Math 203A  Solution Set 3


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1 Math 03A  Solution Set 3 Problem 1 Which of the following algebraic sets are isomorphic: (i) A 1 (ii) Z(xy) A (iii) Z(x + y ) A (iv) Z(x y 5 ) A (v) Z(y x, z x 3 ) A Answer: We claim that (i) and (v) are isomorphic, (ii) and (iii) are isomorphic, and (iv) is not isomorphic to any other algebraic sets We check (i) and (v) are isomorphic Note first that Define the morphisms and Then Z(y x, z x 3 ) = {(t, t, t 3 ) t C} f : A 1 Z(y x, z x 3 ), f(t) = (t, t, t 3 ), g : Z(y x, z x 3 ) A 1, g(x, y, z) = x f g = identity g f = identity Thus f and g are inverse isomorphisms We show (ii) and (iii) are isomorphic Consider the morphisms f : A A, f(x, y) = (x + iy, x iy), ( x + y g : A A, g(x, y) =, x y ) i It is easy to check that Moreover, letting we claim that Indeed, if (x, y) Z, then f g = identity g f = identity Z = Z(x + y ) and W = Z(xy), f(z) W, g(w ) Z f(x, y) = (x iy, x + iy) W since (x iy)(x + iy) = x + y = 0 Similarly, if (x, y) W, then ( x + y g(x, y) =, x y ) ( ) x + y ( ) x y Z, since + = 0 i i Therefore, f and g establish isomorphisms between Z and W We show that (i) and (ii) are not isomorphic Indeed, Z(xy) is the union of two lines, hence it is a reducible affine set On the other hand, A 1 is irreducible, hence it cannot be isomorphic to Z(xy) 1
2 We show (ii) and (iv) cannot be isomorphic This follows by the same argument observing that Z(x y 5 ) is irreducible Indeed, it suffices to prove that the polynomial x y 5 is irreducible Assuming the contrary, write x y 5 = f 1 (x, y)f (x, y) Regarding f 1, f as polynomials in x with coefficients in the integral domain k[y], we conclude that f 1, f can have degree at most with respect to x In fact, it is clear that the combination of degrees (0, ) cannot occur If the degrees are 1, we may assume Then Therefore, f 1 (x, y) = x g(y), f (x, y) = x h(y) x y 5 = x x (g(y) + h(y)) + g(y)h(y) g(y) = h(y), and y 5 = g(y)h(y) = g(y) This is clearly impossible, proving our claim We show that (i) and (iv) cannot be isomorphic Letting t be the coordinate of A 1, we show that there cannot be an ismorphism Indeed, set We must have Φ : k[x, y]/(x y 5 ) k[t] Φ(x) = p, Φ(y) = q p = q 5 This implies that all irreducible factors appearing in q have even exponent, so q = r, p = r 5 for some polynomial r Note that r cannot be constant since otherwise the image of Φ would have to consist in constant polynomials Now since Φ is surjective, there is a polynomial f = i,j a ij x i y j such that Φ(f) = r This means that a ij r 5i+j = r ij In particular, since the left hand side must be divisible by r, we have a 00 = 0 However, since 5i + j for (i, j) (0, 0), the left hand side is in fact divisible by r, so it cannot equal r This contradiction shows that an isomorphism Φ cannot exist Problem Show that X = A \{(0, 0)} cannot be isomorphic to an affine algebraic set: (i) Show that if f : X k is a regular function on X, then f must be a polynomial in k[x, y]
3 (ii) If Φ : Y X is an isomorphism between an affine algebraic set Y and X, consider the composition Ψ = ι Φ, where ι : X A is the inclusion Show that the morphism Ψ induces an isomorphism on coordinate rings Conclude that Ψ must be an isomorphism, hence ι must be an isomorphism, which must be a contradiction Answer: (i) Suppose f = g(x, y)/h(x, y) is a regular function on X and g, h k[x, y], we can assume g(x, y) and h(x, y) has no nonconstant common factors Then h(x, y) must be nonzero at every point in X Therefore the vanishing locus Z(h) {(0, 0)} Taking ideals and using the Nullstellensatz, we obtain (x, y) (h) This implies that x n (h), y m (h) for some n, m, which in turn implies that h divides both x n and y m Therefore, h must be constant and f k[x, y] (ii) Note that g is a regular function on Y g Φ 1 is a regular function on X g Φ 1 k[x, y] by (i) Letting P = g Φ 1, we have g = P Ψ for some polynomial P k[x, y] Thus, Ψ induces an isomorphism between the coordinate ring of Y and k[x, y] proving that Ψ : Y = A is an isomorphism In turn, this means ι : X = A is an isomorphism This is impossible because ι : X A is not bijective Problem 3 Let X A n and Y A m be affine algebraic sets (i) Show that X Y is an affine algebraic set (in such a fashion that its Zariski topology is the subspace topology from A n+m ) (ii) Show that if X and Y are irreducible, then X Y is also irreducible (iii) Show that X Y is a product in the category of affine algebraic sets eg it satisfies the following universal proprety: for every affine algebraic set Z and morphisms f : Z X and g : Z Y, there exists a unique morphism h : Z X Y such that h π X = f and h π Y = g (iv) Conclude that A(X Y ) = A(X) k A(Y ) Answer: (i) Let X A n and Y A m be affine algebraic sets such that X = Z(f 1,, f r ), Y = Z(g 1,, g s ) where f i k[x 1, X,, X n ] and g j k[y 1, Y,, Y m ] We regard f i and g j as polynomials in the variables X 1,, X n, Y 1,, Y m We see that that X Y = Z(f 1,, f r, g 1,, g s ) A n+m 3
