Algebraic Number Theory, PS 2

Bryden Cais

December 10, 2002

1. Let K be a local field of characteristic 0 with residue field k, and let | ∗ | be the normalized valuation of K. Since K is local, k is finite, say #k = q, and hence has characteristic p for some prime p. The image of ℤ in O_{K} then maps to ℤ/(p) in the residue field. An integer n maps to 0 if and only if p|n, moreover, we have p = π^{e} and hence the power of π dividing the image of n in O_{K} is equal to re, where r is the power of p dividing n in ℤ. Recall that | ∗ | is a nonarchimedian valuation, so that the infinite series

∑_{i=0}^{∞} α_{i}

converges if and only if

lim_{i→∞} α_{i} = 0.

We claim that the formal power series

log(1 + x) = x − x^{2}/2 + x^{3}/3 − ⋯
exp(x) = 1 + x + x^{2}/2! + x^{3}/3! + ⋯

converge precisely for |x| < 1 provided that e < p − 1. Indeed, we have

lim_{n→∞} |\frac{x^{n}}{n} | = lim_{n→∞} q^{−(n ord_{π}(x)−e ord_{p}(n))}.

But e ord_{p}(n) ≤ e log_{p}(n). Observe that

lim_{n→∞} n ord_{π}(x) − e ord_{p}(n) = ∞

if and only if ord_{π}(x) > 0, that is, if and only if |x| < 1.

Now it is elementary to determine that the power of p dividing n! is

d_{p}(n) = ∑_{i=0}^{∞} ⌊\frac{n}{p^{i}} ⌋ < ∑_{i=0}^{∞} \frac{n}{p^{i}} = \frac{n}{p − 1}.

It follows that

|x^{n}/n!| < q^{−(n ord_{π}(x)−en/(p−1))}.

But

n ord_{π}(x) − en/(p − 1) = n (ord_{π}(x) − \frac{e}{p − 1}) ,

which tends to ∞ for ord_{π}(x) > e/(p − 1). Supposing e < (p − 1), this shows that the series for exp(x) converges for |x| < 1.

Next, observe that as formal power series we have

log((1 + x)(1 + y)) = log(1 + x) + log(1 + y)  (1)
exp(x + y) = exp(x) exp(y)  (2)
log exp(z) = z  (3)
exp log(1 + z) = 1 + z,  (4)

with equality holding whenever all series in question converge.

Now let

U = {α ∈ O_{K} : |α − 1| < 1}

and

D = {α ∈ O_{K} : |α| < 1}.

Observe that U is a subgroup of O_{K}^{×} and every α ∈ U has the form 1+πx for some x ∈ O_{K}. Moreover, we see that D is an additive subgroup of O_{K} since | ∗ | is nonarchimedian so that if α, β ∈ D then |α + β| < max{|α|, |β|} < 1. In addition, it is clear that if α ∈ D then 1 + α ∈ U and conversely.

We know that the power series defining log(1 + x) converges for x ∈ D, or equivalently, log(x) gives a function

log : U → K.

In fact, for α ∈ U we have

| log(α)| = | log(1 + (α − 1))| = |(α − 1) − \frac{(α − 1)^{2}}{2} + ⋯ | ≤ max \{\frac{|α − 1|^{i}}{i} \} < 1

since |α − 1| < 1. Moreover, by (1), we see that log defines a homomorphism

log : U → D.

Similarly, the series defining exp(x) converges for x ∈ D and a quick calculation gives

| exp(x) − 1| = |∑_{k=1}^{∞} \frac{x^{k}}{k!} | ≤ max \{|\frac{x^{k}}{k!} |\} < 1

provides that ord_{π}(x) > e/(p − 1), so that the image of D under exp lies in U. Thus, by (2) we have a homomorphism

exp : D → U.

Finally, by identities (3) and (4) we see that exp and log are in fact inverses to one another, and hence that

log : U → D and exp : D → U

are mutually inverse algebraic and topological isomorphisms (since exp, log are both continuous functions). Since O_{K} ≃ D via the mapping x → πx we see that

O_{K} ≃ U ⊂ O_{K}^{×}

both algebraically and topologically via the map

x → exp(πx).

