* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Solutions
Survey
Document related concepts
Birkhoff's representation theorem wikipedia , lookup
Factorization of polynomials over finite fields wikipedia , lookup
Gröbner basis wikipedia , lookup
Basis (linear algebra) wikipedia , lookup
Deligne–Lusztig theory wikipedia , lookup
Complexification (Lie group) wikipedia , lookup
Eisenstein's criterion wikipedia , lookup
Laws of Form wikipedia , lookup
Homomorphism wikipedia , lookup
Dedekind domain wikipedia , lookup
Polynomial ring wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Algebraic number field wikipedia , lookup
Transcript
Math 6421 Fall 2009 HW assignment #1 solutions 1. Let k be an algebraically closed field, and let Z ⊆ A3k be the set Z = {(t, t2 , t3 ) | t ∈ k}. Find generators for the ideal I(Z) in k[x, y, z], and show that the ring A(Z) = k[x, y, z]/I(Z) is isomorphic to a polynomial ring in one variable over k. (Z is called the twisted cubic curve in A3 .) Solution: Let J = (y − x2 , z − x3 ). It is clear that J ⊆ I(Z), since y − x2 and z − x3 both vanish identically on Z. For the other inclusion, take f = f (x, y, z) ∈ I(Z), and let g = g(x, y, z) = f (x, x2 , x3 ) ∈ k[x] ⊂ k[x, y, z]. Then f − g ∈ J, since y ≡ x2 and z ≡ x3 (mod J). Since f ∈ I(Z), we have f (t, t2 , t3 ) = 0 for all t ∈ k. Thus f (x, x2 , x3 ) is the zero polynomial in k[x], since it vanishes at every t ∈ k. (Note that k is infinite.) In other words, g = 0, and thus f ∈ J as desired. Now define φ : k[x, y, z] → k[t] by sending x to t, y to t2 , and z to t3 . Clearly φ is surjective, and by definition ker(φ) = I(Z), since f ∈ ker(φ) iff φ(f ) = f (t, t2 , t3 ) = 0 in k[t] iff f vanishes identically on Z. Thus A(Z) = k[x, y, z]/I(Z) ∼ = k[t]. Remark: It is not obvious that J is the kernel of φ, so if your proof of J = I(Z) was based on this observation, make sure that you justify it. 2. Let k be an algebraically closed field, let A be a√finitely generated kalgebra, and let I be an ideal in A. Prove that I is the intersection of all maximal ideals of A containing I. (Compare with Exercise 2(b) below.) Solution: Every finitely generated k-algebra A has the form k[x1 , . . . , xn ]/J for some ideal J. Also, ideals in k[x1 , . . . , xn ]/J correspond to ideals in k[x1 , . . . , xn ] containing J, and this correspondence preserves maximal ideals and radicals. So we may assume without loss of generality that A = k[x1 , . . . , xn ]. 1 Now, it is geometrically obvious that the algebraic set V(I) ⊂ Ank is the union of all points contained in it: [ V(I) = P . P ∈V(I) Taking I√of both sides, we obtain (using Hilbert’s Nullstellensatz to see that I = I(V(I)) and that points in Ank correspond to maximal ideals in k[x1 , . . . , xn ]): \ √ I = I(V(I)) = m. m⊇I 3. If Y ⊆ An is any subset, prove that V (I(Y )) = Y (the closure of Y in the Zariski topology). Solution: From the definitions of V and I, it is rather clear that we always have Y ⊆ V(I(Y )). Now assume V is an arbitrary algebraic set containing Y : V ⊇ Y . Then I(V ) ⊆ I(Y ) and V(I(V )) ⊇ V(I(Y )). Since V is an algebraic set, V = V(I(V )). So V ⊇ V(I(Y )). Therefore V(I(Y )) is the smallest algebraic set containing Y . 4. Let X be a topological space. Prove that the following are equivalent: (a) X is irreducible (i.e., X cannot be written as X = X1 ∪ X2 with X1 , X2 proper closed subsets of X). (b) Any two non-empty open subsets of X intersect. (c) Every non-empty open subset of X is dense in X. Solution: (a) implies (b): open subsets of X. T Let U1 and U2 be two S non-empty c c c Suppose U2 U2 = ∅. Then X = U1 U2 , and U1 and U2c are proper closed subsets of X, a contradiction. (b) implies (c): Recall that a subset of X is dense in X if and only if it meets all nonempty open subsets of X. Thus it follows from (b) that all non-empty open subsets of X is dense in X. 2 (c) implies (a): Suppose for contradiction that X is not irreducible. S Then we can write X as X = X1 X2 with X1 and X2 proper closed subsets of X. Then X1c ⊆ X2 , which means that X1c is a nonempty open subset of X whose closure is not X and thus X1c is not dense in X. 5. Let X be a Noetherian topological space. (a) Prove that X is quasicompact, i.e., every open cover has a finite subcover. (b) If X is Hausdorff, prove that X is a finite set with the discrete topology. (c) Prove or find a counterexample: X has finite Krull dimension. Solution: (a) Suppose for the sake of contradiction that there exists an open cover {Uα }α∈A of X with no finite subcovers. Then we can find a sequence S of non-empty open T sets {Uα1 , · · · , Uαn , · · · } such that Uαi+1 * j6i Uαj for all i. Let Yi = j6i Uαc j . Then Y1 ⊃ Y2 ⊃ · · · does not satisfy the descending chain condition for closed subsets. (b) Recall that we