Math 40510, Algebraic Geometry Problem Set 3 Solutions, Spring 2016 As always, in this problem set k will represent some field. Occasionally we will give a more specific field. 2 1. In x8.1, Definition 1, of CLO, the authors define the projective plane over R, P (R), in a very intuitive way, leaving the more careful definitions to come. They say 2 2 P (R) = R [ fone point at infinity for each equivalence class of parallel linesg 2 Given a line, L in R , they denote by [L]1 the corresponding point at infinity, and they define the associated projective line to be L = L [ [L]1. The collection of all points at infinity (one for each class of parallel lines) is called the line at infinity and is also a projective line. 2 a) Prove that any two distinct points in P (R) determine a unique projective line. [Hint: there are three cases, depending on how many of the points are at infinity.] Solution: 2 Let P1;P2 be two points in P (R). 2 Case 1: If P1 and P2 both lie in R then it is well known that there is a unique affine line, `P1;P2 , joining them. Then adding the point at infinity corresponding to the family of lines parallel to `P1;P2 gives the unique projective line. 2 Case 2: Assume that P1 2 R and P2 lies on the line at infinity. P2 determines a family of parallel lines, and there is exactly one that passes through P1. Case 3: Assume that both P1 and P2 lie at infinity. They certainly both lie on the line at infinity. Suppose that there were some other projective line, `, that also contains P1 and P2. If ` is not 2 the line at infinity then it passes through some point P 2 R . But then P1 and P2 represent two different families of parallel lines, and ` can't be in both. So ` does not exist and there is only one line, namely the line at infinity. 2 b) Prove that any two distinct projective lines in P (R) meet at a unique point. [Hint: you have to divide this into cases too.] Solution: Let `1 and `2 be projective lines. 2 Case 1: Assume that neither line is the line at infinity. If `1 and `2 meet at a point P 2 R then they are not parallel, so they cross the line at infinity at different points. Hence they meet at exactly P and no other point. Case 2: Assume that neither line is the line at infinity. If `1 and `2 are parallel then they do not 2 meet in R , and being parallel they meet at a unique point at infinity (corresponding to that family of parallel lines). Case 3: Assume that `1 is the line at infinity. Then `2 is not the line at infinity, so it meets `1 only at the point corresponding the the family of parallel lines containing `2. 2. Let k be a field. In class we proved that if f 2 k[x0; : : : ; xn] is homogeneous of degree d, and if P = n [a0; : : : ; an] is a point of P such that f(a0; : : : ; an) = 0 then it's also true that f(λa0; : : : ; λan) = 0; n this means that the vanishing of a homogeneous polynomial at a point of P is well-defined, and we say for short that f vanishes at P . We now want to see how far we can push this idea beyond homogeneous polynomials. 1 2 a) If f 2 k[x0; : : : ; xn] is any homogeneous polynomial, we can write it as the sum of its homogeneous parts: f = f0 + f1 + ··· + fd where each fi is homogeneous of degree i. To get things started, let f = 5x4yz2 + 6x3y3 + 7xyz4 + 8y3z + 9xyz + 10y3 + 11y + 8: Tell me what d is, and what each fi is. Solution: d = 7 4 2 f7 = 5x yz 3 3 4 f6 = 6x y + 7xyz f5 = 0 3 f4 = 8y z 3 f3 = 9xyz + 10y f2 = 0 f1 = 11y f0 = 8 n b) Let f = f0 + f1 + ··· + fd 2 k[x0; : : : ; xn] and let P = [a0; a1; : : : ; an] 2 P . Assume that n each fi 2 P vanishes at P in the sense of what I said in the first paragraph of this problem. Prove that then the vanishing of f is well-defined. [There should be some λ's somewhere in your answer.] Solution: We know that P = [a0; a1; : : : ; an] = [λa0; λa1; : : : ; λan] for any λ 6= 0. Then d X j f(λa0; λa1; : : : ; λan) = λ fj(a0; a1; : : : ; an) = 0: j=0 c) Now assume that k is an infinite field. Let f = f0 + f1 + ··· + fd 2 k[x0; : : : ; xn] and let n P = [a0; a1; : : : ; an] 2 P . Assume that f vanishes at P . Prove that then each fi vanishes at P . [Again there should be λ's in your answer, and you should explain what k being infinite has to do with it.] Solution: We have assumed that f(λa0; λa1; : : : ; λan) = 0 for all λ 6= 0. So d d−1 λ fd(a0; : : : ; an) + λ fd−1(a0; : : : ; an) + ··· + λf1(a0; : : : ; an) + f0(a0; : : : ; an) = 0 for all λ 6= 0. But for each j, fj(a0; : : : ; an) is just an element of k, the field. This means that every nonzero λ 2 k is a root of the polynomial d d−1 fd(a0; : : : ; an)x + fd−1(a0; : : : ; an)x + ··· + f1(a0; : : : ; an)x + f0(a0; : : : ; an): Since k is infinite, this is a polynomial of degree d in one variable that has infinitely many roots, which is impossible over a field. 3 3. Homogeneous polynomials satisfy an important relation known as Euler's Theorem. It says the following. For convenience assume that our field is R. Let f 2 R[x0; : : : ; xn] be a homogeneous polynomial of degree d. Then n X @f x = d · f: i @x i=0 i a) Illustrate Euler's theorem by cooking up a homogeneous polynomial, f, having three terms and showing that the theorem is true for your example. Solution: Let f(x; y; z) = x3yz + 4x2yz2 + 5xyz3. Notice that d = 5. Then @f = 3x2yz + 8xyz2 + 5yz3 @x @f = x3z + 4x2z2 + 5xz3 @y @f = x3y + 8x2yz + 15xyz2 @z Then @f @f @f x + y + z @x @y @z = x(3x2yz + 8xyz2 + 5yz3) + y(x3z + 4x2z2 + 5xz3) + z(x3y + 8x2yz + 15xyz2) = (3x3yz + 8x2yz2 + 5xyz3) + (x3yz + 4x2yz2 + 5xyz3) + (x3yz + 8x2yz2 + 15xyz3) = 5(x3yz + 4x2yz2 + 5xyz3): b) Prove Euler's theorem by considering f(λx0; : : : ; λxn) as a function of λ, and differentiating with respect to λ using the chain rule. Solution: We know that d (1) f(λx0; : : : ; λxn) = λ f(x0; : : : ; xn): Then differentiate on both sides with respect to λ. @ (2) f(λx ; : : : ; λx ) = dλd−1f(x ; : : : ; x ): @λ 0 n 0 n Let's look at the left-hand side. For 0 ≤ i ≤ n let ui = λxi. n @ X @f @ui f(λx ; : : : ; λx ) = @λ 0 n @u @λ i=0 i n ! X @f (3) = · xi @xi i=0 xi=ui n X @f = λd−1x i @x i=0 i 4 where we have used the fact that @f is homogeneous of degree d − 1 and applied (1) to the @xi partials. Now substitute the result of (3) into (2) and divide by λd−1 (which is non-zero) to obtain the result. c) Let R = [x; y; z] and let f = xyz. In 2 describe (f), (f ; f ; f ), and the relation between R PR V V x y z these two varieties. (Here fx; fy; fz are the partials with respect to x; y; z respectively.) How is Euler's theorem relevant to this last part? Solution: First note that @f @f @f f = = yz; f = = xz; f = = xy: x @x y @y z @z Now, V(f) = V(xyz) is the union of the three lines defined by x = 0, y = 0 and z = 0. On the other hand, V(fx; fy; fz) is the locus defined by yz = 0 xz = 0 xy = 0: A quick calculation reveals V(fx; fy; fz) = V(x; y) [ V(x; z) [ V(y; z): This is precisely the union of the three points of pairwise intersection of the three lines in V(xyz), that is, the points f[1; 0; 0]; [0; 1; 0]; [0; 0; 1]g. In particular, V(fx; fy; fz) is a subset of V(xyz). And indeed, since Euler's theorem gives, in this case, that x · fx + y · fy + z · fz = 3 · f; if P 2 V(fx; fy; fz) then fx; fy; fz all vanish at P , so Euler's theorem implies that f vanishes at P , so in particular P 2 V(f). d) Let R = [x; y; z] and let f = xyz(x+y +z). In 2 describe (f), (f ; f ; f ), and the relation R PR V V x y z between the two. Again, how is Euler's theorem relevant to this last part? Solution: First note that f = x2yz + xy2z + xyz2, so @f 2 2 fx = @x = 2xyz + y z + yz = yz(2x + y + z) @f 2 2 fy = @y = x z + 2xyz + xz = xz(x + 2y + z) @f 2 2 fz = @z = x y + xy + 2xyz = xy(x + y + 2z): Now, V(f) = V(xyz(x + y + z)) is the union of the four lines defined by x = 0, y = 0, z = 0 and x + y + z = 0. On the other hand, V(fx; fy; fz) is the locus defined by yz(2x + y + z) = 0 xz(x + 2y + z) = 0 xy(x + y + 2z) = 0: Since each of these is a product of three linear forms, each equation is satisfied exactly when one (or more) of the factors is zero.