent line at the singular point.) To simplify our computations, we may as well first make a change of coordinates so that the singular point is at the origin (we can certainly do this with an affine change of variables see the homework). Consider then a plane curve V (f), where f is a polynomial with f(0, 0) = The notation V (f) will be used to mean {(x, y) R 2 : f(x, y) = 0}. More generally, an algebraic variety over the real numbers R will be the locus V (f 1, f 2,..., f k ) = {(x 1,..., x n ) R n : f 1 (x 1,..., x n ) = f 2 (x 1,..., x n ) = = f k (x 1,..., x n ) = 0} of common zeros of finitely many polynomials in n variables f 1,..., f k R[x 1,..., x n ].

4 4 Of course, if the origin were a smooth point, this would be easy: there is only one tangent line, defined by f (0,0) (x 0, y 0) = f x x + f (0,0) y y = 0. (0,0) When the origin is singular, both first partial derivatives are zero and we need to look at higher order terms to find the tangent lines. In particular, let us write f = f m + f m f n where f k is the degree k part of the polynomial f; here n = deg f and f m 0 is the non-zero piece of f of smallest degree. Proposition. If L = V (ax + by) is a tangent line to V (f) at p = (0, 0), then ax + by is a factor of f m. Proof. Let (x(t), y(t)) for t [0, 1) be a local parametrization of a branch of X = V (f) near the origin, with x(0) = y(0) = 0. We assume for convenience that the branch does not have a vertical tangent, i.e. that y(t)/x(t) is bounded as t 0. Then the slope of the tangent line to this branch of X at the origin should be lim y(t) t 0 +. We write x(t) f m (x, y) = a 0 x m + a 1 x m 1 y + + a m 1 xy m 1 + a m y m so that for (x, y) = (x(t), y(t)) with t very small, 0 = f(x, y) = a 0 x m + a 1 x m 1 y + + a m 1 xy m 1 + a m y m + f m+1 (x, y) + + f n (x, y), and dividing through by x m, we get f(x, y) ( y ) ( y ) m 1 ( y ) m 0 = x = a f m+1 (x, y) m 0 + a a m 1 + am f n(x, y) x x x x m x. m Now, since x(t) 0 and y(t) 0 as t 0 + and y(t)/x(t) stays bounded, lim t 0 + x(t)j y(t) m+i j = 0 x(t) m for i > 0 and we find that f m+1 (x, y) lim + + f n(x, y) = 0. t 0 + x m x m This implies that lim t 0 + a 0 + a 1 ( ) ( y(t) y(t) + + a m 1 x(t) x(t) ) m 1 ( ) m y(t) + a m = 0, x(t) so that the slope c of the tangent line is a root of the polynomial a 0 +a 1 z + +a m 1 z m 1 +a m z m. Equivalently, z c is a factor of a 0 + a 1 z + + a m 1 z m 1 + a m z m and y cx is a factor of f m = a 0 x m + a 1 x m 1 y + + a m 1 xy m 1 + a m y m as desired (Why?). If we had been dealing with a vertical tangent, then we could have divided through by y m instead, and we would find that lim t 0 + a 0 ( x(t) y(t) ) m + a 1 ( x(t) y(t) ) m a m 1 ( ) x(t) + a m = 0, y(t) which implies that 0 must be a root of the polynomial a 0 w m + a 1 w m 1 + a m 1 w + a m so that x is a factor of the polynomial f m = a 0 x m + a 1 x m 1 y + + a m 1 xy m 1 + a m y m. It turns out that, at least over the complex numbers, the converse of this proposition is true as well: if ax + by is a factor of f m, then near the origin there is a branch of the curve V (f) having V (ax + by) as its tangent line at the origin. We call that smallest degree m of a non-zero term of f the multiplicity of the curve X = V (f) at the origin. Since V (f m ) is often the union of the tangent lines to branches of the curve X at the

5 2. MULTIPLICITY AND THE TANGENT CONE 5 origin (and at least always contains those lines) we ll give it a name: we call V (f m ) the tangent cone of X at the origin and denote it T C (0,0) (X). We said that we could move a singular point of X = V (f) to the origin by a change of coordinates, but we could also have done everything with Taylor expansions centered at any point p = (a, b): we set f k to be the degree k part of f regarded as a polynomial in x a and y b, as can be computed by Taylor expansion f k = 1 k ( ) k n f k! j x j y k j (x a) j (y b) k j. p j=0 We can then define the multiplicity m as before as the the smallest number so that f m 0, and define the tangent cone to X at p to be T C p (X) = V (f m ). One can check that this agrees with what we d get by instead translating the curve X so that p goes to the origin, computing the tangent cone at the origin as above and then translating back. One might wonder where the term cone is coming from here given that there is no cone in the traditional sense in sight. The answer is that the term cone is often used more generally to refer to any locus traced out by lines through some fixed point (the vertex of the cone ). Here, the base of the tangent cone of our plane curve might be regarded as finitely many points, one for each tangent direction. While this can never really look much like a more familiar cone in the case of plane curves, it is possible for a tangent cone to a singular point on a surface to actually be a cone in the original sense (see Figure 2). Figure 2. Tangent cone at a surface singularity (a) X = V (x 2 + y 2 z 2 x 3 y 3 ) R 3 (b) T C (0,0,0) (X) Exercise. Show that if f(x 1,..., x n ) R[x 1,..., x n ] is a homogeneous polynomial (all of its terms have the same degree), then its zero locus V (f) is a cone with vertex (0,..., 0), i.e. if (c 1,..., c n ) V (f) then (λc 1,..., λc n ) V (f) for all λ R.