Now consider the map

ℤ_{p}^{×} → (ℤ/pℤ)^{×}

given by

x ↦ x mod p.

We have seen that this is a surjection, and the kernel of this map is obviously U. Moreover, this sequence is split, for we have the canonical map

a ↦ w(a),

where w is the Teichmüller character. In other words, since x^{p−1} − 1 splits completely over 𝔽_{p}, (and since it is a separable polynomial), Hensel’s lemma tells us that ℤ_{p} contains the p − 1 st roots of unity, and that modulo p, these map bijectively to 𝔽_{p}. Our splitting homomorphism then maps a ∈ 𝔽_{p} to the unique (p − 1)^{st} root of unity that is congruent to a mod p. It follows that

ℤ_{p}^{×} ≃ (ℤ/pℤ)^{×} × U.

Now (ℤ/pℤ)^{×} ≃ ℤ/(p − 1) (by picking a generator for (ℤ/pℤ)^{×}), while we have shown above that U ≃ ℤ_{p}, with ℤ_{p} considered additively. It follows that

ℤ_{p}^{×} ≃ ℤ/(p − 1) × ℤ_{p}.

2. Let K be a local field of characteristic p. Then we have seen that K is isomorphic to 𝔽_{q}((t)), where q = p^{r} for some r and that π_{K} = t and O_{k}^{×} is the group of formal power series with coefficients in 𝔽_{q} having nonzero constant term. Recall that x → x^{p} is a homomorphism of K. We claim that the index of (O_{K}^{×})^{p} in O_{K}^{×} is infinite. For we can realize O_{K}^{×} as the inverse limit

lim_{\overleftarrow{m}}(𝔽_{q}[t]/t^{m})^{×}

(with the obvious projection maps π_{ij} : (𝔽_{q}[t]/t^{i})^{×} → (𝔽_{q}[t]/t^{j})^{×} for i > j given by reduction modulo t^{j}). Thus, it is enough to show that the index of

((𝔽_{q}[t]/t^{m})^{×})^{p} = (𝔽_{q}[t^{p}]/t^{m})^{×}

in (𝔽_{q}[t]/t^{m})^{×} grows without bound as m → ∞. However, this is easy: The size of (𝔽_{q}[t]/t^{m})^{×} is

(q − 1)q^{m−1}

(since we are considering polynomials with nonzero constant term of maximum degree m − 1), while the size of (𝔽_{q}[t^{p}]/t^{m})^{×} is clearly

(q − 1)q^{⌊(m−1)/p⌋}.

It follows that the index of (𝔽_{q}[t^{p}]/t^{m})^{×} in (𝔽_{q}[t]/t^{m})^{×} is

q^{m−1−⌊(m−1)/p⌋} → ∞

as m → ∞. Our claim follows.

3. The proof given in class that a local field of characteristic 0 has only finitely many extensions breaks down in characteristic p > 0 for the following reason: in characteristic p, not every irreducible polynomial is separable (consider, for example, the polynomial X^{p} −t ∈ 𝔽_{p}((t))[X]). Let us explore where this fundamental fact causes problems in the proof. We set up a correspondence C between the points of

M = 𝔪_{K} × 𝔪_{K} × ⋯ × 𝔪_{K} × (𝔪_{K} \ 𝔪_{K}^{2})

and the Eisenstein polynomials of degree n (here, n is the number of copies of 𝔪_{K}), by mapping a point a = (a_{0}, …, a_{n}) ∈ M to the polynomial C(a) := a_{0}x^{n} + ⋯ + a_{n}. Clearly M is compact since it is a finite product of compact sets. We then used Krasner’s lemma to construct an open set U_{a} about each point a ∈ M such that for every x ∈ U_{a}, the fields cut out by the roots of C(x) are the same as the fields cut out by the roots of C(a). However, Krasner’s lemma crucially requires that the polynomial C(a) be separable in order to construct the open set U_{a} as above. In characteristic 0, this separability is a consequence of irreducibility. As noted above, this is false in characteristic p. Thus, in characteristic p we in general cannot construct an open cover of M. Put differently, the set N ⊂ M of Eisenstein polynomials about which there is an open set V such that every polynomial in V generates the same field extensions is not compact since it is not even closed in M. Observe that N is just the set of separable Eisenstein polynomials, which is the complement of the set N_{s} of non-separable Eisenstein polynomials in M. The set N_{s} is clearly closed since f ∈ N_{s} if and only if the resultant of C(f) vanishes. (So we can view N_{s} as a subvariety of M).

