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Math 203A - Solution Set 1 2 Problem 1 Math 203A - Solution Set 1 2 Problem 1. Show that the Zariski topology on A is not the product of the Zariski topologies 1 1 on A × A . Answer: Clearly, the diagonal 2 Z = f(x; y): x − y = 0g ⊂ A 2 is closed in the Zariski topology of A . We claim this Z is not closed in the product 1 1 2 topology of A × A . Assume otherwise. Then, the complement A n Z is open in the 1 product topology. Therefore, there are nonempty open sets U; V in A such that 2 U × V ⊂ A n Z: Now, the sets U and V have finite complements: 1 U = A n fp1; : : : ; png; 1 V = A n fq1; : : : ; qmg: We obtain a contradiction picking 1 z 2 A n fp1; : : : ; pn; q1; : : : ; qmg: In this case, we clearly have 2 (z; z) 2 U × V; while (z; z) 62 A n 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 quasi-compact i.e., show that any open cover of a Noe- therian space has a finite subcover. n (iv) Show that A is Noetherian in the Zariski topology. Conclude that any affine algebraic set is Noetherian. Answer: (i) If fUigi≥1 is an ascending chain of open subsets of Y , there exists fVigi≥1 open subsets of X such that Vi \ Y = Ui for all i. Note that fVigi≥1 may not be ascending, but let i Wi = [k=1Vk: Then fWigi≥1 is clearly ascending and i i i Wi \ Y = ([k=1Vk) \ Y = [k=1(Vk \ Y ) = [k=1Uk = Ui: Therefore fWigi≥1 is an ascending chain of open subsets of X. Since X is Noe- therian, there exists n0 such that Wn = Wn0 for all n ≥ n0. Therefore, Un = Wn \ Y = Wn0 \ Y = Un0 for all n ≥ n0, i.e. fUigi≥1 eventually stabilizes. This implies Y is Noetherian. 1 2 (ii) Since X is Noetherian and Hausdorff, every subset Y ⊂ X is also Noetherian and Hausdorff. By part (iii), Y is quasi-compact and Hausdorff, hence compact. In particular, X is compact. Since Y is compact, Y is closed. Since Y was ar- bitrary, X must have the discrete topology. In particular, X can be covered by point-sets fxg, for x 2 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 = Uα; α and define the collection U consisting in finite union of the open sets Uα. The collection U must have a maximal element Uα1 [ ::: [ Uαn . We claim X = Uα1 [ ::: [ Uαn : Indeed, otherwise we could find x 2 X n (Uα1 [ ::: [ Uαn ) : Now, x 2 Uβ for some index β. Then, Uβ [ Uα1 [ ::: [ Uαn belongs to the collection U, and strictly contains the maximal element of U, namely Uα1 [ ::: [ Uαn . This is absurd, thus proving our claim. n (iv) It suffices to show that A satisfies the ascending chain condition on open sets. n If fUig be an ascending chain of open sets, let Yi be the complement of Ui in A . Then fYig is a descending chain of closed sets. Since taking ideal reverses inclu- sions, we conclude that I(Yi) is an ascending chain of ideals in k[X1;:::;Xn]. This chain should eventually stabilize, so I(Yn) = I(Yn+1) = :::; for n large enough. Applying Z to these equalities, we obtain Yn = Yn+1 = :::: This clearly implies that the chain fUig stabilizes. 