6 6 If we were working over the complex numbers, the converse would be true as well: if V (f) is a cone with vertex at the origin, then the polynomial f must be homogeneous. Why doesn t this work over the reals? Can you find a counterexample? There s another (closely related) notion of multiplicity, namely the multiplicity of intersection of a curve and a line. Given a plane curve X = V (f) and a line L through a point p = (a, b) X, we can define the multiplicity of their intersection as follows: we choose a linear parametrization of L as α(t) = (a, b) + t(c, d) so that α(0) = p. The the composition f(α(t)) is a polynomial in t with 0 as a root. The intersection multiplicity of X and L at p is defined to be the multiplicity of 0 as a root of f(α(t)), i.e. the power of t in the factorization of f(α(t)). Suppose that p is a smooth point of X. Then we have f(α(t)) = f(a + tc, b + td) = f x (a + tc a) + f p y (b + td b) + f 2 (a + tc, b + td) + + f n (a + tc, b + td) p ( = t c f x + d f ) + t 2 F 2 (c, d) + t 3 F 3 (c, d) + t n F n (c, d), p y p where F k (c, d) = f k (a + c, b + d). 1 We see then that the intersection multiplicity of X with L at the smooth point p is 1 except when c f p + d f = 0, i.e. when L is the tangent line to X at p x y p, in which case the intersection multiplicity is at least 2, and we would need to look at the higher order terms to compute it exactly. The same computation in the case where p is singular shows that the intersection multiplicity of X with L at p is at least 2 for every line L through p. Exercises (1) In the previous exercises, you found that the following curves have only one singularity, at p = (0, 0), and calculated the Taylor series expansions at that point. Now, find the multiplicity of each curve at p and find the tangent cone T C p (X). This should be a matter of interpreting the Taylors series calculations you have already made. Sketch the curves and draw in the tangent cones. (a) f(x, y) = x 4 + y 4 x 2 (b) f(x, y) = x 6 + y 6 xy (c) f(x, y) = y 2 + x 4 + y 4 x 3 (d) f(x, y) = x 4 + y 4 x 2 y xy 2 (e) f(x, y) = x 3 + x 2 y 2 (f) f(x, y) = (x 2 + y 2 ) 2 x 2 + y 2 (2) Use the same methods to find the singularities, the multiplicity at each singularity, and the tangent cones of the following curves. Since these are a bit more complicated, you will probably want to get a computer to do most of the calculations. Sketch a graph of the curve and its tangent cone near each singularity. Depending on what program you use, 1 The fact that the F k (c, d) in our expression above are homogeneous polynomials in c and d also helps show that the intersection multiplicity does not depend on our choice of linear parametrization α for L.

7 3. RATIONAL CURVES 7 you may have to be careful of the behavior near singular points. Use your information from the tangent cone to interpret the behavior near singularities. (a) f(x, y) = 2x 4 3x 2 y + y 2 2y 3 + y 4 (b) f(x, y) = 2y 2 (x 2 + y 2 ) 3y 2 x (c) f(x, y) = 2y 2 (x 2 + y 2 ) 2y 2 (x + y) 2y 2 x 2 + 2x + 2y (d) f(x, y) = (x 2 + y 2 ) 3 4x 2 y 2 (3) One can think of multiplicity as measuring how bad a singularity is. We already showed that for a nonsingular point on curve, most lines intersect that point with multiplicity one. (a) For the curve f(x, y) = x 3 y 2, show that most lines through the origin meet the curve with multiplicity 2. (b) For the curve g(x, y) = x 4 + 2xy 2 + y 4, show that most lines through the origin meet the curve with multiplicity 3. (4) We ve mentioned that we ought to be able to make a simple change coordinates so that, for example, a singular point is moved to the origin. The basic idea we were hinting at is that of affine equivalence. An affine change of coordinates is a map of the form ([ [ [ ] [ [ ] x a b x e a b φ = +, where det y]) c d] y f] c d We can think of this as basically just a change of variables (but one which is allowed to distort angles and distances). Two curves f(x, y) and g(x, y) are affine equivalent if they differ by an affine change of coordinates φ. That is, f(x, y) = g(φ(x, y). Show that the curves f(x, y) = y 2 x 3 x 2 and g(x, y) = x 2 2xy x + y y3 are affine equivalent. (5) Show that multiplicity is invariant under affine equivalence. That is, if φ: C 1 C 2 is an affine equivalence, it maps a point with multiplicity m to a point with multiplicity m. (6) This problem is a little different, and its connection to plane curves or algebraic geometry may not be apparent for a while. (a) What natural numbers n are expressible in the form n = 2x + 3y where x and y are nonnegative integers? What if we allow x or y to be negative? (b) What natural numbers n are expressible in the form n = 4x + 6y where x and y are nonnegative integers? What if we allow x or y to be negative? (c) What natural numbers n are expressible in the form n = 5x + 8y where x and y are nonnegative integers? What if we allow x or y to be negative? Rational curves A rational function in one variable is a function given as a quotient of polynomials f(t) = p(t) q(t) where p(t), q(t) R[t] are polynomials. Note that, despite the name, a rational function isn t a well-defined function at the points where q(t) = 0.

8 8 Given a curve X = V (f), for f R[x, y] an irreducible 4 polynomial, we say that a rational parametrization of X is a pair of rational functions (x(t), y(t)) so that f (x(t), y(t)) = 0 for all values of t where it is defined. If X admits a rational parametrization, we say that X is a rational curve. Proposition. The circle X = V (x 2 + y 2 1) is a rational curve. Proof. To find a rational parametrization of the circle, we use the fact that a non-vertical line through a point ( 1, 0) of the circle meets the circle in exactly one other point. We can thus try to use the slope of the line through ( 1, 0) as a rational parameter for the other point of the circle that the line intersects. The line with slope t through ( 1, 0) is defined by y = tx + t. Combining this with the equation x 2 + y 2 = 1 for the circle, we get Expanding and grouping terms, we get x 2 + (tx + t) 2 = 1. (1 + t 2 )x 2 + 2t 2 x + (t 2 1) = 0, and we see that, as expected, x = 1 is a solution for all t. This allows us to factor the equation as (x + 1) ( (1 + t 2 )x + t 2 1 ) = 0 and we find that the x-coordinate of the point of intersection other than ( 1, 0) is and that x = 1 t2 1 + t 2, y = t 1 t2 1 + t 2 + t = 2t 1 + t 2. 4 Irreducible means it doesn t factor as a product of non-constant polynomials. We need this condition here because if f(x, y) = g(x, y)h(x, y) where g(x, y) defines a rational curve and h(x, y) defines an irrational curve, we wouldn t want to call the whole curve V (f) rational when it consists of both the zero set of g(x, y) (which we can parametrize rationally) and the zero set of h(x, y) which we can t. Even if V (g) and V (h) were both rational, we still wouldn t be able to parametrize all of V (f) by a single rational function.

