Solutions

Math 232 Solutions to problem set #3 ¯ p (the (1) Fix a prime p. For every n ≥ 1 let ζn be a primitive n-th root of unity in Q algebraic closure of Qp ), and let Kn = Qp (ζn ). (a) If p - n, show that OKn = Zp [ζn ]. The discriminant DKn /Qp (Z[ζn ]) = NKn /Qp f 0 (ζn ), where f (x) ∈ Zp [x] is the monic irreducible polynomial for ζn . (Note that f (x) is not in general the cyclotomic polynomial Φn (x) whose roots are all primitive n-th roots of unity, because Φn (x) is not usually irreducible over Qp . In particular [Kn : Qp ] can be less × than ϕ(n).) Since p - n and f (x) | xn − 1, we have f 0 (ζn ) | nζnn−1 ∈ OK . Therefore n DKn /Qp (Zp [ζn ]) = Zp ⊃ DKn /Qp (OKn ), so Zp [ζn ] = OKn . (b) If n is a power of p, show that ζn − 1 is a uniformizing parameter of Kn . (c) Deduce from (b) that if n is a power of p then OKn = Zp [ζn ]. Suppose n = pr . Let r

g(x) = ((x + 1)p − 1)/((x + 1)p

r−1

− 1) =

p−1 X r−1 (x + 1)ip ∈ Zp [x]. i=0

r r−1 xp /xp

r r−1 xp −p ,

Then modulo p we have g(x) ≡ = and g(0) = p, so g(x) is an Eisenstein polynomial. Hence Kn /Zp is totally ramified, the root π = ζn − 1 of g is a uniformizing parameter of Kn , and OKn = Zp [π] = Zp [ζn ]. (d) Deduce from (a) and (c) that for every n ≥ 1, OKn = Zp [ζn ]. Write n = mpr with p - n. Then Km /Zp is unramified, so p is a uniformizing parameter of Km . Hence the polynomial g(x) above is still an Eisenstein polynomial in OKm [x], so Kn /Km is totally ramified and ζpr − 1 is a uniformizing parameter of Kn . Since Kn /Km is totally ramified, their residue fields k are the same. It follows that OKm = Zp [ζm ] and ζpr − 1 generate OKn , so (since ζm , ζpr − 1 ∈ Zp [ζn ]) we have OKn = Zp [ζn ]. (2) For every prime p, exhibit 2 non-isomorphic ramified quadratic extensions of Qp . If p is odd let u be any integer that is not a square modulo p. If p = 2, let u = 5. Then √ √ Qp ( p) and Qp ( up) are (totally) ramified quadratic extensions of Qp . √ ∼ √ √ Suppose Qp ( p) = Qp ( up). Then up is a square in Qp ( p), so u is a square in √ √ Qp ( p). But Qp ( u)/Qp is unramified, and there is no nontrivial unramified extension of √ Qp in Qp ( p), so u is a square in Qp , so u is a square in Zp . But by our choice of u, u is not a square in Zp . √ √ Note: If p ≡ 1 (mod 4), then −1 is a square in Qp , so Qp ( p) = Qp ( −p). For √ every odd prime p, 4 − p is a square in Qp , so p2 − 4p is a square in Q( −p), so the splitting fields of x2 + p and x2 + px + p are the same! If you claimed otherwise, go check your proof.

2

(3) Suppose K is a local field, a ∈ K × , e ≥ 1, and L is the splitting field of f (x) := xe −a. Let α be a root of f . Show that the map σ 7→ σ(α)/α is an injective homomorphism from Gal(L/K(ζe )) to µe . Define φ(σ) = σ(α)/α. Then φ(σ)e = σ(α)e /αe = σ(αe )/αe = σ(a)/a = 1, so φ(σ) ∈ µe . If σ, τ ∈ Gal(L/K(ζe )) then φ(τ σ) = φ(στ ) = στ (α)/α = στ (α)/σ(α) · σ(α)/α = σ(φ(τ )) · φ(σ). Since σ is the identity on K(µe ), σ(φ(τ )) = φ(τ ) and so φ is a homomorphism. Finally, if σ ∈ ker(φ) then σ(α) = α. But L = K(ζe , α), so σ is the identity on L, i.e., σ = 1, Thus φ is injective. (4) Suppose L/K is a finite extension of fields. If α ∈ L, let φα : L → L denote the map multiplication by α. (a) Show that φα is a homomorphism of K-vector spaces. It is immediate that φα preserves addition and multiplication by scalars in K. L ¯ ¯ ∼ (b) Show that L ⊗K K K. (Hint: use that L ∼ = = K[x]/f (x) for an appro¯ σ:L,→K

priate polynomial f (x).) (c) Use (b) to show that X σ(α), trace(φα ) =

det(φα ) =

Y

σ(α).

