exercise of GTM052 1.1

1. 1.1 Guang Hu October 7, 2018 1. (a) Let Y be the plane curve y = x2(i.e. Y is the zero set of the polynomial f = y − x2). Show that A(Y ) is isomorphic to a polynomial ring in one variable over k. Proof. We define the homomorphism from A(Y ) = k[x1,x2] to k[x] by: f0(x) + f1(x)y + ... + fn(x)yn7→ f0(x) + x2f1(x) + ... + x2nfn(x) i.e. ϕ(f) is defined by ϕ(f)(x) = f(x,x2) This is obvious an epimorphism. We need to show that I(Y ) is the kernel of the map: If f(x,y) maps to the zero polynomial in k[x], then f(x,y) = 0 for any y = x2. So f ∈ I(Y ). If f ∈ I(Y ), then f(x,x2) for any x and thus the image of f is the zero polynomial (since k is infinite). (b) Let Z be the plane curve xy = 1. Show that A(Z) is not isomorphism to a polynomial ring in one variable over k. Proof. Suppose that such an isomorphism exists, which is denoted by ϕ. So we have ϕ(xy − 1) = 0 and hence ϕ(x)ϕ(y) = 1. So ϕ(x),ϕ(y) ∈ k. From the fact that x and y generates k[x,y] one can deduce the contradiction that im(ϕ) ⊂ k. (c) Let f be any irreducible quadratic polynomial in k[x,y], and let W be the conic defined by f. Show that A(W) is isomorphic to A(Y ) or A(Z). Which one is it when? 1

2. Proof. We claim that for any non-degenerate (i.e. invertible) affine transfor- mation α, A(Y ) is isomorphic to A(αY ). Indeed, this property based on the fact that we can define fαin the following way: fα(x) := f(α−1x) such that fαis also a polynomial. Once we noticed this, we can define a map from A/I(Y ) to A/I(αY ) by: ¯f 7→¯fα Once can easily check that this is a well-defined isomorphism by definition. The rest things are simply linear algebra. 2. The Twisted Cubic Curve. Let Y ⊂ A3be the set Y = {(t,t2,t3)|t ∈ k}. Show that Y is a affine variety of dimension 1. Find generators for the ideal I(Y ). Show that A(Y ) is isomorphic to a polynomial ring in one variable over k. We say that Y is given by the parametric representation x = t,y = t2,z = t3. Proof. Y is an affine variety since Y = Z(f1)∩Z(f2) where f1= y −x2and f2= z − xy. Choose a nontrivial closed subset of Y , namely Y ∩Z(a) where a is generated by g1,g2,...,gn. Hence we have: Y ∩ Z(a) = Y ∩ (∪n i=1Z(gi)) = ∪n i=1(Y ∩ Z(gi)) For any (t,t2,t3) ∈ Y ∩ Z(gi), we have gi(t,t2,t3) = 0. So t is a solution of the one-variable polynomial gi(x,x2,x3) (which is not the zero polynomial since we cliamed that Y ∩Z(a) is non-trivial), which have only finitely many choice. So Y ∩ Z(gi) = 0 is a finite set and thus Y ∩ Z(a) is a finite set. Now we have proved that any non-trivial closed subset of Y is nothing but a set of finite points. So the only non-trivial irreducible closed sets are sets of a single point, so dim(Y ) = 1. We are going to prove that k[x,y,z]/(y −x2,z −x3) is isomorphic to k[x]. If this is right, then (y − x2,z − x3) is prime and hence equal to I(Y ). Let’s construct a homomorphism of rings: ϕ : k[x,y,z] −→ k[x] f(x,y,z) 7−→ f(x,x2,x3) 2

