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: We obtain a contradiction picking In this case, we clearly have U = A 1 \ {p 1,..., p n }, V = A 1 \ {q 1,..., q m }. 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) Show that a Noetherian space is quasicompact i.e., show that any open cover of a Noetherian space has a finite subcover. (iii) Check that a Noetherian space which is also Hausdorff must be a finite set of points. (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) 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. (iii) Since X is Noetherian and Hausdorff, every subset Y X is also Noetherian and Hausdorff. By part (ii), 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. (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 three coordinate axes. (ii) The union of the (x, y)plane with the zaxis. Answer: (i) The xaxis has the ideal (y, z), the yaxis has the ideal (x, z), and the zaxis has the ideal (x, y). Thus the union of the three axes has ideal (y, z) (x, z) (x, y) = (xy, yz, zx). (ii) 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 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 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. 3
4 4 Problem 5. 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 I(X 1 X 2 ) = I(Z(a) Z(b)) = I(Z(a + b)) = a + b. Therefore, 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). 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).
5 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 ). Problem 6. Find the irreducible components of the affine algebraic set x 2 yz = xz x = 0 in A 3. What is the dimension of this algebraic set? Answer: Let X be the affine algebraic set given by the two equations above. The second equation can be written as Case 1. If x = 0, the first equation becomes xz x = x(z 1) = 0 = x = 0 or z = 1. x 2 yz = yz = 0 = y = 0 or z = 0. Therefore, there are two possibilities: x = y = 0 or x = z = 0. These correspond to two algebraic sets X 1 = the zaxis and X 2 = the yaxis, defined by the equations Case 2. If z = 1, the first equation becomes X 1 = Z(y, z) and X 2 = Z(x, z). x 2 yz = x 2 y = 0. This zero set is a parabola which can be rewritten as According to discussion above, X 3 = Z(z 1, x 2 y). X = X 1 X 2 X 3. We claim that the X i s are irreducible. This can be seen at once since all these sets are homeomorphic to A 1. This shows X 1, X 2, X 3 are irreducible components of the affine algebraic set X. These three sets are homeomorphic to A 1, hence their dimension is 1. Alternatively one may argue as follows. X 1 is irreducible if and only if (y, z) is a prime ideal in C[x, y, z], which is equivalent to C[x, y, z]/(y, z) = C[x] is an integral domain. By the same argument, X 2 is also irreducible. Finally, X 3 = Z(z 1, x 2 y) is irreducible. Indeed (z 1, x 2 y) is a prime ideal in C[x, y, z]. This is the case since is an integral domain. C[x, y, z]/(z 1, x 2 y) = C[x, y]/(x 2 y) = C[x], Problem 7. Find the irreducible components of the affine algebraic set xz y 2 = z 3 x 5 = 0 in A 3. 5
6 6 Answer: If x = 0, then the two equations imply that y = z = 0. Similarly, if y = 0, then x = z = 0. Let us assume that x 0, y 0. Write the first equation as y x = z := t. y Thus, The second equation becomes This implies y = tx, z = t 2 x. z 3 = x 5 t 6 x 3 = x 5 x 2 = t 6 x = ±t 3. y = ±t 4, z = ±t 5. Thus (x, y, z) = (t 3, t 4, t 5 ) or ( t 3, t 4, t 5 ) for some t C. Letting we obtain X 1 = {(t 3, t 4, t 5 ) t C } and X 2 = {( t 3, t 4, t 5 ) t C } X = X 1 X 2. We claim that X 1, X 2 are the irreducible components of X. First, there are polynomial maps f 1 : A 1 A 3 t (t 3, t 4, t 5 ) and f 2 : A 1 A 3 t ( t 3, t 4, t 5 ). Since X 1 = f 1 (A 1 ), X 2 = f 2 (A 1 ) and A 1 is irreducible, the images X 1 and X 2 are irreducible. It remains to explain that X 1 and X 2 are algebraic sets. We claim X 1 = Z(y 3 x 4, z 3 x 5, z 4 y 5 ), X 2 = Z(y 3 + x 4, z 3 x 5, z 4 + y 5 ). Let us check this claim for X 1 only. This follows using the same strategy as before. Indeed, if x = 0 then y = z = 0, so we may assume x 0. Let t = y x. From the first equation we have ( y ) 3 x = = t 3 = y = t 4. x Dividing the last two equations we get z = y5 x 5 = t5 = (x, y, z) = (t 3, t 4, t 5 ), as claimed. The verification for X 2 is entirely similar. Problem 8. Let X be the image of the map A 1 A 3, t (t, t 3, t 5 ). Show that I(X) cannot be generated by fewer than 3 elements.
7 Answer: Let I be the ideal of the set (t 3, t 4, t 5 ) in A 3, as t A 1. Observe that (xz y 2, yz x 3, x 2 y z 2 ) I. We assume that I can be generated by at most two elements f and g. Since I contains no polynomials of degree 0 or 1, it follows that f and g contain only quadratic terms or higher. Writing I (2) for the quadratic terms of the polynomials in I, it follows that f (2) and g (2) generate I (2). However, observe that the part I (2) is a kvector space of dimension at least 3 since it contains (yz, y 2, z 2 ). hence I (2) cannot be generated by 2 polynomials. This contradication shows that I needs at least 3 generators. 7
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