9 3. RATIONAL CURVES 9 ( ) 1 t We ve thus found that 2 2t, is a rational parametrization of the unit circle (which we could 1+t 2 1+t 2 check by substituting back into the implicit equation). Of course, in the case of the circle this may not immediately seem so exciting, since we already have a very convenient (but non-rational) parametrization by (cos t, sin t), but it turns out the rationality (together with the nice integer coefficients) of our parametrization in this example has a fairly interesting corollary: A Pythagorean triple is a triple of three positive integers (a, b, c) such that a 2 + b 2 = c 2. Examples include (3, 4, 5), (5, 12, 13), and (15, 8, 17). Of course, if (a, b, c) is a Pythagorean triple, then so is (da, db, dc), so we may as well restrict our attention to the case where a, b, and c have no common factor. A computation modulo 4 also shows that in this case c must be odd and exactly one of a and b must be even, so we may as well assume b is even. The connection to our parametrization of the circle is that if a 2 + b 2 = c 2, then ( a c ) 2 + ( b c ) 2 = 1, so that ( a, b c c) is a point with rational coordinates on the curve x 2 +y 2 = 1. But our parametrization works just as well over the rational numbers as ( it does over ) the real numbers, and we know that 1 t (aside from ( 1, 0)), every point has the form 2 2t, where t = m is a rational number. This 1+t 2 1+t 2 n gives ( a c, b ) c ) 2 ( ( 1 m n = 1 + ( ) m 2, n 2 ( ) m n 1 + ( m n ) 2 ) = ( ) n 2 m 2 n 2 + m, 2nm. 2 n 2 + m 2 Thus, since we may take n and m to be relatively prime, as long as n and m aren t both odd (which would lead to a triple with b odd instead) we have that a = n 2 m 2, b = 2nm, and c = n 2 + m 2. This gives a sort of parametrization of the Pythagorean triples and in particular makes it easy to show there are infinitely many of them (and not just by multiplying by a constant). Exercises (1) Recall that a rational function x(t) is one of the form x(t) = p(t) q(t), where p and q are polynomials. Show that the following curves are rational by finding non-constant functions x(t) and y(t) such that f(x(t), y(t)) 0. Then use a computer to graph the curve from the implicit function and then from the parametrization to verify that they coincide (at least for some section of the curve). Hint: Try using a substitution such as t = y x or t = y x 2. (a) f(x, y) = y 2 x 3 (b) f(x, y) = x 2 y 2 (x 2y)(x 2 + y 2 ) (c) f(x, y) = x 5 xy 2 + y 3 (d) f(x, y) = 3x 2y y 2 (e) f(x, y) = x 5 x 4 + x 2 y y 2 (f) f(x, y) = x 2 + 2xy + y 2 y (g) f(x, y) = x 2 2x y + 1

10 10 (2) A cardioid is defined by the polar equation r = 1 cos θ. Find an implicit polynomial equation f(x, y) = 0 for the cardioid, and show that (x(t), y(t)) = ((cos t)(1 cos t), (sin t)(1 cos t)) is a (non-rational) parametrization of it. (3) Recall the definition of an affine equivalence from last week. Show that affine equivalence preserves rationality. That is, show that if f(x, y) = g(φ(x, y)) for some affine equivalence φ and V (g) = {(x, y) : f(x, y) = 0} is rational then V (f) is also rational. (4) (a) Show that any nonempty conic is affine equivalent to one with no constant term, i.e. a conic of the form f(x, y) = ax + by + cx 2 + dxy + ey 2. (b) Let f(x, y) = (ax + by) + (cx 2 + dxy + ey 2 ) = f 1 + f 2 be irreducible, where f i is the purely degree i part of the polynomial. Prove that V (f) is rational. (c) Show that any irreducible conic is rational. (d) Now, let f(x, y) by an irreducible degree n polynomial such that f = f n 1 + f n, so that f has no terms of degree less than n 1. Prove that f(x, y) = 0 is a rational curve. (5) On the last homework, we began investigating the solution of equations like n = 4x + 6y and n = 5x + 8y. We discovered that which numbers are expressible in the form ax + by for x, y integers seems to have a lot to do with the the greatest common divisor of a and b. In fact, it turns out that the standard method of computing the g.c.d. d of a and b can help us solve the equation d = ax + by for integers x and y. This computational method is called Euclid s algorithm and works by repeated division with remainder as follows: b = q 1 a + r 1 a = q 2 r 1 + r 2 r 1 = q 3 r 2 + r 3 r n 2 = q n r n 1 + r n r n 1 = q n+1 r n + 0 The algorithm eventually terminates when it gets a zero remainder (since the remainders get smaller at each step). At that point the g.c.d. of a and b is known to be the last non-zero remainder r n. (a) Why does Euclid s algorithm work to find the g.c.d.? [Hint: The common divisors of r 1 and r 2 are the same as the common divisors of r 2 and r 3. (Why?)] (b) How does Euclid s algorithm allow us to write the g.c.d. r n in the form ax + by? Use it to solve 68x + 173y = 1. (c) In the integers modulo 173, what is the multiplicative inverse of 68? 4. Ideals and monomial orders Given an algebraic plane curve f(x, y) = 0, we ve been looking at the problem of finding a rational parametrization (x(t), y(t)) for it, where x(t) and y(t) are rational functions of t. In several examples (and in a couple of general cases) we ve been able to show that curves are rational and exhibit rational parametrizations.