4. Let | ∗ | be a nontrivial valuation on ℚ. We distinguish two cases:

Case 1: There exists some positive integer n such that |n| > 1. Let n_{0} be the least such positive integer. Then since |n_{0}| > 1, there exists some positive real number r such that |n_{0}| = n_{0}^{r}. Let n ∈ ℕ. Then we claim that |n| = n^{r}. We proceed as follows: writing

n = a_{0} + a_{1}n_{0} + ⋯ + a_{k}n_{0}^{k}  (5)

with 0 ≤ a_{i} < n_{0} we have

|n| ≤ |a_{0}| + |a_{1}||n_{0}| + ⋯ + |a_{k}||n_{0}|^{k}

≤ 1 + n_{0}^{r} + … + n_{0}^{rk}

≤ n_{0}^{rk} ∑_{j=0}^{∞} n_{0}^{−j}

≤ Cn^{r}

for some positive constant C independent of n, where we have used the fact that |a_{i}| ≤ 1 since a_{i} < n_{0} and the obvious fact that n_{0} > 1. It follows that for every integer N > 0 we have

|n^{N}| = |n|^{N} ≤ Cn^{rN},

so that

|n| ≤ C^{1/N}n^{r}.

Taking N arbitrarily large shows that

|n| ≤ n^{r}  (6)

for all positive integers n. On the other hand, we obviously have n_{0}^{k+1} > n ≥ n_{0}^{k} by (5), so that

|n| = |n + n_{0}^{k+1} − n_{0}^{k+1}|

≥ |n_{0}^{k+1}| − |n_{0}^{k+1} − n|

= n_{0}^{(k+1)r} − |n_{0}^{k+1} − n|

≥ n_{0}^{(k+1)r} − (n_{0}^{k+1} − n)^{r}

≥ n_{0}^{(k+1)r} (1 − (1 − \frac{n}{n_{0}^{k+1}} )^{r})

≥ n_{0}^{(k+1)r} (1 − (1 − \frac{1}{n_{0}} )^{r}) ,

where we have used the inequality (6) on the third line and the fact that n > n_{0}^{k} on the penultimate line. Setting

D = 1 − (1 − \frac{1}{n_{0}} )^{r} ,

we see that

|n| ≥ Dn^{r}

where D is independent of n. Using the same trick as above, we have

|n|^{N} ≥ Dn^{Nr}

for all positive integers N and hence that

|n| ≥ n^{r}.  (7)

Combining inequalities (6) and (7) we find that

|n| = n^{r}  (8)

for all positive integers n. It follows that |x| = |x|_{0}^{r} for all x in ℚ, where | ∗ |_{0} is the usual absolute value on ℝ. Since ℚ is dense in ℝ, we then have

| ∗ | = | ∗ |_{0}^{r},

so that | ∗ | is equivalent to | ∗ |_{0}.

Case 2: For every positive integer n, one has |n| ≤ 1. Since | ∗ | is nontrivial, there exists some positive integer n with |n| < 1. Let n_{0} be the least positive integer such that |n_{0}| < 1. Then n_{0} must be prime. For if not, say n_{0} = ab with a, b > 1, then we have |n_{0}| = |a||b| = 1 since a, b < n_{0}. Now let p ≠ n_{0} be any prime. Then for any positive integers j, k there exist integers u, v (depending, of course on j, k) such that

p^{j}u + n_{0}^{k}v = 1.

It follows that

1 = |p^{j}u + n_{0}^{k}v|
≤ |p|^{j}|u| + |n_{0}|^{k}|v|
≤ |p|^{j} + |n_{0}|^{k}.