4 4 This shows that X Y is an affine algebraic set (ii) We show that I(X Y ) is prime Let us write x for the collection of the x i, and y for the collection of the y i Let f, g k[x, y] be polynomial functions such that fg vanishes on X Y ; we have to show that either f or g vanishes on all of X Y, ie that X Y Z(f) or X Z(g) Let us assume that there is a point (P, Q) X Y \ Z(f), where P X and Q Y Denote by f(, Q) k[x] the polynomial obtained from f k[x, y] by setting y = Q For all P 0 X \ Z(f(, Q)) we must have Y Z(f(P 0, )g(p 0, )) = Z(f(P 0, )) Z(g(P 0, )) As Y is irreducible and Y Z(f(P 0, )) by the choice of P 0, it follows that Y = Z(g(P 0, )) This is true for all P 0 X \ Z(f(, Q)), so we conclude that (X \ Z(f(, Q)) Y Z(g) But as Z(g) is closed, it must in fact contain the closure of (X \ Z(f(, Q)) Y as well This closure equals X Y as X is irreducible and X \ Z(f(, Q)) is a nonempty open subset of X (iii) It is clear that the morphism h is defined set theoretically as h(p) = (f(p), g(p)) To define h algebraically we will proceed somewhat differently First, the morphism f : Z X corresponds to a kalgebra homomorphism f : k[x 1,, x n ]/I(X) A(Z), denoted by the same letter This in turn is determined by giving the images f i = f(x i ) A(Z) of the generators x i, satisfying the relations of I(X) The same is true for g, which is determined by the images g i = g(y i ) A(Z) Now it is obvious that the elements f i and g i determine a kalgebra homomorphism k[x 1, x n, y 1 y m ]/(I(X) + I(Y )) A(Z) which determines a morphism h : Z X Y (iv) The universal property of X Y corresponds on the level of coordinate rings to the universal property of the tensor product Also, one may directly observe that A(X Y ) = k[x, y]/(i(x) + I(Y )) = k[x]/i(x) k[y]/i(y ) = A(X) k A(Y ) Problem 4 Let n, and let S = {a 1,, a n } be a finite set with n elements in A 1 (i) Show that the quasiaffine set A 1 \ S is isomorphic to an affine set For instance, you may take X to be the affine algebraic set given by the equations X 1 (X 0 a 1 ) = = X n (X 0 a n ) = 1 Show that the projection onto the first coordinate is an isomorphism π : X A 1 \ S, (X 0,, X n ) X 0
5 (ii) Show that the ring of regular functions on A 1 \ {0} is isomorphic to k [ t, 1 t ], the ring of polynomials in t and 1 t (iii) Show that A 1 \ S is not isomorphic to A 1 \ {0} by proving that their rings of regular functions are not isomorphic Answer: (i) If (X 0,, X n ) X we have X i (X 0 a i ) = 1 hence X 0 a i for all 1 i n Therefore π : X A 1 \ S is well defined Let φ : A 1 \ S X, t ( 1 t,,, t a 1 1 t a n Both π and φ are rational maps, regular everywhere It is clear that π φ = identity and φ π = identity Therefore, π and φ are inverse isomorphisms (ii) It is clear that t and 1 t are both regular functions on A1 \ {0}, and so is any polynomial in t and 1 t Conversely, let f K(A1 \ {0}) View f as a rational function on A 1 Consider the ideal of denominators I f of the function f on A 1 Clearly Therefore ) Z(I f ) {0} = t I f = t n I f = t n f k[t] f = g(t) t n m = a i t i n i=0 The last expression is a polynomial in t and 1 t (iii) Assume there is an isomorphism Since it follows that Writing Φ : A(X) k[t, t 1 ] X i (X 0 a i ) = 1 in A(X), Φ(X i )Φ(X 0 a i ) = 1 Φ(X i ) = g i(t), Φ(X 0 a i ) = h i(t), t α i for some polynomials g i, h i, we obtain g i (t)h i (t) = t α i+β i This implies that h i is of the form ct m, or equivalently Φ(X 0 a i ) = c i t m i for some m i Z and c i k Subtracting the relations for i and j, it follows that t β i a j a i = Φ(a j a i ) = c j t m j c i t m i 5
