Q: If R is an integrally closed domain, why is R[x] integrally closed? (PS2, problem 2a)

A: Helpful hints (stolen from Atiyah-Macdonald):

- Let K be the fraction field of R: use "transitivity of integral closure" to reduce to checking that R[x] is integrally closed in K[x].
- Show that if f,g in K[x] are monic and fg is in R[x], then f and g are both in R[x]. (Hint: pass from K to a splitting field for fg.)
- Handy trick: given f in K[x] satisfying an integrality relation over R[x] of some degree, then f + x^n satisfies an integrality relation of the same degree. Take n "large" and write the constant term of the new relation as (f+x^n) times (something else forced to be monic).

Q: How do I get started thinking about PS2, problem 6? (Related question: PS2, problem 2b.)

A: Try thinking about the ideal (x^2,y) in the ring C[x,y]_{(x,y)}: the point is that it is not a power of the maximal ideal M, because it is sandwiched between M and M^2.

Q: Let K be a number field. How do I compute its ring of integers R? (PS2, problem 3)

A: For individual elements of a number field, a good way to test integrality is to compute the minimal polynomial. This may however come out messy when you try to do it with coefficients (e.g., if you try to compute the minimal polynomial of a + b 2^(1/3) + c 2^(2/3) as a polynomial in a,b,c). It may be easiesr to use the following two facts:

- Let b_1, ..., b_n be a basis of K over Q consisting of elements of R. Then R is contained in the Z-span of the dual basis of b_1, ..., b_n for the dual pairing. (See Janusz I.6 or the notes from February 28.)
- R is a ring! That's actually surprisingly useful once you account for the previous item: for instance, if you have something in the Z-span of the dual basis, but its square is not in the span, it can't be in R.

Q: If p,q are primes congruent to 1 modulo 4, how do I find an element of the class group of Q(sqrt(-pq)) of order 2? (PS 5, problem 5a)

A: Construct a rational prime which splits as a product of two primes, but which is not the absolute norm of an element of the ring of integers. (You'll end up using some congruence conditions, and quadratic reciprocity. Oh, and Dirichlet's theorem on primes in arithmetic progressions.) Then each of the two primes lying above is nonprincipal; if you set this up right, the same argument will show that no odd power of these primes is principal either.

Q: How do I prove that any automorphism of Q_p is continuous (and hence trivial)? (PS 7, problem 9(a))

A: One approach is to prove that the set of elements of Q_p^* which have n-th roots for all n coprime to p is a neighborhood of 1. (Hint: analyze convergence of the binomial series.)

Q: Is it really true that a formal power series over Q_p and its derivative have the same radius of convergence? Doesn't the sum x^{p^n} give a counterexample? (PS 8, problem 3)

A: It's really true, as long as you remember that radius of convergence should be defined using the usual calculus formula. The point is that it tells you about convergence when you plug in values of x which lie in

Q: I've convinced myself that every totally tamely ramified extension of a finite extension K of Q_p has the form K(pi^{1/d}), where pi is some uniformizer. How do I use this to show that Q_p(zeta_p) = Q_p((-p)^{1/(p-1)})? (PS 9, problem 7)

A: Check that (1-zeta_p)^(p-1)/(-p) is a (p-1)-st power in Q_p(zeta_p). This amounts to checking that it has image 1 in the residue field, and then verifying (using a binomial series) that any element of Z_p[zeta_p] which is congruent to 1 modulo 1-zeta_p is a (p-1)-st power.

Q: If K is a field of characteristic coprime to n, L = K(zeta_n), and M = L(a^{1/n}) is a Kummer extension, when is M Galois over K? (PS 10, problem 5(b))

A: If and only if every conjugate of a has the form a^m x^n for some integer m coprime to n.

Q: On problem 3 of the final, I have this power series that converges in a certain radius. But how do I know that the limit is actually equal to the rational function that it's formally supposed to be equal to?

A: Because you can check that equality after multiplying by the