3. It is obvious that ϕ is surjective and (y − x2,z − x3) ⊂ kerϕ To show that kerϕ ⊂ (y−x2,z −x3), we only need to expansion f(x,y,z) = f(x,y −x2+x2,z −x3+x3) by regarding y −x2z −x3as variables, leading to the follwing equation: f(x,y,z) =¯f + i(x,y,z) where¯f ∈ k[x]. If 0 = ϕ(f) = ϕ(¯f) + ϕ(i) = ϕ(¯f), we have¯f = ϕ(¯f) = 0, hence f ∈ (y − x2,z − x3). 1.3 Let Y be the algebraic set in A3defined by the two polynomials x2−yz and xz−x. Show that Y is a union of three irreducible components. Describe them and find their prime ideals. Proof. One can easily check that Y = Z(a1)∪Z(a2)∪Z(a3), where a1= (x,y), a2= (x,z), a3= (y − x2,z − 1). It’s easy to show that k[x,y,z]/ai∼= k[x]. Hence aiis prime and Z(ai) is irreducible. 1.4 If we identify A3with A1×A1in the natural way, show that the Zariski topology on A2is not the product topology of the Zariski topologies on the two copies of A1. Proof. The non-trivial closed set in the product topology of A1×A1is noth- ing but finite union of {(x,y) : x = c} or {(x,y) : y = c}. So we can see {(x,y) : x = y} is closed in the Zariski topology of A2, while not closed in the product topology. 1.5 Show that a k−algebra B is isomorphic to the affine coordinate ring of some algebraic set in An, for some n, if and only if B is finitely generated k−algebra with no nilpotent elements. Proof. We have known it in Remark 1.4.6 that any k− algebra B is isomor- phic to the affine coordinate ring of some algebraic set if and only if B is finitely generated and is a domian. Further more, write B as the quotient of a polynomial ring k[x − 1,...,xn]/a. If B have no nilpotent elements, then fm∈ a implies that f ∈ a. So√a = a and hence A/I(Z(a)∼= A/a = B. 3

4. 1.6 Any nonempty open subset of an irreducible topological space is dense and irreducible. If Y is a subset of a topological space X, which is irreducible in its induced topology, then the closure¯Y is also irriducible. Proof. For any two open subsets U1,U2of an irreducible topological space X, Uc Suppose that U1 ⊂ V1∪ V2 where V1 and V2 are both closed. We have (Uc to say, U1⊂ V1or V2= X ⇒ U1⊂ V2. So U1is also irreducible. Now we come to the second statement. Suppose that¯Y ⊂ V1∪ V2where V1,V2is closed in X. So Y ⊂ V1∪ V2and hence Y ⊂ V1or Y ⊂ V2. So¯Y ⊂ V1of¯Y ⊂ V2by the universal property of closure. 2are not the whole space, thus Uc 2̸= X. So U1∩ U2̸= ∅. 1,Uc 1∪ Uc 1∪V1= X or V2= X. That is 1∪V1)∪V2= X. Since X is irreducible, Uc 1.7 (a) Show that the following conditions are equivalent for a topological space X: (i) X is noetherian; (ii) every nonempty family of closed subsets has a minimal element; (iii) X satisfies the ascending chain condition for open subsets; (iv) every nonempty family of open subsets has a minimal element. Proof. (i)⇒(ii) If there is exist a nonempty family S of closed subsets without a minimal element, i.e. for any subset in the family S, there is a smaller element in S. So we can find a descending chain which is not eventually stationary by induction. (ii)⇒(i) The family of subsets in the descending chain has a minimal element, so the chain is stationary after that element. It is alternative that (i) is equivalent to (iii) and (ii) is equivalent to (iv) (b) A noetherian topological space is quasi-compact, i.e. every open cover has a finite subcover. Proof. Suppose that ∪i∈IUiis an open cover of the topological space. Let S be the collection of all the finite union for some Ui. From a maximal element in S, we get a finite subcover we want. (c) Any subset of a noetherian topological space is noetherian in its induced topology. 4

5. Proof. Let X be a noetherian topological space and Y its subspace. A de- scending chain of closed set in Y is Y1∩ Y ⊃ Y2∩ Y ⊃ ... Since we have a descending chain of closed set: Y1⊃ Y1∩ Y2⊃ ... There is a minimal element in it, denoted by ∩n i=1Yi. So for any m > n we have ∩n Yn∩ Y = ∩n i=1Yi⊂ Ym. Hence i=1(Yi∩ Y ) = (∩n i=1Yi) ∩ Y ⊂ Ym∩ Y (d) A noetherian space which is also Hausdorff must be a finite set with the discrete topology. Proof. Form proposition 1.5 we know that a noetherian space X itself can be expressed as a finite union X = Y1∪Y2...∪Ynof irreducible closed subsets.For any i = 1,2,...,n, if Yicontains more than 1 elements, then they can sepa- rated by two open sets in X, namely U1,U2. Hence (U1∩Yi)∩(U2∩Yi) ̸= ∅ while neither of them is empty, that’s a contradiction. So every Yihas only 1 element and thus X is finite. So the topology in X is discrete since X is Hausdorff. 1.8. Let Y be an affine variety of dimension r in An. Let H be a hypersurface in An, and assume that Y ⊈ H. Then every irreducible component of Y ∩H has dimension r − 1 Proof. Suppose that H is the hypersurface defined by f = 0. And we may also assume that Y ∩H ̸= ∅, hence f +I(Y ) is not a unit in A(Y ) = A/I(Y ). Since Y ⊈ H, we have f / ∈ I(Y ) and hence f +I(Y ) is not the zero in A(Y ). Since Y is irreducible, I(Y ) is prime and hence f is not a zero divisor in A(Y ). Then let’s consider an irreducible components U of Y ∩ H. Define J(U) := {g + I(Y ) ∈ A(Y ) : g(U) = 0}. 5

