Math 254A Corrigenda

Problem Set 10 Corrections

• Cyclotomic Fields 3: The hint should state that every prime divisor of phi_n(m) either divides n or is congruent to 1 modulo n.
• Fermat 2: retain the assumption from Problem 1, that x, y and x^p+y^p are not divisible by p.

Problem Set 9 Corrections

• Ramification in Galois Extensions 2: replace "b is a prime ideal which is unramified over L" with "b is a prime ideal of L which is unramified over K".
• Ramification in Galois Extensions 3: Replace the second sentence with "If the prime ideal p of K does not ramify in L (that is, every prime above it is unramified over K), and has two prime factors in L of inertia degree 1, then p is totally split." Also, the degree of a permutation group is the number of objects it is acting on.
• Ramification in Galois Extensions 4: "p is totally split" means every prime above p has ramification index and inertia degree equal to 1.
• Ramification in Galois Extensions 6 and 7: n should be prime (else the group theory won't work out) and P should be monic.

Problem Set 8 Comments/Corrections

• Extension of Valuations 1: if it's less confusing, take the field to be Q[x]/(x^n - 2). (The point is, it's an "abstract" number field, without a fixed choice of a particular n-th root of 2 in the complex numbers.)
• Relative Discriminant 3: replace Q[x]/(P(x)) by Q(\sqrt{D})[x]/(P(x)). Also, assume that P is still irreducible when viewed as a polynomial over Q(\sqrt{D}) (so that the new thing I wrote down is still a field).
• Relative Discriminant 4: p is supposed to be prime. (We'll discuss the general case when we do cyclotomic fields, probably next week.)

Problem Set 7 Comments

• Local Fields 2: this one is worked out completely in the text. To make it more interesting, I recommend that you drop the assumption that n is not divisible by p. (All the results you need will still be available in Neukirch, but the exact statement is not there.)
• Local Fields 4: the exponential valuation on power series is given by the order of vanishing at 0. That is, if the first nonzero coefficient is the coefficient of z^m, then the series has exponentatial valuation m.

Problem Set 6 Corrections

• p-adic Numbers 5: x_n should be x_i.
• Completions 4: "finite extension" should be "finite unramified extension". Since this sentence is in parentheses anyway (and I haven't defined ramified or unramified), you may ignore it if you prefer.
• More on Valuations 2: replace Z by either Z_p or the localization of Z at p. Then interpret "integer" and "rational" accordingly. (E.g., if you chose Z_p, then "integer" becomes "p-adic integer" and "rational" becomes "p-adic number".)
• Completions 1: r_n-p should be r_n-r.
• Completions 5: make the same assumptions as in Completions 4.

Problem Set 5 Clarifications

• Dirichlet 3: The Magma function UnitGroup, when applied to a number field, actually returns two answers: an abstract abelian group isomorphic to the units group, and a homomorphism from that abstract group into the number field that maps to the units group. You can retrieve both answers by a statement like U, f := UnitGroup(K).
• Valuations 2: the sentence beginning with "Suppose" is not as clear as it ought to be. To clarify, replace the second clause with "but an odd number of sides of the triangle have one of their own vertices colored red and the other colored blue." The first clause remains as is; that is, no side of a triangle or the square can contain vertices of all three colors, including both its own vertices and those of other triangles.
• Valuations 4-5: you may modify the coloring if you prefer, as long as you can still deduce the final result. (I.e., you can change from a strict to a nonstrict inequality or vice versa, if that's how things work out for you. The coloring I gave will definitely work, though.)

  Also, Ulrich noted that the odd number can only be equal to 1. True enough. The reason I said "an odd number" is that the statement is still true if the square is replaced by an arbitrary convex polygon.

Problem Set 4 Corrections

• Title: should be Class Numbers, not Lattices.
• Quadratic Fields: in the definition of a binary quadratic form, assume that if $D < 0$, then $a > 0$. Also, remove the plus/minus sign in the definition of equivalence. And do not allow 1 to be a fundamental discriminant. (That way, the quadratic field $Q(\sqrt{D})$ is actually a quadratic field and not just the rationals.)
• Quadratic Fields 1 and 2: you may assume $D<0$, or you may assume $D>0$. For $D<0$, try proving that the inverse map is given as follows: given an ideal, choose a basis $a,b$ of the ideal as Z-module, which is positively oriented in a fixed embedding of the quadratic field. (That is, in the order $a, b$, they span a parallelogram of positive area. Or, the counterclockwise angle from $a$ to $b$ is less than 180 degrees.) Then show that some scalar times $f(x,y) = Norm(ax+by)$ is a binary quadratic form of the right discriminant.
• Quadratic Fields 3: the last inequality in the first sentence should be $-a < b \leq a$ (that is, remove the 2s). If you prefer, you may prove the inequality $|b| \leq |a| \leq |c|$ instead of the given ones.
• Quadratic Fields 4: assume that the reduced form also satisfies the optional clause of problem 3. The condition should read "if and only if $b=0$ or $a=b$". Also, "order 2" should be "order at most 2", since the order could also be 1. And even if you didn't do problems 1 or 2, you may use without proof that the inverse of $ax^2+bxy+cy^2$ is $ax^2-bxy+cy^2$.
• Quadratic Fields 5: should read "Prove that the subgroup of the class group of $Q(\sqrt{D})$ of elements of order at most 2 has order at most $2^n$, where $n$ is the number of prime divisors of $D$." Remember that we're assuming $D$ is a fundamental discriminant (see top of the section).
• Quadratic Fields 7: run only over $D$ prime; the statistics will be more interesting that way.

Problem Set 3 Comments/Corrections

• Lattices 2: First of all, let's make things simpler by assuming a,b,c are all odd. (The problem gets a bit messier otherwise.) Then the condition should state that there exists a solution to ax^2+by^2+cz^2==0 modulo |abc| such that gcd(x,y,c) = gcd(y,z,a) = gcd(z,x,b) = 1. And a hint: define L by imposing a congruence condition modulo each prime power dividing 2|abc|.
• Lattices 3: disregard the norm entirely if you wish. You can use the usual norm on $\RR^3$ instead.
• LLL problems: the inequality (2) should be a greater-than-or-equal, not a less-than-or-equal. Oops!

Problem Set 2 Comments

• CNT 2: This problem will not be counted for credit (it's too hard). Sorry about that.
• Ideals 3: I didn't get to the definition of the ideal class group in lecture. If you haven't read ahead to that, prove that the product of any two nonprincipal ideals in this ring is principal, or that the square of any nonprincipal ideal is principal.
• Ideals 5: Again, if you haven't read ahead to the definition of ideal classes, replace the second sentence with "Prove that for every ideal A, there exists an element x of the fraction field of the Dedekind domain such that xA is an integral ideal and xA + m is the unit ideal."

Problem Set 1 Corrections

• Gaussian Integers 5: you may invoke without proof the characterization of primes modulo which 2 is a square (or use quadratic reciprocity).
• Number Theory 1: read "mod 4" instead of "mod 8".
• Number Theory 3 and 4: the polynomial should be irreducible.
• Number Theory 5: n should be at least 5 (since the assertion breaks down for n equal to 2 or 4).