If |p| < 1 then we may choose j, k sufficiently large so that |p|^{j} + |n_{0}|^{k} < 1, a contradiction. It follows that |p| = 1 for all primes p ≠ n_{0}. Therefore, for any integer n we have

|n| = |n_{0}|^{ord_{p}(n)},

where ord_{n_{0}}(n) denoted the highest power of n_{0} dividing n. As before, this means that

|x| = |x|^{ord_{n_{0}} (x)},  (9)

for all x ∈ ℚ, where we have extended ord_{n_{0}} to ℚ in the usual way. Let r = log |n_{0}|/ log n_{0}. Then we have

|x| = n_{0}^{r ord_{n_{0}} (x)} = |x|_{n_{0}}^{r},

where | ∗ |_{n_{0}} is the usual n_{0}-adic valuation on ℚ. In short, this shows that | ∗ | is equivalent to | ∗ |_{n_{0}} and we are done.

5. Let K be a local field equipped with a valuation v and π a uniformizer of K. Let ṽ be an extension of

v to K^{sep}. Suppose that

f(X) = X^{n} + a_{1}X^{n−1} + ⋯ + a_{n}

with a_{i} ∈ K^{sep} is a monic polynomial with roots

α_{1}, α_{2}, …, α_{n} ∈ K^{sep}

numbered so that we have

ṽ(α_{1}) = ṽ(α_{2}) = … = ṽ(α_{l_{1}}) = m_{1}
ṽ(α_{l_{1}+1}) = ṽ(α_{l_{1}+2})= … = ṽ(α_{l_{2}}) = m_{2}
⋮  ⋮
ṽ(α_{l_{r}+1}) = ṽ(α_{l_{r}+2})= … = ṽ(α_{n}) = m_{r+1}

and

m_{1} < m_{2} < m_{3} < … < m_{r+1}.

Recall that we have

a_{i} = s_{i}(α_{1}, …, α_{n}),

where s_{i} is the i^{th} symmetric polynomial in the α_{j}. We then have

v(a_{1}) = ṽ(s_{1}(α_{1}, …, α_{n})) ≥ min{ṽ(α_{i_{1}})}  = m_{1}
v(a_{2}) = ṽ(s_{2}(α_{1}, …, α_{n})) ≥ min{ṽ(α_{i_{1}}α_{i_{2}})}  = 2m_{1}
⋮  ⋮
v(a_{l_{1}−1}) = ṽ(s_{l_{1}−1}(α_{1}, …, α_{n})) ≥ min{ṽ(α_{i_{1}} ⋯ α_{i_{l_{1}−1}})}= (l_{1} − 1)m_{1}

while since ṽ(α_{1} ⋯ α_{l_{1}}) is strictly less than ṽ(α_{i_{1}} ⋯ α_{i_{l}}) for any other set {i_{1}, …, i_{l}}, it follows that in fact

v(a_{l_{1}}) = min{ṽ(α_{i_{1}} ⋯ α_{i_{l_{1}}} )} = l_{1}m_{1}

Continuing in this manner, we have

v(a_{l_{1}+1}) = ṽ(s_{l_{1}+1}(α_{1}, …, α_{n})) ≥ min{ṽ(α_{i_{1}} ⋯ α_{i_{l_{1}+1}})}  = l_{1}m_{1} + m_{2}
v(a_{l_{1}+2}) = ṽ(s_{l_{1}+2}(α_{1}, …, α_{n})) ≥ min{ṽ(α_{i_{1}} ⋯ α_{i_{l_{1}+2}})}  = l_{1}m_{1} + 2m_{2}
⋮  ⋮
v(a_{l_{2}−1}) = ṽ(s_{l_{2}−1}(α_{1}, …, α_{n})) ≥ min{ṽ(α_{i_{1}} ⋯ α_{i_{l_{2}−1}})}= l_{1}m_{1} + (l_{2} − l_{1} − 1)m_{1}

while again, since ṽ(α_{1} ⋯ α_{l_{2}}) is strictly less than ṽ(α_{i_{1}} ⋯ α_{i_{2}}) for any other set {i_{1}, …, i_{2}}, it follows that we have

v(a_{l_{2}}) = min{ṽ(α_{i_{1}} ⋯ α_{i_{l_{2}}} )} = l_{1}m_{1} + (l_{2} − l_{1})m_{2}.