11 4. IDEALS AND MONOMIAL ORDERS 11 It s natural to ask the reverse question as well: given a parametric rational curve (x(t), y(t)), can we find a polynomial f(x, y) R[x, y] so that f(x(t), y(t)) = 0? (One of the homework problems asks you to do this for a non-rational parametrization of the cardioid.) Proposition. Given rational functions x(t) and y(t), there exists a polynomial f(x, y) R[x, y] such that f(x(t), y(t)) 0. Proof. We can write the given rational functions as x(t) = a(t) q(t) and y(t) = b(t) q(t) for some polynomials a(t), b(t), q(t) R[t]. For some large degree N, we ll try to find a polynomial f(x, y) R[x, y] of degree N so that f(x(t), y(t)) 0, or equivalently so that q(t) N (f(x(t), y(t)) 0. Let n = deg q(t) and m = max{deg a(t), deg b(t)}. Then q(t) N f(x(t), y(t)) is a polynomial in t of degree at most Nn+Nm, whose coefficients are homogeneous linear functions of the coefficients of f. Thus setting all of its coefficients equal to zero give at most Nn + Nm + 1 homogeneous linear equations for the coefficients of f, and if we can pick N so that there are at least as many variables as there are equations (i.e. f has at least Nn + Nm + 1 coefficients), then we can solve the system and find an f with q(t) N f(x(t), y(t)) 0, and we ll be done. The degree k part of f is f k (x, y) = c k,0 x k + c k 1,1 x k 1 y + + c 0,k y k, which has k + 1 coefficients. Overall then, f has (N + 1) = N(N+1) 2 coefficients, and since this is quadratic in N and Nn + Nm + 1 is linear, for sufficiently large N we have N(N+1) 2 Nn + Nm + 1 as desired. While this proves that it is always possible to find a polynomial vanishing on a rationally parametrized curve, there are a few things that we might not like so much ( about it. ) For one thing, 1 t N may be bigger than it has to be; for example, in the parametrization 2 2t, for the circle, 1+t 2 1+t 2 the proof would use the value N = 8 to find a degree 8 polynomial f(x, y) vanishing on the circle, i.e. the solutions to the system would be the f(x, y) = (x 2 + y 2 1)g(x, y) for arbitrary polynomials g(x, y) of degree 6. This isn t really a big problem though, since we could always try to solve the systems for lower degrees first. The bigger complaint we might have is that even for small degree examples, this involves solving a big system of linear equations in a big number of variables. Of course, computers are pretty good at solving systems of linear equations, but this certainly isn t something we d want to do by hand, and even computer algebra systems might have some trouble dealing with huge numbers of variables. In fact, there is a nicer way to do the computation, and it involves thinking about( the problem ) more geometrically. We can think of the graph of the rational function (x(t), y(t)) = a(t), b(t) in p(t) q(t) R 3 = R 2 R as being the common zeros of the two polynomials g(x, y, t) = a(t) xp(t) and h(x, y, t) = b(t) yq(t) in R[x, y, t]. Our goal then is to find a polynomial f(x, y) in only the variables x and y so that whenever g(x 0, y 0, t) = h(x 0, y 0, t) = 0 for some value of t, we have f(x 0, y 0 ) = 0. If we could write some polynomial f(x, y) R[x, y] in the form f(x, y) = a(x, y, t)f(x, y, t) + b(x, y, t)g(x, y, t),

12 12 for polynomials a(x, y, t), b(x, y, t) R[x, y, t], then f(x, y) would certainly have this property, and thus be a polynomial vanishing on the parametrized curve. For example, x 2 + y 2 1 = ( 1 txy + 1ty x y2 1 ) ( 1 t 2 (1 + t 2 )x) ) ( 1 2 tx2 + tx t xy y) ( 2t (1 + t 2 )y ). Over the next few weeks, we ll start learning about a computational tool called Gröbner bases that will tell us how to find such an f, and will more generally allow us to study the question of, given polynomials h 1,... h k R[x 1,..., x n ], which polynomials f R[x 1,..., x n ] can be written in the form f = q 1 h 1 + q k h k for some q 1,..., q k R[x,..., x n ]? We ll begin with some terminology. If f 1,..., f k R[x 1,..., x n ] are polynomials then the ideal generated by f 1,..., f k is the set f 1,..., f k = {q 1 f q k f k : q 1,..., q k R[x 1,..., x n ]}. More generally, a non-empty subset I R[x 1,..., x n ] is defined to be an ideal if g 1 f 1 + g 2 f 2 I for every f 1, f 2 I and g 1, g 2 R[x 1,..., x n ]. Theorem (Hilbert Basis Theorem). Every ideal I R[x 1,..., x n ] is generated by finitely many polynomials, so that I = f 1,..., f k for some f 1,..., f k I. We probably won t prove this. We should note that this theorem doesn t say anything about the size of the smallest generating set of I, so k here could be much bigger than n. When dealing with polynomials in one variable, a polynomial always has a clear leading term, namely the term of highest degree. For polynomials in several variables, there are many different ways we might want to order the monomials. For convenience, if α = (a 1,..., a n ) is an n-tuple of non-negative integers, then we will write x α = x a 1 1 x a 2 2 x an n as an abbreviated notation for the corresponding monomial. Although there are many orderings on the monomials to choose from, we want them to respect the algebraic structure. For example if x α divides x β, then we would like x α to be smaller than x β. A monomial order for R[x 1,..., x n ] is a total order 5 on the monomials such that if x α < x β then x γ x α < x γ x β for all monomials x γ which is a well-ordering 6. Example: Lexicographic order Probably the simplest monomial ordering is the lexicographic (or dictionary ) ordering. In this ordering, the power of the first variable is used to determine the order, with powers of the second variable only looked at when the first variable appears to the same power in two monomials. 5 This means that: (1) it is never the case that both x α < x β and x β < x α, and (2) if x α < x β and x β < x γ, then x α < x γ. 6 Well-ordering means that if S is any subset of monomials, then S has a least element according to the ordering. This implies that 1 is the least monomial, since if x α < 1 were the least monomial, then x 2α < x α would be even smaller, a contradiction.

