Math 160a - Problem Set 2 Evan Dummit

2 Math 160a - Problem Set 2 solutions.nb
3.17a. One has f HQ » 2L = f HQ » PL ÿ f HP » 2L. Since f HQ » 2L = 11 as 2 11 ª 1 mod 23, and f HP » 2L § 2 = @K : �D, it follows that
f HQ » PL must also be 11 since it is an integer.
3.17b. The minimal polynomial of q is f HxL = x 2 - x + 6 so the minimal polynomial of q -2 is f Hx + 2L = x 2 + 3 x + 8, so
NHq -2L = 8. Therefore, the principal ideal Hq -2L is the product of 3 primes lying over 2. It is clearly divisible by P, since it
contains q-2 e P. Then it cannot be divisible by P' = H2, q-1L, the other prime ideal lying above 2, as then it would contain P',
implying that Hq - 2L - Hq - 1L = 1 e P3 which is not true. Thus all three primes in the product must be P so that Hq - 2L = P3 .
If P were principal -- say, with P = HaL, then taking norms would imply that NHaL = 2, but there are no elements of K of norm 2 -explicitly,
NJa + bJ1 + -23 Ní2N = a2 + ab+ 6 b2 ¥ Ha + b ê 2L 2 + 5 b2 ; the latter is greater than 2 for positive b, and if b = 0
then NHaL = a 2 which can also not equal 2. Thus P is nonprincipal.
3.17c. If Q were principal -- say, Q = HbL -- then its norm in K would also be principal, hence
N LêKHQL = N LêKHH bLL = IN LêKHbLM = P11 by part (a). Then P = IP3M -7 ÿ IP11M 2 would be principal, since P3 is principal by part (b).
But P is not principal, contradiction. Hence Q is also not principal.
3.17d. If 2 =ab in L then H2L = HaLHbL. But H2L = H2, qL H2, q - 1L so since H2, qL»H2L = HaL HbL and H2, qL is prime, H2, qL»HaL or
H2, qL»HbL. Similarly H2, q-1L»HaL or H2, q-1L»HbL. Hence either one of HaL or HbL equals H2, qL -- impossible since H2, qL = Q is
not principal, by part (c) -- or both H2, qL and H2, q-1L contain HaL or HbL -- suppose it is HaL. Then HaL Œ H2, qL›H2, q-1L = H2L
so that HaLHbL ΠH2L with equality possible only if HbL = L and HaL = H2L, so that b is a unit.
3.30a. If f H0L = 0 then the result is obviously true so suppose f H0L ∫ 0, and define gHxL = f Hx ÿ f H0LL ê f H0L; note gH0L = 1. Suppose
that the set S of primes for which f had a root mod pi e S were finite; then the same would also hold for g. So let
q = maxHSL. For any polynomial hHxL, hHaL ª hHbL mod p if a ª b mod p. Thus, for all k e �, gHk ÿ q!L ª f H0L ª 1 mod pi for each
pi e S. Since gHk ÿ q!L can have no prime other divisors other than the pi by assumption, it follows that gHk ÿ q!L = 1 for all k e �.
Then the polynomial gHx ÿ q !L - 1 has infinitely many roots hence must be constant, so g and hence also f is constant. So the
only polynomials for which the set S is finite are constants.
3.30b. Let K = �HaL where a is scaled so as to be an algebraic integer, and let f be the minimal polynomial of a. By part (a)
there exist infinitely many primes pi such that f HrL ª 0 mod pi for some r. Then take p to be any such prime such that p does not
divide n = @�K : �@aDD; since the latter index is finite there are still infinitely many such p. We claim that the ideal P = Ha -r, pL
is prime in �K of degree 1.
To see this, first note that �K ê P @ �@aD ê P' where P' = �@aD›P -- since �K ê �@aD is a finite abelian group of order n,
Lagrange's theorem applied to the element 1 implies n e �@aD. Then as gcdHp, nL = 1, p e P, and n e �@aD it follows
P + �@aD = �k. The second isomorphism theorem then implies �K ê P @ �@aD ê HP › �@aDL = �@aD ê P'.
Then �K ê Xa -r, p\ @ �@aD ê Xa -r, p\ @ �@xD ê H f HxL, x - r, pL = �@xD ê Hx - r, pL @ � ê p �, so P is prime of index p hence of
degree 1, as required. Thus, for each such P, f HP » pL = 1; there are infinitely many choices for the prime p, so there are infinitely
many P. (Note that this construction of prime ideals is the same one used in problem 3.13.)
3.30c. Applying the result from (b) to K = �HzmL where zm is a primitive mth root of unity implies that there are infinitely many
primes of inertial degree 1 in K. But by theorem 26, the degree f of a prime p in �HzmL not dividing m is the smallest positive f
such that p f ª 1 mod m. Since there are only finitely many primes p dividing m, it follows that there are infinitely many primes
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