# 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 quasi-compact 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 quasi-compact 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 point-sets {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 3-dimensional 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 z-axis. Answer: (i) The x-axis has the ideal (y, z), the y-axis has the ideal (x, z), and the z-axis 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 z-axis 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 z-axis and X 2 = the y-axis, 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 k-vector 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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