Math 203A  Solution Set 1


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1 Math 203A  Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in the Zariski topology of A 2. We claim this Z is not closed in the product topology of A 1 A 1. Assume otherwise. Then, the complement A 2 \ Z is open in the product topology. Therefore, there are nonempty open sets U, V in A 1 such that U V A 2 \ Z. Now, the sets U and V have finite complements: U = A 1 \ {p 1,..., p n }, V = A 1 \ {q 1,..., q m }. We obtain a contradiction picking In this case, we clearly have z A 1 \ {p 1,..., p n, q 1,..., q m }. (z, z) U V, while (z, z) A 2 \ Z. Problem 2. A topological space X is said to be Noetherian if it satisfies the ascending chain condition on open sets, i.e. any ascending chain of open sets eventually stabilizes. (i) Check that any subset Y X of a Noetherian space is also Noetherian in the subspace topology. (ii) Check that a Noetherian space which is also Hausdorff must be a finite set of points. (iii) Show that a Noetherian space is quasicompact i.e., show that any open cover of a Noetherian space has a finite subcover. (iv) Show that A n is Noetherian in the Zariski topology. Conclude that any affine algebraic set is Noetherian. Answer: (i) If {U i } i 1 is an ascending chain of open subsets of Y, there exists {V i } i 1 open subsets of X such that V i Y = U i for all i. Note that {V i } i 1 may not be ascending, but let W i = i k=1 V k. Then {W i } i 1 is clearly ascending and W i Y = ( i k=1 V k) Y = i k=1 (V k Y ) = i k=1 U k = U i. Therefore {W i } i 1 is an ascending chain of open subsets of X. Since X is Noetherian, there exists n 0 such that W n = W n0 for all n n 0. Therefore, U n = W n Y = W n0 Y = U n0 for all n n 0, i.e. {U i } i 1 eventually stabilizes. This implies Y is Noetherian. 1
2 2 (ii) Since X is Noetherian and Hausdorff, every subset Y X is also Noetherian and Hausdorff. By part (iii), Y is quasicompact and Hausdorff, hence compact. In particular, X is compact. Since Y is compact, Y is closed. Since Y was arbitrary, X must have the discrete topology. In particular, X can be covered by pointsets {x}, for x X. Since X is compact, X must be finite. (iii) Order the open sets in X by inclusion. It is clear that any collection U of open sets in X must have a maximal element. Indeed, if this was false, we could inductively produce a strictly increasing chain of open sets, violating the fact that X is Noetherian. Now, consider an open cover X = α U α, and define the collection U consisting in finite union of the open sets U α. The collection U must have a maximal element U α1... U αn. We claim Indeed, otherwise we could find X = U α1... U αn. x X \ (U α1... U αn ). Now, x U β for some index β. Then, U β U α1... U αn belongs to the collection U, and strictly contains the maximal element of U, namely U α1... U αn. This is absurd, thus proving our claim. (iv) It suffices to show that A n satisfies the ascending chain condition on open sets. If {U i } be an ascending chain of open sets, let Y i be the complement of U i in A n. Then {Y i } is a descending chain of closed sets. Since taking ideal reverses inclusions, we conclude that I(Y i ) is an ascending chain of ideals in k[x 1,..., X n ]. This chain should eventually stabilize, so I(Y n ) = I(Y n+1 ) =..., for n large enough. Applying Z to these equalities, we obtain Y n = Y n+1 =.... This clearly implies that the chain {U i } stabilizes. Problem 3. Let A 3 be the 3dimensional affine space with coordinates x, y, z. Find the ideals of the following algebraic sets: (i) The union of the (x, y)plane with the zaxis. (ii) The image of the map A 1 A 3 given by t (t, t 2, t 3 ). Answer: (i) The ideal of the (x, y)plane is (z). The ideal of the zaxis is (x, y). By Problem 6, the ideal we re looking for is (z) (x, y) = (xz, yz).
3 (ii) Let I be the ideal of the set (t, t 2, t 3 ) in A 3, as t A 1. Observe (z x 3, y x 2 ) is clearly contained in I. We claim that I = (z x 3, y x 2 ). We need to show every f I is in the ideal (z x 3, y x 2 ). Indeed, let us consider the image of f in the quotient ring A = k[x, y, z]/(z x 3, y x 2 ). We would like to show that the image of f is 0 in A. Now, in the ring A the relations z = x 3, y = x 2 are satisfied. Thus f(x, y, z) = f(x, x 2, x 3 ) in A. Consider the polynomial g(x) = f(x, x 2, x 3 ). Since f(t, t 3, t 5 ) = 0, we see g(t) = 0 for all t. Therefore g 0 as a polynomial in x. (This uses that the ground field is infinite). Therefore, f(x, y, z) = 0 in A = C[x, y, z]/(z x 3, y x 2 ), which is what we wanted. Problem 4. Let f : A n A m be a polynomial map i.e. f(p) = (f 1 (p),..., f m (p)) for p A n, where f 1,..., f m are polynomials in n variables. Are the following true or false: (1) The image f(x) A m of an affine algebraic set X A n is an affine algebraic set. (2) The inverse image f 1 (X) A n of an affine algebraic set X A m is an affine algebraic set. (3) If X A n is an affine algebraic set, then