Math 145 - Solution Set 1 Problem 1. 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). Check that any subset Y ⊂ X is also Noetherian in the subspace topology. Answer: If fUigi≥1 is an ascending chain of open subsets of Y , by the definition of in- duced topology, 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 Noetherian, 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. Problem 2. (i) Show that if A is a Noetherian ring and f : A ! B is a surjective ring homomorphism, then B is also Noetherian. (ii) Show that if I is any ideal in a polynomial ring k[X1;X2;:::;Xn], then the quo- tient ring k[X1;X2;:::;Xn]=I is Noetherian. (Terminology: such quotients are called finitely generated k-algebras.) Answer: (i) If J is an ideal in B, we will show that J is finitely generated. It is easy to check that I := f −1(J) = fα 2 Ajf(α) 2 Jg is an ideal in A. Since A is Noetherian, I is finitely generated. Pick a set of generators α1; α2; : : : αn. To complete the proof, we will show that f(α1); : : : ; f(αn) generate J. Indeed, for all β 2 J, there exists an element α 2 I such that f(α) = β: Then α can be written as a linear combination α = a1α1 + a2α2 + : : : anαn for some ak 2 A. Then β = f(α) = f(a1α1 + a2α2 + : : : anαn) = f(a1)f(α1) + f(a2)f(α2) + : : : f(an)f(αn): Therefore J is spanned by f(α1); f(α2); : : : ; f(αn). (ii) By the Hilbert Basis Theorem, k[X1;:::;Xn] is Noetherian. The ring homomor- phism k[X1;:::;Xn] ! k[X1;:::;Xn]=I sends elements of I is surjective. By (ii), k[X1;:::;Xn]=I must be Noetherian. 1 2 Problem 3. Let A be a Noetherian ring and f : A ! A a ring homomorphism. (i) Show that the kernels In of fn = f ◦ f ◦ · · · ◦ f form an ascending chain of ideals of A. (ii) Show that if f is surjective then f is bijective. Answer. (i) Since In is the kernel of the morphism fn, it must be an ideal. If a 2 In, n n+1 n f (a) = 0. Then, f (a) = f(f (a)) = f(0) = 0 implies a 2 In+1. Therefore In ⊂ In+1. (ii) We check that f is injective. Because A is Noetherian and fIngn≥1 is an ascend- ing chain of ideals of A, there exists m such that Im = Im+1: Let a 2 A be in the kernel of f. Because f is surjective, there exists a1 2 A and m f(a1) = a. Inductively we can find faigi=1 ⊂ A and f(ai) = ai−1 m hence f (am) = a. Since f(a) = 0, m+1 m f (am) = f(f (am)) = f(a) = 0: Thus am 2 Im+1. But Im = Im+1, so am 2 Im: Therefore m a = f (am) = 0: This proves that f is injective. Thus f is bijective. Problem 4. If A[X] is Noetherian, is it true that A has to be Noetherian? Answer. Yes. Note the ring homomorphism φ : A[X] ! A; φ(f) = f(0) for every polynomial f 2 A[X]. Since φ(a) = a for every a 2 A, φ is surjective. Because A[X] is Noetherian and φ is surjective, by Problem 2 (i), A is also Noetherian. .