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 {Ui}i≥1 is an ascending chain of open subsets of Y , by the definition of in- duced topology, there exists {Vi}i≥1 open subsets of X such that

Vi ∩ Y = Ui for all i. Note that {Vi}i≥1 may not be ascending, but let i Wi = ∪k=1Vk.

Then {Wi}i≥1 is clearly ascending and i i i Wi ∩ Y = (∪k=1Vk) ∩ Y = ∪k=1(Vk ∩ Y ) = ∪k=1Uk = Ui.

Therefore {Wi}i≥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. {Ui}i≥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) = {α ∈ A|f(α) ∈ J} 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 ∈ J, there exists an element α ∈ I such that f(α) = β. Then α can be written as a linear combination

α = a1α1 + a2α2 + . . . anαn

for some ak ∈ 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 ∈ In, n n+1 n f (a) = 0. Then, f (a) = f(f (a)) = f(0) = 0 implies a ∈ In+1. Therefore In ⊂ In+1. (ii) We check that f is injective. Because A is Noetherian and {In}n≥1 is an ascend- ing chain of ideals of A, there exists m such that

Im = Im+1.

Let a ∈ A be in the kernel of f. Because f is surjective, there exists a1 ∈ A and m f(a1) = a. Inductively we can find {ai}i=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 ∈ Im+1. But Im = Im+1, so am ∈ 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 ∈ A[X]. Since φ(a) = a for every a ∈ A, φ is surjective. Because A[X] is Noetherian and φ is surjective, by Problem 2 (i), A is also Noetherian.