Math 160a - Problem Set 2 Evan Dummit

Math 160a - Problem Set 2
Evan Dummit
3.9a. Clearly it suffices to show that vPHIL § vPHJL for any prime ideal P in R, for factoring into prime ideals will imply I » J. So
fix a prime P in R and suppose PS = P1 e1 ÿÿÿPg eg for primes P1, …,Pg e S. Then we may write
IS= HPa ÿ IxL S = P1 e1 a
ÿÿÿPg eg a
HIx SL where Ix is relatively prime to P (so that Ix S is relatively prime to the Pi); similarly we
may write JS= IPb JxM S = P1 e1 b
ÿÿÿPg eg b
HJx SL where Jx is relatively prime to P (so that Jx S is relatively prime to the Pi). Then
since IS» JS it follows that e1 a § e1 b so a § b and hence vPHIL § vPHJL as claimed.
3.9b. Let J = IS› R. Then clearly I Œ J so J » I in R. Also, by the definition J SŒ IS. Thus I S» JS hence by (a), I » J so
I = J.
3.9c. Let K Œ S; then if K = HK › RL S then K is the extension of some ideal (namely, I = K › R) in R to S. Conversely, if
K = IS for some I Œ R then HK › RL S = HI S› RL S = IS= K by part (b), so every such extension ideal satisfies the property.
Thus, the ideals K with this property are precisely those of the form K = IS for some ideal I ΠR.
3.11. Suppose aeI; then J = HaL is a principal ideal contained in I so the third isomorphism theorem implies that
HR ê JL ê HI ê JL @ HR ê IL. Taking degrees yields †R ê J§ ê †I ê J§ = †R ê I§ hence †R ê I§ = ∞I¥ divides †R ê J§ = ∞J¥. But by theorem 22c,
∞J¥ = N K HaL since J is principal, so ∞I¥ divides N K HaL for all aeI, which shows the first part. Also, equality holds if and only if
†I ê J§ = 1; i.e., when I = J = HaL, which shows the second part.
3.13. Let A = H23, a-10L 2 ÿ H23, a-3L. Then A is generated by the elements x1 = 23 3 , x2 = 23 2 Ha -3L, x3 = 23 2 Ha -10L,
x4 = 23 Ha -3L Ha -10L, x5 = 23 Ha -10L 2 , and x6 = Ha -10L 2 Ha -3L; each generator is clearly inside 23 S -- the last following
from x6 = Ha -10L 2 Ha -3L =a3- 23 a2 + 160 a - 300 = 23 I-a2 + 7 a - 13M. Thus A Π23 S.
Conversely, it is straightforward if somewhat boring to verify that 23 2 =-3 x1 + 10 x2 - 10 x3 and that
23 - 23 2 ÿ 88 = 221 x4 - 102 x5 + 119 ÿ x6 hence 23 = 264 x1 + 880 x2 - 880 x3 + 221 x4 - 102 x5 + 119 ÿ x6, so 23 S Œ A. Hence
23 S = A.
Also, 1 = 7 Ha -3L - 7 Ha -10L - 3 H23L so H23, a-10, a-3L = S and thus H23, a-10L and H23, a-3L are relatively prime ideals.
Equivalently, one could do this with less computation: for f HxL = x 3 - x - 1, �@aD @ �@xD ê H f HxLL, so
�@aD ê H23L = H�@xD ê H f HxLLL ê H23 �@xD ê H f HxLLL @ �@xD ê H23, f HxLL @ �23@xD ë I f HxLM , where the bar denotes projection into �23@xD
(i.e., f HxL is f HxL viewed as a polynomial with coefficients in �23). The Chinese Remainder Theorem implies that the latter is
isomorphic to the direct product ¤�23@xD ê Hx -aiL, where ai runs through the roots of f . Since the discriminant of f is
-23 ª 0 mod 23, f has a multiple root; taking the gcd with f ' HxL = 3 x2 - 1 shows that the multiple root is 10 hence that the other
root is 3. So �@aD ê H23L @ �23@xDëHx - 10L 2 μ �23@xD ê Hx - 3L. Rewriting these in a manner similar to the above shows that
�23@xD ê Hx - 10L @ �@aD ê H23, a-10L and �23@xD ê Hx - 3L @ �@aD ê H23, a-3L; the two ideals are then prime since the quotients are
just isomorphic to the field �23. The result in the problem above then follows from the isomorphisms given. Alternatively, this
could be used to derive NHH23, a-3LL = NHH23, a-10LL = 23 so since A Π23 S and NHAL = 23 3 = NHSL it would then follow that
A = S and that H23, a-10L and H23, a-3L were relatively prime, prime ideals.
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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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p ª 1 mod m.
Math 160a - Problem Set 2 solutions.nb 3
3.30d. Let M be the Galois closure of L over �. Then let a be an algebraic integer such that M = �HaL, let f be the minimal
polynomial of a, and let p be a prime not dividing n = @�M : �@aDD, not dividing the discriminant of f HxL, and such that
f HrL = 0 mod p for some r; by part (a) the set of such primes is infinite. Since M is Galois, f splits completely into a product of
distinct linear factors in M , yielding n distinct prime ideals over p. Thus each corresponding prime ideal P in �K above p splits
completely in �L (i.e., has g = @L : KD), as desired.
3.30e. Let f e R@xD be nonconstant, monic, and irreducible, and let L = RHaL for some root a of f HxL. Then, for all primes not
dividing @�L : �@aDD, the Chinese Remainder Theorem implies that gHxL splits mod P the same way as P splits in �L. By part (d),
there exist infinitely many primes P which split completely, hence for each of those primes not dividing @�L : �@aDD, gHxL also
splits completely.
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