¯ L,→K

¯ L,→K

Let L = K(β) for some β, and let f (x) ∈ K[x] be the monic irreducible polynomial for β. Then L ∼ = K[x]/f (x) (with an isomorphism that sends β to x) so ¯ ∼ ¯ = K[x]/f ¯ ¯ ¯ L⊗K (x) ∼ − σ(β)) ∼ = (K[x]/f (x)) ⊗ K =⊕ =⊕ ¯ K[x]/(x ¯ K. σ:L,→K

σ:L,→K

Note that this isomorphism sends g(β) ⊗ z to (. . . , σ(g(β))z, . . .) for every g(x) ∈ K[x] and ¯ Thus if we denote this isomorphism L ⊗ K ¯ → ⊕ ¯ by ρ, then we have a z ∈ K. ¯K σ:L,→K commutative diagram ρ /⊕ ¯ ¯ ¯K L⊗K σ:L,→K φα



¯ L⊗K

ρ



λ

/⊕ ¯ ¯K σ:L,→K

where λ is the map (. . . , zσ , . . .) 7→ (. . . , σ(α)zσ , . . .). In other words, with respect to the ¯ that has 1 in the σ component and 0 elsewhere, basis ρ−1 (eσ ), where eσ is the element of ⊕K the matrix for φα is diagonal with diagonal entries σ(α). Now (c) follows. ¯ ∼ ¯ n . But this Note: as vector spaces we trivially have L ∼ = K n , so L ⊗ K = K isomorphism is not compatible with the action of multiplication by L on both sides, so this isomorphism is not useful in computing trace(φα ) and det(φα ) (5) Use problem (1) to compute the different and discriminant of Qp (ζn )/Qp for every n.

3

Write n = mpr with p - m. Since p - m, Kn = Kpr (µm ) is unramified over Kpr , so DKn /Kpr = OKpr . Thus [K :Q ]

[K :Q ]

DKn /Qp = NKpr /Qp (DKn /Kpr )DK mr /Qpp = DK mr /Qpp . p

p

We computed DKpr /Qp in class, and got p

(pr−1 (pr−r−1)

DKn /Qp = p[Km :Qp

. Thus

](pr−1 (pr−r−1)

.

Note that [Km : Qp ] is not in general ϕ(m). (6) Compute the sequence of higher ramification groups G−1 ⊃ G0 ⊃ G1 ⊃ · · · for the √ extension Qp ( d)/Qp , for every d and every p. Pay special attention to the case p = 2. After dividing by a power of p2 , we may assume that either d = u or d = pu with u ∈ Z× p. √ Let K = Qp ( d), and let G = Gal(K/Qp ). Recall that G−1 = G, G0 is the inertia group, and G1 is the Sylow p-subgroup of G0 . √ Case 1: d = u is a square in Qp . Then Qp ( d) = Qp , so Gi = {1} for every i. Case 2: d√= u is not a square in Qp , and either p > 2 or p = 2 and d ≡ 1 (mod 4). In this case Qp ( d)/Qp is an unramified quadratic extension, so G−1 = G and Gi = {1} for i ≥ 0. √ Case 3: p > 2 and d = pu. In this case Qp ( d)/Qp is a tamely ramified quadratic extension, so G−1 = G0 = G and Gi = {1} for i ≥ 1. √ Case 4: p = 2 and d = u ≡ −1 (mod 4). √ In this case Q2 ( d)/Q2 is (wildly) ramified, so G−1 = G0 = G√ 1 = G, K = Z2 [ d]. Let τ be the nontrivial element of G. √ and Oi+1 Then τ ∈ G ⇔ τ ( d) − d ∈ p , where p is the√ prime√ of K (so p2 = 2OK ). Since i √ √ √ √ × τ ( d) − d = −2 d and d ∈ OK , we have ordp (τ ( d) − d) = 2 so τ ∈ G1 but τ ∈ / G2 . Hence Gi = {1} if i ≥ 2. √ √ Case 5: p = 2 and d = 2u. As in Case√4, OK = Z2 [ d] and τ ∈ Gi ⇔ −2 d ∈ pi+1 . We still have ordp (2) = 2, but now ordp ( d) = ord2 (d) = 1, so τ ∈ G2 but τ ∈ / G3 . Thus G−1 = G0 = G1 = G2 = G, and Gi = {1} if i ≥ 3. √ √ (7) Suppose p is a prime, p 6= 3. Let K = Q( 3 p). Show that OK = Z[ 3 p] and that the discriminant DK/Q = −33 p2 . √ The monic irreducible polynomial for 3 p is f (x) = x3 − p, so √ √ √ √ DK/Q (Z[ 3 p]) = −NK/Q f 0 ( 3 p) = −NK/Q (3 3 p2 ) = −27NK/Q ( 3 p)2 = −27p2 √ Thus we need only prove that OK = Z[ 3 p]. √ Let t = [OK : Z[ 3 p]]. Then DK/Q = −27p2 /t2 . Since p is ramified in K/Q (f is an Eisenstein polynomial at p), p divides DK/Q , so p - t. It turns out that this problem is false in general: if p ≡ ±1 (mod 9), then t = 3, √ DK/Q = −3p2 , and OK 6= Z[ 3 p]. Oops.