13 4. IDEALS AND MONOMIAL ORDERS 13 Similarly, we only look at the third variable when the first two are tied, and so on. For example, in the lex order for R[x, y, z] with x > y > z, we have x 4 > x 3 y 2 z > x 3 yz 7 > x 3 yz 4 > x 2 yz 5 > xy 3 z 2 > xy > xz 2 > x > y 6 > y 5 z 3 > yz 6 > y > z 3 > 1. More formally, given two monomials x α and x β in R[x 1,..., x n ], we say that x α > lex x β if in the difference of vectors α β, the leftmost non-zero entry is positive. One can check that this does in fact define a monomial order. 7 Example: Graded lexicographic order One thing we might not like about lex order is that it doesn t respect degrees (e.g. xy > y 3 z 4 ). We can define a new order, called graded lexicographic order by saying that higher degree monomials are bigger and using lex order to break ties. For example, x 7 > z 7 > x 2 y 2 z 2 > x 2 yz 3 > xy 5 > y 3 z 3 > yz 5 > x 5 > x 4 y > x 3 y 2 > x 3 yz > x 3 z 2. More formally, we say that x α > grlex x β if deg x α > deg x β or if deg x α = deg x β and x α > lex x β. Since the partial ordering by degree and the lexicographic ordering both have the property that x α < x β = x γ x α < x γ x β for all monomials x γ, the graded lexicographic order has this property as well. Since any graded order (an order in which degree is used first and then something else is used as a tie-breaker) satisfies well-ordering automatically (because there are only finitely many monomials of each degree), we see that grlex is a term order. 7 See section 2.2 of Cox, Little, and O Shea for more details about term orderings, including proofs that the well-ordering property holds, etc.

14 14 Example: Graded reverse lexicographic order Perhaps one of the most frequently used term orders in practice (because it tends to result in faster computations) is graded reverse lexicographic or grevlex order. This one is perhaps a little more confusing. As the name suggests, graded reverse lexicographic order uses degree first, and uses reverse lexicographic order to break ties. If we reverse the lexicographic order however, so that x α > revlex x β if x α < lex x β, the result isn t a monomial ordering. It fails well-ordering, because there are infinite descending sequences, e.g. in revlex order we have 1 > revlex z > revlex z 2 > revlex z 3 > revlex > revlex yz > revlex yz 2 > revlex yz 3 > revlex. However, the reverse of an order preserved under multiplication by x γ is at least still preserved under multiplication by x γ, so while reverse lexicographic order isn t a monomial order by itself, we can still use it to break ties in a graded order (for which well-ordering is automatic). There s one final issue to defining grevlex: when we reverse the lex order, it reverses the order of the variables, but we still want to get an order with x 1 > x 2 > > x n in the end. Thus we start with a lex order with x n > > x 1, so that when we reverse it we get a reverse lexicographic order with x 1 > revlex x 2 > revlex > revlex x n. We then say that x α > grevlex x β if deg x α > deg x β or if deg x α = deg x β and x α > revlex x β. For example, in the graded reverse lexicographic order on R[x, y, z] with x > y > z, we have y 2 z 2 > x 3 > xy 2 > xyz > y 2 z > xz 2 > x 2 > xy > y 2 > xz > yz > z 2 > x. Basically, we order by degree first, and break ties by saying a monomial is bigger if it has a smaller power of the least significant variable. More formally, we say that x α > x β if deg x α > deg x β or if deg x α = deg x β and in the vector difference α β, the rightmost non-zero entry is negative. Exercises (1) In each part, determine whether the polynomial f R[x] is in the given ideal I R[x]. Notice that determining if f lies in the ideal g is equivalent to determining if g divides f. How do we use the same idea in (c) and (d), where I = g 1, g 2? (a) f(x) = x 2 3x + 2, I = x 2 (b) f(x) = x 5 4x + 1, I = x 3 x 2 + x (c) f(x) = x 2 4x + 4, I = x 4 6x x 8, 2x 3 10x x 8 (d) f(x) = x 3 1, I = x 9 1, x 5 + x 3 x 2 1 (2) Find an ideal I R[x] in which every element f I is divisible by x, but such that x I. (3) (a) Show that x y 2, xy, y 2 = x, y 2. (b) Is x y 2, xy = x 2, xy? (4) Rewrite each of the following polynomials, ordering the terms first with the lex order, then the graded lex order, and finally the graded reverse lex order, provided that x > y > z. (a) f(x, y, z) = 2x + 3y + z + x 2 z 2 + x 3 (b) 2x 2 y 8 3x 5 yz 4 + xyz 3 xy 4 (c) 7x 2 y 4 z 2xy 6 + x 2 y 2 (5) Ideals make sense in the ring of integers Z just as they do in polynomial rings like R[x]. For example, in Z the ideal I = a, b consists of all integers xa + yb for x, y Z. (a) Is 10 in the ideal I = 3? (b) Is 2 in the ideal I = 5, 8?

15 (c) Is 6 in the ideal I = 12, 22? (d) Is 3 in the ideal I = 68, 173? 5. THE DIVISION ALGORITHM The division algorithm We would like to generalize the division algorithm to polynomials in several variables. In the single variable case, it suffices to be able to divide by a single polynomial g(x) = q(x)f(x) + r(x) with deg r(x) < deg f(x), in the sense that even if we want to determine, given g(x), f 1 (x), f 2 (x), whether it s possible to write g(x) = q 1 (x)f 1 (x) + q 2 (x)f 2 (x) we can use Euclid s algorithm to find the gcd f(x) of f 1 (x) and f 2 (x) and write it as f(x) = a 1 (x)f 1 (x) + a 2 (x)f 2 (x), and then simply divide g(x) by f(x). Another way of saying this is that every ideal f 1 (x),..., f k (x) R[x] is in fact generated by a single element f 1 (x),..., f k (x) = f(x), and both computing this element (using Euclid s algorithm) and testing whether g(x) is divisible by it can be carried out by dividing by single polynomials. The situation is much more complicated when dealing with polynomials in several variables. For example, the ideal x, y R[x, y] can not be generated by a single element, and while x and y have no common factors, it is not possible to write 1 = q 1 (x, y)x + q 2 (x, y)y. As such, we won t be able to limit ourselves to dividing by a single polynomial. Another condition that we ll have to change in the multivariable case is our requirement that deg r(x) < deg f(x). For example, if we tried to write x 2 y + xy 2 + z 3 = q 1 (x, y)x + q 2 (x, y)y + r(x, y), it seems like whatever choices of q 1 and q 2 we make we ll always be stuck with a z 3 term in the remainder, which has a larger degree than the polynomials x and y that we re dividing by. The division algorithm in several variables We now describe what we can do over R[x 1,..., x n ] by essentially just following the usual division algorithm for single variable polynomials. Since the single variable division algorithm involves the leading terms of various polynomials, we fix a monomial order on R[x 1,..., x n ] and use it to define LT (f) to be the leading term of a polynomial f according to that monomial order. Now suppose we are given a polynomial g that we are trying to divide by polynomials f 1,..., f k R[x 1,..., x n ] to write g = q 1 f q n f n + r. If LT (g) is divisible by LT (f i ), then we can add to q LT (g) LT (f i ) i and subtract f LT (f i ) i from g to cancel out the leading term of g, leaving it with a a smaller leading term. In this way, we eventually arrive at a polynomial whose leading term is not divisible by any of the LT (f i ), and so we must move that leading term to the remainder. We continue in this way: we either cancel out LT (g) if it is divisible by some LT (f i ) or we just move it to the remainder if it isn t. Since the leading term decreases at each step, this process must terminate eventually with no terms left of g, and at that point every term of the remainder we re left with will not be divisible by any of the LT (f i ). LT (g)