We therefore see that the Newton polygon of f has precisely the vertices

(0, 0), (l_{1}, l_{1}m_{1}), (l_{2}, l_{1}m_{1} + (l_{2} − l_{1})m_{2}), …, (l_{r+1}, l_{1}m_{1} + (l_{2} − l_{1})m_{2} + ⋯ + (l_{r+1} − l_{r})m_{r+1}).

Observe that the j^{th} line segment in the Newton polygon is

\frac{(l_{1}m_{1} + (l_{2} − l_{1})m_{2} + ⋯ + (l_{j} − l_{j−1})m_{j}) − (l_{1}m_{1} + (l_{2} − l_{1})m_{2} + ⋯ + (l_{j−1} − l_{j−2})m_{j−1})}{l_{j} − l_{j−1}} = m_{j}

and that the x length of this segment is l_{j} − l_{j−1}. Thus we see that the Newton polygon of f contains a line segment of x-length m and slope s if and only if f has precisely m roots with valuation s.

6. Let L/K be a finite separable extension of local fields with corresponding residue field extension l/k.

Recall that

ϕ : I(L/K) → l^{×}

defined by

ϕ(σ) = a_{σ}  (10)

where

σ(π_{L}) = (a_{σ} + a_{1}π_{L} + ⋯)π_{L}.  (11)

induces an injection

I^{t}(L/K) ↪ l^{×}.

We claim that this can be an isomorphism. Consider, for example, the extension

L/K = ℚ_{p}(ζ_{p^{m}})/ℚ_{p},

where ζ_{p^{m}} is a primitive root of unity of order p^{m} and m > 1. We have seen that this is a totally ramified extension of degree

φ(p^{m}) = (p − 1)p^{m−1}

with Galois group (ℤ/p^{m}ℤ)^{×}. It is not difficult to see from this that the maximal tamely ramified extension of ℚ_{p} in ℚ_{p}(ζ_{p^{m}}) is K^{t} = ℚ_{p}(ζ_{p}); it is a totally tamely ramified extension of degree p − 1 with Galois group (ℤ/pℤ)^{×}. It follows that L/K^{t} is a totally wildly ramified extension of degree p^{m−1}. We therefore have

#I(L/K) = # Gal(L/K^{ur})  = # Gal(L/K) = (p − 1)p^{m−1}
#I_{p}(L/K) = # Gal(L/K^{t})  = [ℚ_{p}(ζ_{p^{m}}) : ℚ_{p}(ζ_{p})] = p^{m−1}
#I^{t}(L/K) = #I(L/K)/I_{p}(L/K)= (p − 1)p^{m−1}/p^{m−1} = (p − 1).

Since L/K it totally ramified, we have [l : k] = 1 and hence l = k = 𝔽_{p}. Now I^{t}(L/K) injects into 𝔽_{p}, but since both have cardinality p − 1, we have a surjection. It follows that

ϕ : I^{t}(ℚ_{p}(ζ_{p^{m}})/ℚ_{p}) ⟶ 𝔽_{p}

is an isomorphism (since we have seen in class that it is a homomorphism).

Next, using (10) and (11) we find

στ(π_{L}) = (a_{στ} + ⋯)π_{L}
= σ((a_{τ} + ⋯)π_{L})
= (σ(a_{τ}) + cdots)σ(π_{L})
= (σ(a_{τ}) + ⋯)(a_{σ} + ⋯)π_{L}

so that

a_{στ} = σ(a_{τ})a_{σ}.  (12)

Using (12) we find

a_{τστ^{−1}} = τσ(a_{τ^{−1}})a_{τσ}
= τσ(a_{τ^{−1}})τ(a_{σ})a_{τ} ,

but we are now viewing τ, σ as elements of Gal(l/k) since a_{σ}, a_{τ} etc. are elements of l. Since Gal(l/k) is abelian (l, k are finite fields) we have

a_{τστ^{−1}} = στ(a_{τ^{−1}})τ(a_{σ})a_{τ} .  (13)

Now observe that by (12) we have

a_{ττ^{−1}} = τ(a_{τ^{−1}})a_{τ}

= a_{id} = 1, so that

τa_{τ^{−1}} = \frac{1}{a_{τ}} .