3 Problem 3. Let A be the 3-dimensional 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 z-axis. 1 3 2 3 (ii) The image of the map A ! A given by t ! (t; t ; t ). Answer: (i) 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 2 3 3 1 3 2 (ii) Let I be the ideal of the set (t; t ; t ) in A , as t 2 A . Observe (z − x ; y − x ) is clearly contained in I. We claim that I = (z − x3; y − x2). We need to show every f 2 I is in the ideal (z − x3; y − x2). Indeed, let us consider the image of f in the quotient ring A = k[x; y; z]=(z − x3; y − x2): We would like to show that the image of f is 0 in A. Now, in the ring A the relations z = x3; y = x2 are satisfied. Thus f(x; y; z) = f(x; x2; x3) in A: Consider the polynomial g(x) = f(x; x2; x3). Since f(t; t3; t5) = 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, 3 2 f(x; y; z) = 0 in A = C[x; y; z]=(z − x ; y − x ); which is what we wanted. n m n Problem 4. Let f : A ! A be a polynomial map i.e. f(p) = (f1(p); : : : ; fm(p)) for p 2 A , where f1; : : : ; fm are polynomials in n variables. Are the following true or false: m n (1) The image f(X) ⊂ A of an affine algebraic set X ⊂ A is an affine algebraic set. −1 n m (2) The inverse image f (X) ⊂ A of an affine algebraic set X ⊂ A is an affine algebraic set. n n+m (3) If X ⊂ A is an affine algebraic set, then the graph Γ = f(x; f(x)) : x 2 Xg ⊂ A is an affine algebraic set. Answer: (i) False. Consider the following counterexample which works over infi- nite ground fields. Consider the hyperbola 2 X = f(x; y): xy − 1 = 0g ⊂ A : Clearly, X is an algebraic set. Let 2 1 f : A ! A ; f(x; y) = x 1 be the projection onto the first axis. The image of f in A is fx : x 6= 0g. This 1 is not an algebraic set because the only algebraic subsets of A are empty set, 1 finitely many points or A . (ii) True. Suppose m X = Z(F1;:::;Fs) ⊆ A ; where F1;:::;Fs are polynomials in m variables. Define Gi = Fi(f1(X1;:::;Xn); : : : ; fm(X1;:::;Xn)); 1 ≤ i ≤ s: These are clearly polynomials in n variables. In short, we set Gi = Fi ◦ f; 1 ≤ i ≤ s We claim −1 n f (X) = Z(G1;:::;Gs) ⊆ A ; which clearly exihibits f −1(X) as an algebraic set. This claim follows from the remarks −1 p 2 f (X) () f(p) 2 Z(F1;:::;Fs) () F1(f(p)) = F2(f(p)) = ::: = Fs(f(p)) = 0 4 () G1(p) = ::: = Gs(p) () p 2 Z(G1;:::;Gs): (iii) True. Assume that X is defined by the vanishing of the polynomials F1;:::;Fs in the variables X1;:::;Xn. Set G1 = Y1 − f1(X1;:::;Xn);:::;Gm = Ym − fm(X1;:::;Xn): We regard the F and G’s as polynomials in the n+m variables X1;:::;Xn;Y1;:::;Ym. We claim that the graph Γ is defined by the equations n+m Γ = Z(F1;:::;Fs;G1;:::;Gs) ⊆ A : Indeed, (x; y) 2 Z(F1;:::;Fs;G1;:::;Gs) () F1(x) = ::: = Fs(x) = 0; y1 = f1(x); : : : ; ym = fm(x) () x 2 X; y = f(x) () (x; y) 2 Γ: 3 Problem 5. Let X be the union of the three coordinate axes in A . 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 X i j k f = aikjx y z (i;j;k) vanishes on the x-axis, then f(x; 0; 0) = 0; hence ai00 = 0; for i ≥ 0: Similarly, ai00 = a0j0 = a00k = 0; for all i; j; k ≥ 0: This means that f 2 (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 2 (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 A1;B1 such that yz = A1f + B1g: Look at the degree 2 terms of the above equality. Letting a1; b1 be the free terms in A1;B1, we obtain yz = a1f~+ b1g:~ 5 Similarly, xz = a2f~+ b2g;~ xy = a3f~+ b3g:~ To reach a contradiction, consider the V the three dimensional vector space of poly- nomials of degree 2 which contain the monomials xy; yz; zx only.
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