16 16 Theorem (Division algorithm in R[x 1,..., x n ]). Fix a monomial order. Let f 1,..., f k R[x 1,..., x n ] be given. Then every g R[x 1,..., x n ] can be expressed as g = q 1 f q k f k + r with q 1,..., q k, r R[x 1,..., x n ] and either r = 0 or every term of r is not divisible by the leading term of any of the f i. Also, the leading monomial of each q i f i is no greater than that of f. The division algorithm is useful, but it doesn t give us everything that we want by itself. For example, the polynomial y 2 y 3 is in the ideal x y 2, x y 3, but if we try to divide y 2 y 3 by x y 2 and x y 3 in lex order, nothing happens: we get q 1 = q 2 = 0 and r = y 2 y 3, so the division algorithm is not telling us that it is in the ideal. For now, we ll solve this problem by defining it away. For an ideal I R[x 1,..., x n ], we say that f 1,..., f k I are a Gröbner basis for I if LT (I) = LT (f 1 ),..., LT (f k ) where LT (I) = LT (f) : f I is the monomial ideal generated by the leading terms of all the polynomials in I. In our example above, {x y 2, x y 3 } is not a Gröbner basis for I = x y 2, x y 3 since y 2 = LT (y 2 y 3 ) is in LT (I) but not in LT (x y 2 ), LT (x y 3 ). For a Gröbner basis though, the division algorithm does what we want it to. Proposition. Fix a monomial order. Suppose f 1,..., f k R[x 1,..., x n ] are a Gröbner basis for the ideal I = f 1,..., f n that they generate and suppose g R[x 1,..., x n ]. Let r be the remainder upon division of g by (f 1,..., f k ). Then (1) g = q 1 f q k f k has a solution (q 1,..., q k ) R[x 1,..., x n ] k if and only if r = 0, and (2) r is unique in the sense that if g = f + r with f I and no term of r is divisible by any LT (f i ), then r = r. In particular, changing the order of the f i does not change r. Proof. The first statement is the special case of the second when r = 0. To prove the second, we just note that r r I, so that if r r, then LT (r r) LT (I). But every term of r r, and in particular the leading term, is not divisible by any of the LT (f i ), so that LT (r r) LT (f 1 ),..., LT (f k ), which contradicts our assumption that f 1,..., f k is a Gröbner basis. Thus we now know how to determine whether a given polynomial g is in the ideal I: we find a Gröbner basis for I and then just use the division algorithm. At this point though, we ve never even shown that any particular set of polynomials is a Gröbner basis (you re asked to show this in a very simple example on the homework). What we really want to be able to do is start with an arbitrary (finite) set of generators for an ideal and find a Gröbner basis for it. 8 Exercises (1) Determine whether x 2 4 x 3 + x 2 4x 4, x 3 x 2 4x + 4, x 3 2x 2 x A Gröbner basis will always exist because the Hilbert basis theorem tells us that LT (I) is generated by finitely many elements LT (f i ). This doesn t tell us anything about how to find one though.

17 6. BUCHBERGER S ALGORITHM FOR COMPUTING GRÖBNER BASES 17 (2) (a) Compute the remainder on division of the polynomial f = x 7 y 2 + x 3 y 2 y + 1 by the set {xy 2 x, x y 3 } with respect to the grlex order on R[x, y] with x > y. (b) Repeat, using the lex order. (3) If I = x α(1),..., x α(s) is a monomial ideal, prove that a polynomial f is in I if and only if the remainder of f on division by x α(1),..., x α(s) is zero. (4) For the ideal I = 2xy 2 x, 3x 2 y y 1 with grlex order, show that 2xy 2, 3x 2 y LT (I). (5) (a) Show that {x + z, y z} is a Gröbner basis for lex order. (b) Divide xy by the ordered set (y z, x + z). (c) Now divide xy by (x + z, y z). How can your reconcile the different quotients? (6) Show that {x y 37, x y 38 } is not a Gröbner basis with respect to lex order. 6. Buchberger s algorithm for computing Gröbner bases Given an ideal I = f 1,..., f k, we ve seen several examples now of how it is possible that f 1,..., f k may not be a Gröbner basis for I, i.e. we may have LT (f 1 ),..., LT (f k ) LT (I). Essentially, the problem is that it may be possible to cancel out the leading terms of some of the f i to get new elements of I with smaller leading terms. Let s look at an example. Consider the ideal I = f 1, f 2 R[x, y, z] in graded lex order, with f 1 = x 2 y + y 2 z and f 2 = xy 2 + z 2. We can try to cancel out the leading terms of f 1 and f 2 in hopes of getting a polynomial in I with a new leading monomial: f 3 = yf 1 xf 2 = y(x 2 y + y 2 z) x(xy 2 + z 2 ) = y 3 z xz 2. To potentially find more new leading terms of I that aren t in x 2 y, xy 2, y 3 z, we might, for example try to attempt the same sort of cancellation on the leading terms of f 1 and f 3, giving y 2 zf 1 x 2 f 3 = y 2 z(x 2 y + y 2 z) x 2 (y 3 z xz 2 ) = y 4 z 2 + x 3 z 2, whose leading term is already known to be in LT (I) since it is divisible by LT (f 3 ). However, we can divide it by f 3 (along with f 1 and f 2 ) to possibly get a remainder in I with a new leading term y 4 z 2 + x 3 z 2 = yzf 3 + (x 3 z 2 + xyz 3 ), and we see that f 4 = x 3 z 2 + xyz 3 I so that x 3 z 2 LT (I) is a new leading term which is not divisible by any of x 2 y, xy 2, y 3 z. What we re doing is looking at pairs of polynomials in our current list of generators, f i and f j and cancelling out their leading terms, and then dividing the result by our current list of generators to potentially get an element of the ideal with a new leading term. It turns out that if we keep doing this until we no longer get anything new in this way out of any pair (f i, f j ) of our current generators, then we can stop this process and our current list of generators is a Gröbner basis. More precisely, given monomials x α and x β, with α = (α 1,..., α n ) and β = (β 1,..., β n ) in R[x 1,..., x n ], the least common multiple of x α and x β is x γ where γ i = max{α i, β i }. If x γ is the LCM of the leading monomials of f and g, then we say that the S-polynomial of f and g is the polynomial S(f, g) = xγ LT (f) f xγ LT (g) g. This is more precisely what we mean above by cancelling the leading terms of f i and f j.