6 6 Comparing degrees, we see that this implies m i = m j = 0, as a i a j for i j In turn, we obtain Φ(X 0 a i ) = c i Since Φ(c i ) = c i, we contradicted the injectivity of Φ This shows that Φ cannot be an isomorphism completing the proof Problem 5 Let n Consider the affine algebraic sets in A : Z n = Z(y n x n+1 ) and W n = Z(y n x n (x + 1)) Show that Z n and W n are birational but not isomorphic (i) Show that f : A 1 Z n, f(t) = (t n, t n+1 ) is a morphism of affine algebraic sets which establishes an isomorphism between the open subsets A 1 \ {0} Z n \ {(0, 0)} Similarly, show that g : A 1 W n, g(t) = (t n 1, t n+1 t) is a morphism of affine algebraic sets Find open subsets of A 1 and W n where g becomes an isomorphism (ii) Using (i), explain why Z n and W n are birational Write down a birational isomorphism between Z n W n (iii) Assume that there exists an isomorphism h : Z n W n such that h((0, 0)) = (0, 0) Observe that this induces an isomorphism between the open sets Z n \ {(0, 0)} W n \ {(0, 0)} Use part (i) and the previous problem to conclude this cannot be true if n (iv) Repeat the argument above without the assumption that h sends the origin to itself You may need to prove a stronger version of Problem 4 (v) Show that Z 1 and W 1 are isomorphic Write down an isomorphism between them Answer: (i) It is clear that f : A 1 Z n, t (t n, t n+1 ) is a well defined morphism Consider the morphism given by f 1 : Z n \ {0} A 1 \ {0} (x, y) y x A direct computation shows that f 1 is the inverse morphism of f Similarly, g : A 1 W n, t (t n 1, t n+1 t)
7 7 is a well defined morphism Its inverse morphism is g 1 : W n \ {0} A 1 \ S, (x, y) y x Here, S is the set of all n roots of unity The two morphisms g and g 1 establish an isomorphism between A 1 \ S W n \ {0} (ii) Part (i) shows that both Z n and W n are birational to A 1 so they are birational to each other An explicit birational isomorphism is A direct computation shows ( y g f 1 n (x, y) = g f 1 : Z n W n x n, yn+1 x n+1 y x (iii) If h : Z n W n is an isomorphism sending the origin to itself, then h : Z n \ {0} W n \ {0} is also an isomorphism By part (i), g 1 h f induces an isomorphism between the quasiaffine sets A 1 \ {0} A 1 \ S Such an isomorphism cannot exist by Problem 1 (v) Let F : Z 1 W 1, (x, y) (x, y + x) This is an isomorphism between Z 1 and W 1 with inverse F 1 (x, y) = (x, y x) Problem 6 Let λ k \ {0, 1} Consider the cubic curve E λ A given by the equation y x(x 1)(x λ) = 0 Show that E λ is not birational to A 1 In fact, show that there are no nonconstant rational maps F : A 1 E λ (i) Write ( f(t) F (t) = g(t), p(t) ) q(t) where the pairs of polynomials (f, g) and (p, q) have no common factors Conclude that p f(f g)(f λg) = q g 3 is an equality of fractions that cannot be further simplified Conclude that f, g, f g, g λg must be perfect squares (ii) Prove the following: Lemma: If f, g are polynomials in k[t] without common factors and such that there is a constant λ 0, 1 for which f, g, f g, f λg are perfect squares, then f and g must be constant )
8 8 Answer: (i) We have p f(f g)(f λg) = q g 3 Since p, q are relatively prime, the right hand side cannot be further simplified Similarly, f, g, f g, f λg cannot have any common factors Indeed, a common factor for instance of f and f g, will necessarily have to divide f (f g) = g as well But this is impossible since f and g are coprime Therefore the right hand side cannot be further simplied as well Thus, for some constant a k, we must have ap = f(f g)(f λg), aq = g 3 The exponents of the irreducible factors of the left hand sides of the above equations must be even Therefore, the same must be true about the right hand side This immediately implies that g is a square But since f, f g, f λg have no common factors, it follows that the exponents of the irreducible factors of each f, f g and f λg must be even as well Thus f, f g, f λg must be squares (ii) Pick f and g such that max(deg f, deg g) is minimal among all pairs (f, g) which satisfy the requirement that f, g, f g, f λg are squares for some λ 0, 1 We may assume that f, g are coprime since otherwise we can reduce their degree by dividing by their gcd which is also a square Write where u, v are coprime Then f = u, g = v, f g = (u v)(u + v) is a square Note that u v and u + v cannot have common factors since such factors will have to divide both 1 ((u v) + (u + v)) = u and 1 ((u v) (u + v)) = v which is assumed to be false Thus u v and u + v are coprime, and since their product f g is a square, it follows that u v, u + v must be square The same argument applied to f λg = (u λv)(u + λv) shows that u λv, u + λv are squares Let ũ = 1 + λ λ 1 (u + v), ṽ = (u v) Clearly, ũ, ṽ are squares A direct computation shows that ũ ṽ = u + λv, which is also a square Finally, ( 1 + ) λ ũ 1 ṽ = 1 + λ λ 1 λ (u λv)
9 is a square Setting ( 1 + ) λ µ = 1, λ we see that ũ, ṽ, ũ ṽ, ũ µṽ are squares Note that µ 0, 1 for λ 0, 1 Furthermore, max(deg ũ, deg ṽ) = 1 max(deg f, deg g) Unless f and g are irreducible, this contradicts the assumption that f, g are of minimal degree 9
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