Math 40510, Algebraic Geometry

Transcription

1 Math 40510, Algebraic Geometry Problem Set 1, due February 10, Let k = Z p, the field with p elements, where p is a prime. Find a polynomial f k[x, y] that vanishes at every point of k 2. [Hint: Recall from MATH that for any a k (including 0), we have a p = a in k. We did a problem similar to this one in class.] f(x, y) = x p + (p 1)x + y p + (p 1)y. 2. Let k be any field. If V 1,..., V s are (finitely many) affine varieties in k n, prove that V 1 V s is again an affine variety. (FYI: It happens to be true that in this problem we can replace finite by infinite, but we need a little more background before we can prove that.) By induction. We proved in class that the intersection of two affine varieties is again an affine variety. Now assume that the result is true for any N varieties and let V 1,..., V N+1 be affine varieties. Then V 1 V N V N+1 = (V 1 V N ) V N+1. By induction the first is an affine variety, and so by the case for two varieties we get that V 1 V N V N+1 is an affine variety. Thus the result holds for any finite number of affine varieties. then Alternatively, if 3. Let k be any field. V 1 = V (f 1,1,..., f 1,t1 ) V 2 = V (f 2,1,..., f 2,t2 ). V s = V (f s,1,..., f s,ts ) V 1 V s = V (f 1,1,..., f 1,t1, f 2,1,..., f 2,t2,..., f s,1,..., f s,ts ). a) Prove that every (single) point (a 1,..., a n ) k n is an affine variety. (a 1,..., a n ) = V (x 1 a 1,..., x n a n ). b) Let V 1,..., V s be (finitely many) affine varieties in k n. Prove that V 1 V s is again an affine variety. [Hint: Using Lemma 2 of section 2 (which we proved in class) and induction, you can actually do this problem without giving explicit polynomials.] By induction. We proved in class that the union of two affine varieties is again an affine variety. Now assume that the result is true for any N varieties and let V 1,..., V N+1 be affine varieties. Then V 1 V N V N+1 = (V 1 V N ) V N+1. By induction the first is an affine variety, and so by the case for two varieties we get that V 1 V N V N+1 is an affine variety. Thus the result holds for any finite number of affine varieties. 1

2 c) Prove that any finite set of points in k n is an affine variety. [Hint: there is a reason why this is the third part of this problem. This should take less than one line.] Immediate from parts a) and b). d) Give an example of an infinite set of points in R 2 whose union is an affine variety. [Hint: this can also be done in less than one line.] Let V be the union of all the points on the line V (x), i.e. V = {(0, t) t R}. 4. Now we ll see that part d) of Problem 3 is not true for all infinite unions of affine varieties (in fact it s almost never true for infinite unions). Let k = R. Let Z be the set of points in R 2 with integer coordinates. a) Let f(x, y) be a polynomial vanishing at every point of Z. Prove that f(x, y) must be the zero polynomial. [Hint: if f(x, y) vanishes at every point of Z, what can you say about f(x, 0)?] In particular we have f(n, 0) = 0 for all n Z. But g(x) = f(x, 0) is a polynomial, and the first sentence means that g(x) has infinitely many zeros. So g(x) is the zero polynomial. This means that plugging in y = 0 into f(x, y) gives the zero polynomial, so f(x, y) contains no terms that are pure powers of x. In a similar way we can show that f(x, y) contains no terms that are pure powers of y. Now consider f(x, 1). Since each term of f(x, y) contains both powers of x and of y, f(x, 1) converts each term of f(x, y) into a term involving only x. Now the fact that f(x, 1) has infinitely many zeros means that it, too, is the zero polynomial, so all its terms are zero. This means that all terms of f(x, y) are zero, so f is the zero polynomial. b) Conclude that Z is not an affine variety. From a), if f I(Z) then f is the zero polynomial. If Z were an affine variety then we would have Z = V (f 1,..., f s ) for some polynomials f 1,..., f s that (by definition) vanish on Z. But any polynomial vanishing on Z is the zero polynomial, so the smallest variety containing Z is R 2. In particular, Z is not an affine variety. 5. Let V k n and W k m be affine varieties. Let V W = {(a 1,..., a n, b 1,..., b m ) k n+m (a 1,..., a n ) V and (b 1,..., b m ) W }. (This set is called the Cartesian product of V and W.) Prove that V W is an affine variety in k n+m. [Hint: this is problem 15 (d) of section 2 in CLO, and they give a nice hint.] We work in the polynomial ring R = k[x 1,..., x n, y 1,..., y m ]. Let V = V(f 1,..., f s ) k n and W = V(g 1,..., g t ) k m, where f 1,..., f s k[x 1,..., x n ] and g 1,..., g t k[y 1,..., y m ]. Think of the f i and the g j as being in R. We claim (1) V W = V(f 1,..., f s, g 1,..., g t ). A picture might be helpful here. Say m = n = 1 and k = R. Let and let V = {1, 2, 3} = V((x 1)(x 2)(x 3)) W = {1, 2} = V(y 1)(y 2).