6. J(U) is well-defined since g ∈ I(Y ) ⇒ g(U) ⊂ g(Y ) = 0 We claim that J(U) is a minimal prime ideal containing f + I(Y ) in A(Y ). In fact, from f(U) ⊂ f(H) = 0 we know that f + I(Y ) ∈ J(U). If we have (g1+ I(Y ))(g2+ I(Y )) := g1g2+ I(Y ) ∈ J(U) while neither of them is in J(U). That is to say, g1g2∈ I(U). Since U is irreducible, I(U) is prime. So g1∈ I(U) or g2∈ I(U), namely g1+ I(Y ) ∈ J(U) or g2+ I(Y ) ∈ J(U). So J(U) is prime. To see that J(U) is minimal of f + I(Y ), suppose that there is a prime ideal p such that f + I(Y ) ∈ p ⊊ J(U), and define the retraction of ideals: pc:= {g ∈ A : g + I(Y ) ∈ p}. It obvious that f ∈ pc⊊ I(U). Recall that the retraction of a prime ideal is also prime, we can deduce that U = Z(I(U)) ⊊ Z(pc) ⊂ Z(f) = H It’s trivil that I(Y ) ⊂ pcsince 0 ∈ p. So Z(pc) ⊂ Z(I(Y )) = Y . Hence Z(pc is an irreducible subset of Y ∩ H strictly containing U, this contradicts to the assumption that U is an irreducible component. Now use Krull’s Hauptidealsatz, J(U) has height 1. From theorem 1.8A (b) we can deduce that dimA(Y )/J(U) = dimA(Y ) − 1 = r − 1. The final step is to prove that A(Y )/J(U)∼= A/I(U). After constructing the following homomorphism of rings: A −→ A(Y )/J(U) f 7−→ (f + I(Y )) + J(U) we can easily find that the kernel of the homomorphism is I(U), since f + I(Y ) ∈ J(U) means that f(U) = 0, namely f ∈ I(U), by definition. 1.9. Let a be an ideal which can be generated by r elements. Then every irreducible component of Z(a) has dimension ≥ n − r. Proof. Use the induction on r, if a can be generated by r + 1 elements, namely f1,f2,...fr,fr+1. Since Z(a) = Z(f1,...,fr)∩Z(fr+1), any irreducible component Y in Z(a) is in Z(f1,...,fr), thus in some irreducible component 6