18 18 Thus to find a Gröbner basis, we claim that all we need to do is start with some generating set of polynomials G = {f i }, and then keep computing S(f i, f j ) for pairs of polynomials in our set and dividing the S-polynomial by the set G, and adding the remainder to G if it isn t zero (and hence its leading term isn t in LT (G) yet). Eventually this process will stop 9 when division of S(f i, f j ) by the set G yields a zero remainder for every pair f i, f j G, and once that happens, G is a Gröbner basis. This algorithm for computing a Gröbner basis is called Buchberger s Algorithm, and it relies on the following theorem (for a proof, see Cox, Little, and O Shea, 2.6): Theorem (Buchberger s criterion). Let G = { f 1,... f k } R[x 1,..., x n ] be a set of polynomials and I = G be the ideal they generate. Then G is a Gröbner basis for I if and only if for every pair 1 i, j k, the remainder on division of S(f i, f j ) by G is zero. In our above example, it turns out though that we would have been better off first trying to cancel the leading terms of f 2 and f 3 : the polynomial f 5 = yzf 2 xf 3 = yz(xy 2 + z 2 ) x(y 3 z xz 2 ) = x 2 z 2 + yz 3 has a leading term which properly divides that of f 4, and in fact we see that once we have f 5 I, the polynomial f 4 is redundant since f 4 = xf 5. We must then check the S-polynomials of f 5 with f 1, f 2, and f 3 (this is enough since we ve already looked at every pair from f 1, f 2, f 3 ): S(f 1, f 5 ) = z 2 (x 2 y + y 2 z) y(x 2 z 2 + yz 3 ) = 0, S(f 2, f 5 ) = xz 2 (xy 2 + z 2 ) y 2 (x 2 z 2 + yz 3 ) = y 3 z 3 + xz 4 = z 2 f 3, S(f 3, f 5 ) = x 2 z(y 3 z xz 2 ) y 3 (x 2 z 2 + yz 3 ) = y 4 z 3 x 3 z 3 = yz 2 f 3 xzf 5, and we see that the remainders are all zero, so {f 1, f 2, f 3, f 5 } is a Gröbner basis for I. 10 Elimination of variables We may want to try to find elements of an ideal I R[x 1,..., x n ] which only involve some of the( variables) x i,..., x n. For example, to find an implicit equation for the curve given parametrically by a(t), b(t), we would like to find an element of the ideal p(t)x a(t), q(t)y b(t) that only p(t) q(t) involves x and y and not t. It turns out that to do this, all we need to do is find a Gröbner basis for the ideal in Lex order. This is because in Lex order, if the leading term of a polynomial only involves the variables x i,..., x n, then in fact all of its terms involve only x i,..., x n. Thus we have LT (I R[x i,..., x n ]) = LT (I) R[x i,..., x n ] and if G is a Gröbner basis for I in Lex order, then G R[x i,..., x n ] is a Gröbner basis (and hence a generating set!) for I R[x i,..., x n ] This process must stop eventually, because each time a polynomial is added to G, the ideal LT (G) gets bigger, and it is impossible for there to be an infinite ascending chain I 1 I 2 I 3 of ideals in R[x 1,..., x n ]. This fact is equivalent to the claim that any ideal of R[x 1,..., x n ] is finitely generated. (Hint: Consider I = I n.) 10 The set {f 1, f 2, f 3, f 4, f 5 } is also a Gröbner basis for this ideal, but f 4 is redundant, so we may as well leave it out. 11 See 3.1 in Cox, Little, O Shea for more details.

19 6. BUCHBERGER S ALGORITHM FOR COMPUTING GRÖBNER BASES 19 Thus, to eliminate variables from our ideal (i.e. find the polynomials in the ideal which only involve the other variables), we just need to put the variables to be eliminated first in Lex order and find a Gröbner basis for the ideal. This is something very special about Lex order: none of the other orders we ve looked at are elimination orders in this sense. Sample Code Here is a sample session in Macaulay 2 computing a Gröbner basis for I = x 2 y+y 2 z, xy 2 +z 2 R[x, y, z] in graded lex order: Macaulay 2, version 1.2 with packages: Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra, SchurRings, TangentCone i1 : R=QQ[x,y,z,MonomialOrder=>GLex] o1 = R o1 : PolynomialRing i2 : f=x^2*y+y^2*z 2 2 o2 = x y + y z o2 : R i3 : g=x*y^2+z^2 2 2 o3 = x*y + z o3 : R i4 : I=ideal(f,g) o4 = ideal (x y + y z, x*y + z ) o4 : Ideal of R i5 : gens gb I o5 = xy2+z2 x2y+y2z y3z-xz2 x2z2+yz3 1 4