3 (y 1)(y 2) W = 2 points V W = 6 points V = 3 points (x 1)(x 2)(x 3) Now we want to prove (1). We ll prove the two inclusions. To prove, let P V W. So P = (a 1,..., a n, b 1,..., b m ) k n+m where (a 1,..., a n ) V and (b 1,..., b m ) W. Since (a 1,..., a n ) V we have f i (a 1,..., a n ) = 0 for 1 i s. So thinking of f i as a polynomial in n+m variables, we have f i (a 1,..., a n, b 1,..., b m ) = 0 (since the b i don t do anything). Similarly we have g j (a 1,..., a n, b 1,..., b m ) = 0 for 1 j t. Hence P V(f 1,..., f s, g 1,..., g t ). Finally we ll prove in (1). Let P = (a 1,..., a n, b 1,..., b m ) V(f 1,..., f s, g 1,..., g t ). So f 1,..., f s, g 1,..., g t all vanish at P. Since the f i only involve x 1,..., x n, we have f i (a 1,..., a m ) = 0 for 1 i s, so (a 1,..., a n ) V. Similarly we see (b 1,..., b m ) W, so by definition P V W. 6. Recall we said that if f 1, f 2 R = k[x 1,..., x n ] and if f f 1, f 2 then there can be more than one way to find polynomials h 1 and h 2 so that f = h 1 f 1 + h 2 f 2. Let s explore that a bit. Let R = k[x, y] and let f = x 5 + 2x 4 y + 3x 3 y 2 + 4x 2 y 3 + 5xy 4 + 6y 5. a) If f 1 = x 3 and f 2 = y 3, prove that there are unique polynomials h 1 and h 2 of the form h 1 = a 1 x 2 + a 2 xy + a 3 y 2, h 2 = a 4 x 2 + a 5 xy + a 6 y 2 (where the a i are elements of k, i.e. scalars) so that f = h 1 f 1 + h 2 f 2. In particular, find a 1, a 2, a 3, a 4, a 5, a 6. We want to solve the equation x 5 + 2x 4 y + 3x 3 y 2 + 4x 2 y 3 + 5xy 4 + 6y 5 = (a 1 x 2 + a 2 xy + a 3 y 2 ) x 3 + (a 4 x 2 + a 5 xy + a 6 y 2 ) y 3 We immediately get = a 1 x 5 + a 2 x 4 y + a 3 x 3 y 2 + a 4 x 2 y 3 + a 5 xy 4 + a 6 y 5 a 1 = 1 a 2 = 2 a 3 = 3 a 4 = 4 a 5 = 5 a 6 = 6

4 Since these are uniquely determined, we are done. b) If f 1 = x 3 and f 2 = y 2, find all possible pairs (h 1, h 2 ) of the form h 1 = a 1 x 2 + a 2 xy + a 3 y 2, h 2 = a 4 x 3 + a 5 x 2 y + a 6 xy 2 + a 7 y 3 (again the a i are in k) so that f = h 1 f 1 + h 2 f 2. In a similar way we want to solve the equation x 5 + 2x 4 y + 3x 3 y 2 + 4x 2 y 3 + 5xy 4 + 6y 5 = (a 1 x 2 + a 2 xy + a 3 y 2 ) x 3 + (a 4 x 3 + a 5 x 2 y + a 6 xy 2 + a 7 y 3 ) y 2 We immediately get = a 1 x 5 + a 2 x 4 y + a 3 x 3 y 2 + a 4 x 3 y 2 + a 5 x 2 y 3 + a 6 xy 4 + a 7 y 5 a 1 = 1 a 2 = 2 a 3 + a 4 = 3 a 5 = 4 a 6 = 5 a 7 = 6 So the set of all possible (h 1, h 2 ) = (a 1 x 2 + a 2 xy + a 3 y 2, a 4 x 3 + a 5 x 2 y + a 6 xy 2 + a 7 y 3 ) consists of all ordered pairs where the coefficients satisfy the above equations. (As we mentioned in class, this problem is related to the topic of syzygies, which hopefully we will talk about later.) 7. Let R = k[x 1,..., x n ]. Let I be an ideal in R. We define an ideal to be radical if the following condition holds: If f m I for some positive integer m then f I. (Keep this definition in mind because we ll come back to it in class.) a) Prove that if f I then f m I for all positive integers m. If f I we certainly have f m 1 R, so f m = f m 1 f I by definition of an ideal. b) If X k n is any set, prove that I(X) is always a radical ideal. If f m I(X) then 0 = f m (P ) = f(p ) m for all P X, so f(p ) = 0 for all P X, i.e. f I(X). c) Give an example of an ideal that is not radical, and prove that it s not radical. Let I = x 2. Note that f(x) = x satisfies f 2 I. But every element of x 2 is a multiple of x 2 and clearly x is not a multiple of x 2, so f / I. Thus I is not radical. (Keep this definition in mind because we ll come back to it in class.) 8. Let I and J be ideals in k[x 1,..., x n ]. We define I J = {f R f I and f J}. We define IJ to be the set of polynomials that can be written as finite sums in the following way: { m } IJ = f i g i f i I, g i J.

