Notes 17: Parametrically Defined Varieties

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1 Notes 17: Parametrically Defined Varieties Our goal is to find the implicit equations defining a set of point given parametrically For example, if we have the curve defined by x = 2t 1 + t 2, (1) y = 1 t2 1 + t 2, (2) then we have seen that this parametrizes the unit circle x 2 + y 2 = 1 To reach our goal we use the Closure Theorem, but we need to reorient this result We have viewed theorem as describing which partial solutions to a set of equations can be lifted We now look at the Closure theorem as a description of the image of a variety under a special kind of map, a projection We restate the theorem in this optic: Theorem 1 Let I k[x 1, x 2, x m+n ] be an ideal and X = V (I) Let π be the projection onto the last n variables Let I m be the m-th elimination ideal, that is, I m = I k[x m+1, x m+2,, x m+n ] Then i The image of X under the projection π is contained in V (I m ) ii The variety V (I m ) is the smallest variety containing π(x) iii The variety V (I) = π(x) W where W is an affine variety properly contained in V (I m ), that is, W V (I m ) Our first step is to formulate the question about parametrized varieties as a question about describing the image of k m under a polynomial map F : k m k m+n We then change this into a question about the describing the image of a variety under a very special polynomial map, the projection map We can apply the Closure Theorem to that question Polynomial Parametrizations Question/Problem: Given a set of all points X = {(x 1, x 2,, x n ) k n such that x 1 = f 1 (t 1, t 2,, t m ), x 2 = f 2 (t 1, t 2,, t m ), x n = f n (t 1, t 2,, t m ) with x i (t) polynomials}, we can ask: Is this an affine variety? If so, how do we find the equations defining the affine variety? We answer these questions in two steps: 1

2 1 We reformulate the question/problem in terms of finding the image of a projection map 2 We employ the Closure theorem First we reformulate the question in terms of a projection map We define three maps: Let F : k m k n, (t 1, t 2,, t m ) (f 1 (t), f 2 (t),, f n (t)) We wish to describe the mage of F To do this we introduce the graph of F Let i : k m k m+n, the image of i is just the graph of the map F (t 1,, t m ) (t 1,, t m, f 1 (t), f 2 (t), f n (t)); Set π m : k m+n k n, (t 1, t 2,, t m, x 1, x 2, x n ) (x 1, x 2,, x n ) This gives us the diagram of maps: k n+m i π m k m F k n Observations The image of i is the graph of F The composition π m i = F The image of F is the image of i(k m ) under the projection map π m The image of i is the variety V defined by the polynomials x 1 f 1 (t) x 2 f 2 (t) x n f n (t) We are in a position to use the Closure Theorem This result tells us what the image of a variety looks like under a projection map Theorem 2 Let k = C We use the notation above The image of F is π m (V ) Let I be the ideal defining the variety V k n+m and let I m be the elimination ideal associated to eliminating the m variables t 1, t 2,, t m from the ideal I Then V (I m ) k n is the smallest variety containing the image of F Indeed V (I m ) = im(f ) W where W is an affine subvariety properly contained in V (I m ) 2

3 Example 1 We parametrize the surface consisting of the union of the tangent lines to the twisted cubic by (t, u) (x = t + u, y = t 2 + 2tu, z = t 3 + 2t 2 u) We aim to find the equation of this surface We use the lexicographic order t > u > x > y > z The Groebner basis is g 1 = t + u + x, g 2 = u 2 x 2 + y, g 7 = x 3 z (3/4)x 2 y 2 (3/2)xyz + y 3 + (1/4)z 2 We see that the elimination ideal I 2 =< g 7 > From this we know that the surface consists of all of the points of V (g 7 ) except possible those in a proper subvariety of V (g 7 ) Using the extension theorem we can show that actually every point in V (g 7 ) lifts We do this in two steps First we lift (x, y, z) to a partial solution (u, x, y, z) We are looking for u so that (u, x, y, z) is a zero of all the polynomials in I 1 = k[u, x, y, z] I From our Groebner basis calculation we know that I 1 =< g 2, g 3,, g 7 > We look at these polynomials as polynomials in the variable u with coefficients in k[x, y, z] We look at the leading coefficients From the extension theorem we know that a partial solution (x, y, z) lifts provided at least one of these coefficients is not zero at the point (x, y, z) In this case one of the coefficient polynomials is the polynomial 1 (The coefficient of u 2 in g 2 is 1) This means that every partial solution (x, y, z) lifts to a (still partial) solution (u, x, y, z) The second step is to see if the solution (u, x, y, z) lifts to a solution (t, u, x, y, z) Again using the extension theorem and noting that the coefficient of t in g 1 is 1 we see that every solution lifts Rational Parametrizations We often write t = (t 1, t 2, t m ) and x = (x 1, x 2,, x n ) Consider a parametrization of the form x 1 = f 1(t 1, t 2,, t m ) g 1 (t 1, t 2,, t m ), x 2 = f 2(t) g 2 (t), x n = f n(t) g n (t), with all of the f i, g i polynomials in k[t] 3