20 20 o5 : Matrix R <--- R i6 : h=y^4*z^2+x^3*z^ o6 = y z + x z o6 : R i7 : h % I o7 = 0 o7 : R i8 : h // gens gb I o8 = {3} -xyz {3} y2z {4} 0 {4} x 4 1 o8 : Matrix R <--- R Here is some code computing the same Gröbner basis in Singular: SINGULAR / A Computer Algebra System for Polynomial Computations / version < by: G.-M. Greuel, G. Pfister, H. Schoenemann \ Mar 2009 FB Mathematik der Universitaet, D Kaiserslautern \ > ring R = 0,(x,y,z),Dp; //Dp = glex, lp=lex, dp=grevlex > poly f=x^2*y+y^2*z; > poly g=x*y^2+z^2; > ideal I=f,g; // An "ideal" for Singular is just a list of polynomials. > I; // This line is just to display I. I[1]=x2y+y2z I[2]=xy2+z2 > ideal gi=groebner(i); > gi; gi[1]=xy2+z2 gi[2]=x2y+y2z gi[3]=y3z-xz2 gi[4]=x2z2+yz3 > poly h=y^4*z^2+x^3*z^2; // Next we will divide h by the list of polynomials I, which > reduce(h,i); // gives a warning since we aren t dividing by a Groebner basis. // ** I is no standard basis

21 y4z2+x3z2 > reduce(h,gi); 0 6. BUCHBERGER S ALGORITHM FOR COMPUTING GRÖBNER BASES 21 Here is some code computing the same Gröbner basis in Sage: Sage Version 4.5.2, Release Date: Type notebook() for the GUI, and license() for information sage: R.<x,y,z> = PolynomialRing(QQ,3,order= deglex ) # or degrevlex, lex, etc. sage: f = x^2*y+y^2*z; sage: g = x*y^2+z^2; sage: I = (f,g)*r sage: I Ideal (x^2*y + y^2*z, x*y^2 + z^2) of Multivariate Polynomial Ring in x, y, z over Rational sage: gi = I.groebner_basis(); gi [x^2*z^2 + y*z^3, y^3*z - x*z^2, x^2*y + y^2*z, x*y^2 + z^2] sage: h = y^4*z^2+x^3*z^2; sage: h.mod(i) 0 sage: h.mod(gi) # I m not sure how to get Sage to just do the division algorithm 0 Exercises You may wish to use a computer to do most of the work in the following calculations. 12 (1) (a) Determine whether of not f = xy 3 z 2 + y 5 z 3 is in the ideal I = x 3 + y, x 2 y z. (b) Determine whether or not f = x 3 z 2y 2 is in the ideal I = xz y, xy + 2z 2, y z. (2) (a) Find the points on the variety V ( x 2 + y 2 + z 2 1, x 2 + y 2 + z 2 2x, 2x 3y z ). (b) Find the points on the variety V ( x 2 y z 3, 2xy 4z 1, z y 2, x 3 4zy ). (3) (a) Find an implicit equation for the surface parametrized by: x = ut y = 1 u z = u + t ut 12 One advantage of Macaulay 2 is the getchangematrix command to express the generators of the Gröbner basis as linear combinations of the original generators, something I m not sure how to do in the other programs.

22 22 (b) Find an implicit equation for the surface parametrized by: Last time we found that x = t + u y = t 2 + 2tu z = t 3 + 3t 2 u 7. Reduced Gröbner bases {x 2 y + y 2 z, xy 2 + z 2, y 3 z xz 2, x 3 z 2 + xyz 3, x 2 z 2 + yz 3 } was a Gröbner basis for I = x 2 y + y 2 z, xy 2 + z 2 R[x, y, z] in grlex order with x > y > z. Of course, we d like to be able to say our Gröbner bases are unique. As a first step, we noticed last time that one element of the Gröbner basis was redundant: x 3 z 2 + xyz 3 = x(x 2 z 2 + yz ), so we could remove it and still have a Gröbner basis {x 2 y + y 2 z, xy 2 + z 2, y 3 z xz 2, x 2 z 2 + yz 3 }. More generally, if G is a Gröbner basis with f, g G and LT (f) is a multiple of LT (g), then f will be redundant and can be removed, i.e. the set G {f} is still a Gröbner basis for the same ideal. (Why?) Of course, multiplying any element of a Gröbner basis by a scalar will give a different Gröbner basis, so if we want uniqueness, we should require that each leading coefficient be 1. Since a monomial ideal certainly has a unique minimal monomial generating set, we might hope that forcing constant leading coefficients and removing redundant elements would be enough to get a Gröbner basis which is unique, but that is not quite the case. The problem is that we could still add a multiple of one generator to another. For example, for any a, b R {x 2 y + axy 2 + y 2 z + az 2, xy 2 + z 2, y 3 z xz 2, x 2 z 2 + by 3 z + yz 3 bxz 2 } is another Gröbner basis for the same ideal as above. To avoid non-uniqueness arising in this way, we say that a reduced Gröbner basis G is a Gröbner basis where the leading coefficient of every f G is 1 and no term of any f G is divisible by the leading term of any g G with g f. Starting with a Gröbner basis, we can get a reduced Gröbner basis by multiplying by constants to clear any leading coefficients, throwing away any elements whose leading term is a proper multiple of another leading term, and then replacing each polynomial by the remainder upon dividing it by the rest (to clear out any terms divisible by any of the other leading terms). Moreover, reduced is all we need to impose to make our Gröbner bases unique: Theorem. Fix a term order on R[x 1,..., x n ]. Then every ideal I R[x 1,..., x n ] has a unique reduced Gröbner basis. Proof. To prove uniqueness, suppose that G and G are two different reduced Gröbner bases for I. The set of leading terms of both G and G must simply by the minimal set of monomial generators of LT (I), so if G G, it is because there is some f G and f G with LT (f) = LT ( f) but f f. Then f f I, so the remainder of f f upon division by G is zero, since G is a Gröbner basis. However, since G and G are both reduced Gröbner bases, no non-leading term of f or f is divisible by any leading term in G. Since the leading terms of f and f cancel in f f, no term of f f is divisible by a leading term of G, so we see that no actual division occurs, and the remainder is f f, so f f = 0, a contradiction.