say X has the discrete topology if every subset of X is open (closed). Now let X be a Noetherian Hausdorff space. Suppose X has infinitely many points. Then pick two distinct points x1 and y1 , and we can find disjoint neighborhoods U1 3 x1 and V1 3 y1 . Either X \ U1 or X \ V1 is infinite. WLOG, we assume X \ U1 is infinite. Let U10 = U1 . Now X \ U1 is also Hausdorff and we can find U2 ⊆ XS\ U1 which is non-empty and open relative to X \ U1 such that X \ (U1T U2 ) is infinite. Then there exists an open subset U20 of X such that U20 (X \ U1 ) = U2 . In this way, we can continue S S to find Ui such that S UiSis nonempty, open relative to X \ (U1 · · · Ui−1 ) and X \ (U1 · · · Ui ) is infinite. Correspondingly, there exist open subsets Ui0 S of X such that T S S S 0 0 Ui (X \ (U1 · · · Ui−1 )) = Ui . Let Zi = U1 · · · Ui0 . Then by construction, Ui0 * Zi−1 , which means the ascending chain of open sets Z1 ⊂ Z2 ⊂ · · · never stabilizes, contradicting the assumption that X is Noetherian. Thus X is finite. And since every point of a Hausdorff space is closed, X has the discrete topology. (c) A Noetherian space can have infinite Krull dimension. Let us build an infinite dimensional Noetherian space. Let X be the unit interval 3 [0, 1]. Let X and ∅ be closed. Let the subintervals [1/n, 1] be closed for n ∈ Z+ . Then X is a Noetherian topological space and all closed subsets are irreducible. But now we get an ascending chain of distinct irreducible closed subsets of X as [1, 1] ⊂ [1/2, 1] ⊂ · · · ⊂ [1/n, 1] ⊂ · · · . It follows that X has infinite Krull dimension. 6. Let A be an integral domain with field of fractions K. Prove that the integral closure of A in K is the intersection of all valuation rings of K containing A. [Hint: Use the Extension Theorem.] Solution: It suffices to show that x ∈ K is integral over A iff x belongs to every valuation ring of K containing A. Suppose first that x is integral over A, and let B be a valuation ring of K containing A. Then either x ∈ B or x−1 ∈ B. If x ∈ B, we are done. If x−1 ∈ B, then by the integrality of x we have xn + a1 xn−1 + · · · + an−1 x + an = 0 with ai ∈ A, and thus x = −(a1 + a2 x−1 + · · · + an (x−1 )n−1 ) ∈ B as desired. Conversely, suppose x ∈ K is not integral over A, and let A0 = A[x−1 ]. Then x 6∈ A0 by an argument similar to the one above. In particular, x−1 is a non-unit in A0 , so is contained in some maximal ideal m0 of A0 . Let k be an algebraic closure of A0 /m0 . The natural map φ from A to A0 to A0 /m to k sends x−1 to 0, and by the Extension Theorem, φ can be extended to a homomorphism Φ : B → k for some valuation ring B of K containing A. As Φ(x−1 ) = 0, we must have x 6∈ B. 4 Supplementary commutative algebra exercises 1. (?) (Localization) Let A be a commutative ring with identity, and S a multiplicative subset of A. (a) If I is an ideal of A, prove that S −1 I (the image of I under the natural homomorphism from A to S −1 A) is a proper ideal iff I ∩ S = ∅. (b) Prove that every ideal of S −1 A has the form S −1 I for some ideal I of A. (c) Prove that there is a bijection between prime ideals of S −1 A and prime ideals of A disjoint from S. (d) If S = A\p with p a prime ideal of A, prove that Ap := S −1 A is a local ring, i.e., that it has a unique maximal ideal. Solution: See, e.g., pages 708,709, and 718 of [DF04] or pages 41,42 of [AM69]. 2. (?) (Radical of an ideal) Let A be a commutative ring with identity. (a) If f ∈ A is not nilpotent, prove that there is a prime ideal of A which does not contain f . [Hint: Since S = {1, f, f 2 , . . .} does not contain 0, the localization S −1 A = Af is not the zero ring, so it has a maximal (and hence prime) ideal.] √ (b) If I is an ideal in A, show that I is the intersection of all prime ideals of A containing I. [Hint: The image of f ∈ A in A/I is nilpotent iff f m ∈ I for some m ≥ 1.] Solution: See, e.g., page 71 of [Eis95]. For a proof of (b) without localization, see e.g. page 9 of [AM69]. 3. (?) (Integral closure) (a) Let A ⊆ B be rings, let C be the integral closure of A in B, and let S be a multiplicative subset of A. Prove that S −1 C is the integral closure of S −1 A in S −1 B. In particular, every localization of an integrally closed domain is again integrally closed. (b) Prove that a UFD is integrally closed. Solution: See page 61 of [AM69] and page 693 of [DF04]. 5 References [AM69] M. F. Atiyah and I. G. Macdonald. Introduction to commutative algebra. Addison-Wesley Publishing Co., Reading, Mass.-LondonDon Mills, Ont., 1969. [DF04] D. S. Dummit and R. M. Foote. Abstract algebra. John Wiley & Sons Inc., Hoboken, NJ, third edition, 2004. [Eis95] D Eisenbud. Commutative algebra, volume 150 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1995. With a view toward algebraic geometry. 6