Substituting this relation in (13), we have

a_{τστ^{−1}} = \frac{a_{τ}}{σa_{τ}} τ(a_{σ}).  (14)

Thus, if σ ∈ I(L/K) = Gal(L/K^{ur}) then σ fixes K^{ur}, and since

Gal(K^{ur}/K) ≃ Gal(l/k),

we see that σ projects to the identity element of Gal(l/k), and hence fixes a_{τ} ∈ l. Thus, if σ ∈ I(L/K) and if τ ∈ Gal(L/K) projects to a Frobenius element, then we have

a_{τστ^{−1}} = \frac{a_{τ}}{σa_{τ}} τ(a_{σ})  (15)
= τ(a_{σ})  (16)
= a_{σ}^{#k},  (17)

as required.

7. Let K be a local field of characteristic 0 and let T be the maximal abelian, tamely ramified extension of K. Then certainly T ⊃ K^{ur} since K^{ur}/K as abelian. Since T/K^{ur} is a tamely ramified extension, we have

I_{p}(T/K^{ur}) = {1},

so that

I(T/K^{ur}) = I^{t}(T/K^{ur})

and we have (as in problem 6) an injection

ϕ : I(T/K^{ur}) ↪ (k^{sep})^{×},

where k^{sep}/k is the associated residue field extension of T/K^{ur}. Now there is some τ ∈ Gal(T/K^{ur}) that projects to the Frobenius automorphism of Gal(k^{sep}/k). Since Gal(T/K^{ur}) is abelian, problem 6 tells us that we have

ϕ(τστ^{−1}) = ϕ(σ) = a_{σ}^{#k} = α_{σ}

for any σ ∈ I(T/K^{ur}). We conclude that in fact

ϕ : I(T/K^{ur}) ↪ k^{×}.

This at any rate gives

deg(T/K^{ur}) ≤ #k.

How to obtain equality?

Well, in the special case that k = 𝔽_{p}, we know that K^{ur}(ζ_{p})/K^{ur} is a totally tamely ramified abelian extension of degree p−1, where ζ_{p} is a primitive p^{th} root of unity. This shows that deg(T/K^{ur}) ≥ p−1. I have been trying to find a similarly explicit argument for the general case, but to no avail yet.

8. Let K be a local field of characteristic 0 and K̄ an algebraic closure of K. Suppose further that the residue field k of K has characteristic p. Let ζ_{n} be a primitive n^{th} root of unity and let m_{n} = n!p^{− ord_{p}(n!)}; that is, m_{n} is n! with the highest power of p removed. Set

a_{N} = ∑_{n=1}^{N} ζ_{m_{n}}π_{K}^{m_{n}}.  (18)

Obviously {a_{i}} forms a Cauchy sequence. Suppose that this sequence converges to some

α = ∑_{n=1}^{∞} ζ_{m_{n}}π_{K}^{m_{n}} ∈ K̄.

We will obtain a contradiction. Let σ ∈ Gal(K̄/K). We claim that for (i, p) = 1 we have |ζ_{i} − σ(ζ_{i})| = 1. For

β = ∏_{σ(z_{i})≠z_{i}}(ζ_{i} − σ(ζ_{i}))

divides

∏_{\begin{matrix}j|i \\ σ(ζ_{i})≠ζ_{i} \end{matrix}}(ζ_{i} − σ(ζ_{j})) = \frac{d}{dx}(x^{i} − 1)|_{x=ζ_{i}} = \frac{i}{ζ_{i}} ,

and since (p, i) = 1 we see that |\frac{i}{ζ_{i}} | = 1 and since β ∈ O_{K}, we must have |β| = 1. It follows that

|σ(ζ_{i}) − ζ_{i}| = 1.