the graph Γ = {(x, f(x)) : x X} A n+m is an affine algebraic set. Answer: (i) False. Consider the following counterexample which works over infinite ground fields. Consider the hyperbola Clearly, X is an algebraic set. Let X = {(x, y) : xy 1 = 0} A 2. f : A 2 A 1, f(x, y) = x be the projection onto the first axis. The image of f in A 1 is {x : x 0}. This is not an algebraic set because the only algebraic subsets of A 1 are empty set, finitely many points or A 1. (ii) True. Suppose X = Z(F 1,..., F s ) A m, where F 1,..., F s are polynomials in m variables. Define G i = F i (f 1 (X 1,..., X n ),..., f m (X 1,..., X n )), 1 i s. These are clearly polynomials in n variables. In short, we set G i = F i f, 1 i s We claim f 1 (X) = Z(G 1,..., G s ) A n, which clearly exihibits f 1 (X) as an algebraic set. This claim follows from the remarks p f 1 (X) f(p) Z(F 1,..., F s ) F 1 (f(p)) = F 2 (f(p)) =... = F s (f(p)) = 0 3
4 4 G 1 (p) =... = G s (p) p Z(G 1,..., G s ). (iii) True. Assume that X is defined by the vanishing of the polynomials F 1,..., F s in the variables X 1,..., X n. Set G 1 = Y 1 f 1 (X 1,..., X n ),..., G m = Y m f m (X 1,..., X n ). We regard the F and G s as polynomials in the n+m variables X 1,..., X n, Y 1,..., Y m. We claim that the graph Γ is defined by the equations Indeed, Γ = Z(F 1,..., F s, G 1,..., G s ) A n+m. (x, y) Z(F 1,..., F s, G 1,..., G s ) F 1 (x) =... = F s (x) = 0, y 1 = f 1 (x),..., y m = f m (x) x X, y = f(x) (x, y) Γ. Problem 5. Let X be the union of the three coordinate axes in A 3. Determine generators for the ideal I(X). Show that I(X) cannot be generated by fewer than 3 elements. Answer: We claim that I(X) = (xy, yz, zx). Clearly, xy, yz, zx vanish on X. Conversely, if a polynomial f = a ikj x i y j z k (i,j,k) vanishes on the xaxis, then f(x, 0, 0) = 0, hence a i00 = 0, for i 0. Similarly, a i00 = a 0j0 = a 00k = 0, for all i, j, k 0. This means that f (xy, yz, zx). Now, I(X) cannot be generated by one polynomial f since then f will have to divide xy, yz, zx which have no common factors. Assume that I(X) is generated by two elements (f, g). Since f (xy, yz, zx), we see that f = xyp + yzq + zxr for some polynomials P, Q, R. Letting f be the degree 2 component of f, and p, q, r be the free terms in P, Q, R, we obtain that f = pxy + qyz + rzx only contains the monomials xy, yz, zx. The similar remark applies to the degree 2 piece g of the polynomial g. Furthermore, there must be polynomials A 1, B 1 such that yz = A 1 f + B 1 g. Look at the degree 2 terms of the above equality. Letting a 1, b 1 be the free terms in A 1, B 1, we obtain yz = a 1 f + b1 g.
5 Similarly, xz = a 2 f + b2 g, xy = a 3 f + b3 g. To reach a contradiction, consider the V the three dimensional vector space of polynomials of degree 2 which contain the monomials xy, yz, zx only. We observed that f, g V. Moreover, the last three equations above show that f and g span the three dimensional vector space V. This is a contradiction showing that I(X) cannot be generated by fewer than 3 elements. Problem 6. Let X 1, X 2 be affine algebraic sets in A n. Show that (i) I(X 1 X 2 ) = I(X 1 ) I(X 2 ), (ii) I(X 1 X 2 ) = I(X 1 ) + I(X 2 ). Show by an example that taking radicals in (ii) is necessary. Answer: (i) If f I(X 1 X 2 ), f vanishes on X 1 X 2. In particular, f vanishes on X 1. Thus, f I(X 1 ). Similarly, we have f I(X 2 ). Then f I(X 1 ) I(X 2 ). We proved that I(X 1 X 2 ) I(X 1 ) I(X 2 ). On the other hand, if f I(X 1 ) I(X 2 ), then f I(X 1 ) and f I(X 2 ). Therefore, f vanishes on both X 1 and X 2, and then also on X 1 X 2. This implies that f I(X 1 X 2 ). Hence I(X 1 ) I(X 2 ) I(X 1 X 2 ). (ii) We assume the base field is algebraically closed. Because X 1, X 2 be affine algebraic sets, we can assume X 1 = Z(a), X 2 = Z(b). Furthermore, we may assume a and b are radical. Otherwise, if for instance a is not radical, we can replace a by a. This doesn t change the algebraic sets in question as X 1 = Z(a) = Z( a). By Hilbert s Nullstellensatz, I(X1 ) + I(X 2 ) = a I(Z(a)) + I(Z(b)) = + b = a + b and Therefore, I(X 1 X 2 ) = I(Z(a) Z(b)) = I(Z(a + b)) = a + b. I(X 1 X 2 ) = I(X 1 ) + I(X 2 ). Note: In the above, we made use of the equality Z(a) Z(b) = Z(a + b). 5
6 6 This can be argued as follows. If x Z(a) Z(b), then f(x) = g(x) = 0 for all f a and g b. Thus x Z(a + b) proving that Z(a) Z(b) Z(a + b). Conversely, if x Z(a + b), then (f + g)(x) = 0 for every f a and g b. In particular, picking g = 0, we get f(x) = 0 for every f a, so x Z(a). Similarly x Z(b). Therefore Z(a + b) Z(a) Z(b). Counterexample: Consider the following two algebraic subsets of A 2 : X 1 = (y x 2 = 0), X 2 = (y = 0). Graphically, these can be represented by a parabola and a line tangent to it. It is clear that X 1 X 2 = {0}, so I(X 1 X 2 ) = (x, y). On the other hand, Therefore I(X 1 ) + I(X 2 ) = (y x 2 ) + (y) = (y x 2, y) = (y, x 2 ). I(X 1 X 2 ) I(X 1 ) + I(X 2 ).
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