4 We ask what is the resulting set of points? Is it affine? If so, what are its defining equations? These questions are trickier in this situation Example 2 Consider the surface described by the rational parametrization x = u2 v, y = v2 u, z = u We attempt to find the implicit equation defining this surface using the techniques above We look at the variety defined by V (f 1 = xv u 2, f 2 = uy v 2, f 3 = z u) k 5 We find its projection to (xyz) space by finding a Groebner basis with respect to the order u > v > x > y > z The method we used for polynomial parametrized surfaces says that the surface should be most of the variety defined by I 2 The Groebner basis is {z(x 2 y z 3 ), xyz + vz 2, v 2 yz, u z} The ideal I 2 = I k[x, y, z] =< z(x 2 z 3 ) > The surface V (z(x 2 z 3 )) is the union V (z) V (x 2 y z 3 ) Observe that any point of the form x = u2, y = v2, z = u) lies on the surface V v u (x2 y z 3 ) Hence the only points that arise from the parametrization are the ones on this surface None of the points on the surface z = 0 arise from the parametrization Our method is not adequate When we cleared denominators from the rational parametrization we added too many solutions Start with a rational parmetrization x i = f i(t) g i (t) i = i,, n Our problems arise at the points t where one or more of the g i vanish We wish to require that none of the g i (t) vanish To get rid of these points we add a restriction: g 1 (t)g 2 (t) g n (t)y 1 = 0 with y a new variable This equation has a solution at a point t iff none of the g i (t) = 0 This little idea does the job We now formally work through the algorithm Let g(t) = i=1,,n g i(t) Let W = V (g) Our goal is to find the equations that identify the image of the map F : (k m W ) k n, (t 1,, t m ) ( f 1(t) g 1 (t),, f n(t) g n (t) ) We define two more maps: j : (k m W ) k n+m+1, j : (t 1, t 2, t m ) ( 1 g(t), t 1,, t m, f 1(t) g 1 (t),, fn(t) g n(t)), π : k m+n+1 k n, π : (y, t 1, t m, x 1, x m ) (x 1,, x m ) 4

5 1 Observe that we have a commutative diagram (k m W ) k n+m j π F k n This implies that the image of F is the image of the map π j In particular,the image of F is the image of π acting on j(k m W ) 2 We identify j(k m W ) as a variety We claim that j(k m W ) is the variety defined by the ideal < gy 1, g 1 x i f 1,, g n x n f > Proof We show that j(k m W ) V (gy 1, g 1 x i f 1,, g n x n f) This follows from the formula for j We have j(t 1,, t m ) = ( 1 g(t), t 1,, t m, f 1(t) g 1 (t),, f n(t) ) g n (t) We show that V (gy 1, g 1 x 1 f 1, g n x n f n ) j(k m W ) Note that gy 1 = 0 implies that g(t) = g i (t) 0 This gives two facts First we see that g i x i f i = 0 implies that x i = f i g i We also get that gy 1 = 0 implies that y = 1 Thus we g(t) obtain our set inclusion Example 3 We return to the example that started the section on rational parametrizations Let W = V (uv) and set F : (k 2 W ) k 3, (u, v) (x = u2 v, y = v2 u, z = u) We employ the new algorithm that we have developed The product of the denominators in the parmetrization is g(u, v) = uv We set j : k 2 W k 3+1, (u, v) (t = 1 uv, x = u2 v, y = v2 u, z = u) We want to describe the image of the projection of V (tuv 1, xv u 2, uy v 2, z u) The Groebner basis of this ideal includes the element x 2 y z 3 Thus we do find the implicit equation of the parametrized surface 5