7. of Z(f1,...,fr), namely Y′. So Y ⊂ Y′∩Z(fr+1), and in fact Y is a irreducible component of Y′∩ Z(fr+1). Then use 1.8, and the induction assumption. If Y′⊂ Z(fr+1), Y′= Y and hence dimY′= dimY ≥ n − r. If Y′⊈ Z(fr+1), then dimY = dimY′− 1 ≥ n − (r + 1) When r = 0, Z(a) =An, so dimZ(a) = n ≥ n − r. 1.10. (a) If Y is any subset of a topological space X, then dimY ≤ dimX. Proof. If Y0 ⊂ Y1 ⊂ ... ⊂ Yn is a sequence of distinct closed irreducible subsets of Y , then¯Y0 ⊂¯Y1 ⊂ ... ⊂¯Yn is a sequence of closed irreducible subsets of X by 1.6. For any i = 0,1,...,n, Yi= V ∩ Y for some V closed in X. V ∩ Y ⊂¯Yiand V ∩Y ⊂ Y together implies that V ∩Y ⊂¯Yi∩Y . At the same time, Yi⊂ V implies that¯Yi⊂ V , thus¯Yi∩ Y ⊂ V ∩ Y . So¯Yi∩ Y = V ∩ Y = Yi. If¯Yi =¯Yj for some i ̸= j, then Yi =¯Yi =¯Yj = Yj, contradiction. So ¯Y0⊂¯Y1⊂ ... ⊂¯Ynis a distinct sequence. (b) If X is a topological space which is covered by a family of open subsets{Ui}, then dimX = supdimUi. Proof. dimX ≥ dimUifor all i by (a), so dimX ≥ supdimUi. If X0⊂ X1⊂ ... ⊂ Xnis a sequence of distinct closed irreducible subsets of X, we choose a U (equal to some Ui) such that X0∩ U ̸= ∅. So X0∩ U ⊂ X1∩U ⊂ ... ⊂ Xn∩U is a sequence of distinct closed subsets of U. From 1.6 we know that Xi∩U is irreducible and dense in Xi. Since Xc open subset in Xi, (Xi∩U)∩(Xc not contained in Xi−1and hence the sequence X0∩U ⊂ X1∩U ⊂ ... ⊂ Xn∩U is distinct. i−1∩Xi̸= ∅ is an i−1̸= ∅. So Xi∩U is i−1∩Xi) = (Xi∩U)∩Xc (c) Give an example of a topological space X and a dense open subset U with dimU < dimX. Let X = 0,1 the topology of X is defined by given the open sets: ∅,X,U = {0}. Then U is dense in X while dimX = 2,dimU = 1. (d)If Y is a closed subset of an irreducible finite-dimensional topologi- cal space X, and if dimY = dimX then Y = X. 7

8. Proof. If Y ̸= X, and dimY = n, then for any sequence of distinct closed irreducible subsets: Y0 ⊂ Y1 ⊂ ... ⊂ Yn, Y0 ⊂ Y1 ⊂ ... ⊂ Yn ⊂ X is a sequence of disdinct closed irreducible subsets of X, so dimX > dimY , a contradiction. (e) Give an example of a noetherian topological space of infinite dimen- sion. The cofinite topology in any infinite set. 1.11. Let Y ⊂ A3be the curve given parametrically by x = t3,y = t4,z = t5. Show that I(Y ) is a prime ideal of height 2 in k[x,y,z] which cannot be gen- erated by 2 elements. We say Y is not a local complete intersection–cf.(Ex. 2.17). Proof. Considerating the following ring homomorphism from k[x,y,z] to k[t]. f(x,y,z) 7−→ f(t3,t4,t5) The kernel of this homomorphism is precisely I(Y ), so A(Y ) is isomorphic to a subring of k[x], hence the transcendence degree of its quotient field (as a intermediate filed of k and k(x) is 0 or 1. However, t3is not in k, so the transcendence degree is 1, thus the height of I(Y ) is 2. Suppose that I(Y ) is generated by 2 elements, f and g,we need to deduce a contradiction. First we need to prove that for any polynomial h ∈ I(Y ), the coefficient of x,x2,y,z are all equal to zero. Note that is obvious that h has no constant term. i) Denote h by h = ax+x2h1+yh2+zh3, then from h(t3,t4,t5) = 0 we know that at3+ t6h1+ t4h2+ t5h3= 0, the only terms of t3appears at the first so a = 0. ii) For h = ax2+ x3h1+ yh2+ zh3, we have at6+ t9h1+ t4h2+ t5h3= 0, and hence a = 0. Note that we use the same symble because we use them as disposable. iii) h = ay+y2h1+xh2+zh3, then at4+t8h1+t3h2+t5h3= 0, hence a = 0. iv) h = az+z2h1+xh2+yh3, then at5+t10h1+t3h2+t4h3= 0, hence a = 0. Then, since it is obvious that x3− yz,y2− xz,z2− x2y are all in I(Y ), we have  u v    [f ] x3− yz y2− xz z2− x2y p q r  ·   s = g 8

9. where p,q,r,s,u,v ∈ k[x]. Suppose that the coefficients of terms x3,y2,z2in f and g are respectively fx,fy,fzand gx,gy,gz, and the constant term of p,q,r,s,u,v are respectively p0,q0,r0,s0,u0,v0. It easy to verify that  u0 v0    [fx fy ] p0 r0 q0 s0 1 0 0 0 1 0 0 0 1 fz gz  ·   = gx gy which is a contradiction. 1.12. Give an example of an irreducible polynomial f ∈ R[x,y], whose zero set Z(f) in A2 Ris not irreducible (cf. 1.4.2). f = (x2− 1)2+ y2. 9