5 a) Prove that I J is an ideal. We use the fact that both I and J are ideals. Since 0 I and 0 J, we have 0 I J. If f, g I J then f and g are both in I and both in J, so f + g I J. If f I J and h R then hf I and hf J so hf I J. b) Prove that IJ is an ideal. 0 I and 0 J so 0 = 0 0 IJ. Assume f = m f ig i for some f i I, g i J and g = m f i g i for some f i I, g i J. Then f + g = m m f i g j + f ig j IJ. Finally, if f = m f ig i for some f i I, g i J and h R then m m hf = h f i g i = (hf i )g i IJ since hf i I (because I is an ideal). c) Show that IJ I J (as ideals). It s enough to prove that each generator of IJ is in I J. If I = f 1,..., f s and J = g 1,..., g t then the generators of IJ have the form f i g j for 1 i s and 1 j t. But then f i g j I (since f i I) and also f i g j J (since g j J) so f i g i I J. d) Give an example to show that IJ is not necessarily equal to I J. Justify your answer! For example take I = x and J = x. Then I J is clearly equal to x, while IJ = x 2. We have already seen that these two ideals are not equal. e) If I and J are ideals in k[x 1,..., x n ], prove that V (IJ) = V (I) V (J). [Hint: this is closely related to our proof that V (I) V (J) is again an affine variety.] Let s prove the two inclusions. : Let P V (IJ). We want to show that P V (I) V (J). If P V (I) then we re done, so assume P / V (I); we want to show that then P V (J). Since P / V (I), there is some f I such that f(p ) 0. But fg IJ for all g J; hence (fg)(p ) = 0 for all g J. Thus P V (J) as desired. : Let P V (I) V (J). So either P V (I) or P V (J) (or both). Assume without loss of generality that P V (I). Then f(p ) = 0 for all f I. Let g IJ, so m g = f i g i f i I and g i J. Then we get g(p ) = m f i(p )g i (P ) = 0. Hence g(p ) = 0 for all g IJ, and so P V (IJ).

6 f) If I and J are ideals in k[x 1,..., x n ], prove that V (I J) = V (I) V (J). Combined with the previous part, conclude that V (IJ) = V (I J). Again we prove the two inclusions. : Let P V (I J), so h(p ) = 0 for all h I J. Suppose that P / V (I). We want to show P V (J), i.e. we want to show that g(p ) = 0 for all g J. Since P / V (I), there is some f I such that f(p ) 0. Then for any g J, we know that fg I J so (fg)(p ) = 0. Sincef(P ) 0, this forces g(p ) = 0 for all g J, so P V (J) as desired. : Let P V (I) V (J), so either P V (I) or P V (J) or both. Let f I J. Since f is in both I and J, we must have f(p ) = 0. So P V (I J). g) It happens to be true that if P = (0, 1) k 2 then I(P ) = x, y 1. (You can accept this without proof.) Let Q = (0, 0). We saw in class that I(Q) = x, y. You can also accept without proof the fact that I(P ) I(Q) = x, y(y 1). Show that I(P ) I(Q) = I(P ) I(Q) = I(P Q). We ll show I(P ) I(Q) I(P ) I(Q) I(P Q) I(P ) I(Q). (1) (2) (3) (1) It s always true for two ideals I and J that IJ I J (we showed this in part c) above), so the first inclusion is immediate. (2) If f I(P ) I(Q) then f vanishes on both P and Q, so f I(P Q). In fact it s convenient to observe here that the reverse is true: if f I(P Q) then f vanishes on each of P and Q so f I(P ) I(Q). (I.e. I(P ) I(Q) = I(P Q).) (3) We know from the statement of the problem that I(P ) = x, y 1, I(Q) = x, y and I(P Q) = x, y(y 1) = x, y 2 y. (The last one also uses what we observed in (2) above.) To show the inclusion (3), it s enough to show that each generator of I(P Q) is in I(P ) I(Q). So we have to show that x I(P ) I(Q) and that y(y 1) I(P ) I(Q). The second one is obvious since y 1 I(P ) and y I(Q). So let s look at the first one. We have to show that x x, y 1 x, y. But this is true because x = (y 1)( x) + (x)(y). (The point of d) and g) is that sometimes IJ = I J and sometimes IJ I J.)