Since

σ(a_{N}) − a_{N} = ∑_{n=1}^{N}(σ(ζ_{m_{n}}) − ζ_{m_{n}})π_{K}^{m_{n}}

for any σ ∈ Gal(K̄/K), we see that since |(σ(ζ_{m_{n}}) − ζ_{m_{n}})π_{K}^{m_{n}}| = |π_{K}|^{−m_{n}} if σ does not fix ζ_{m_{n}} and since |α + β| = min{|α|, |β|} if |α| ≠ |β|, we must have

|σ(a_{N}) − a_{N}| = |π_{K}|^{−m_{n}},

where n is the smallest positive integer j such that σ(ζ_{m_{j}}) ≠ ζ_{m_{j}}. In particular, if σ(ζ_{m_{N}}) ≠ ζ_{m_{N}} then n ≤ N whence |σ(a_{N}) − a_{N}| ≠ 0 so that σ(a_{N}) ≠ a_{N}. Since K(a_{N}) ⊆ K(ζ_{m_{N}}) by our construction of a_{N}, and since any σ ∈ Gal(K(ζ_{m_{N}})/K) acts nontrivially on a_{n} we in fact have equality. We have seen that the degree of K(ζ_{m_{N}}) is f, where f is the smallest positive integer such that q^{f} ≡ 1 mod m_{N}. It is clear that as m_{N} → ∞ we also have f → ∞. Now we use Krasner’s lemma to show that K(a_{N}) ⊆ K(α). Indeed, by our discussion above, we clearly have

|a_{N} − α| = |π_{K}|^{−m_{N+1}}.

On the other hand, if σ ∈ Gal(K̄/K) does not fix a_{N} then since K(a_{N}) = K(ζ_{m_{N}}) it cannot fix ζ_{m_{N}}, whence |a_{N} − σ(a_{N})| ≥ |π_{K}|^{−m_{N}}. It follows that K(a_{N}) ⊆ K(α) so that K(α)/K has degree at least as large as the degree of K(a_{N})/K. But we have shown above that as N → ∞, the degree of K(a_{N})/K grows without bound. Hence, α ∉ K̄ since otherwise the degree of K(α)/K would be finite. It follows that K̄ is not complete.

9. Let K be a local field with valuation v and L an extension of K with valuation w extending v. Suppose that L/K has ramification index e so that |π_{L}|^{e} = |π_{K}|. It follows that the index of the valuation ring w(L) in the ring v(K) is e. Thus, if L/K is any unramified extension, we see that

w(L) = v(K).

It follows that the valuation ring of K^{ur} is that of K since the valuation ring of any finite subextension L/K of K^{ur}/K is v(K). Since v(K) is discrete, it follows that the extension of v to K^{ur} is discrete. Observe that we are critically using the fact that K^{ur}/K is unramified. For example, even though any finite extension L/K has discrete valuation ring (if v(K) is discrete) the extension K̄/K, for example, does not have discrete valuation ring. Now we claim that K^{ur} is not complete. Indeed, if (n, p) = 1 (where p is the characteristic of K/π_{K}) we see that x^{n} − 1 is separable over the residue field of K so that by Hensel’s lemma, the irreducible polynomial of ζ_{n} (a primitive n^{th} root of unity), which divides x^{n} − 1 and is hence separable, remains irreducible over the residue field. It follows that the extension l/k of the residue fields of the extension K(ζ_{n})/K has the same degree as that of K(ζ_{n})/K so that K(ζ_{n})/K is unramified. (Indeed, we can say more: since the residue field of K^{ur}/K is the separable closure of k, which is generated by adjoining the roots of x^{n} − 1 for all (n, p) = 1 to k—this follows directly from the theory of finite fields—we see that K^{ur}/K is generated by adjoining all roots of unity of order prime to p). Thus, the cauchy sequence (18) is actually a cauchy sequence in K^{ur}. Since K^{ur} ⊂ K̄ and the sequence (18) does not even converge to a limit in K̄ (c.f. problem 8), we see that K^{ur} is not complete.

10. Let K = ℚ(\sqrt{n}, \sqrt{m}) for m, n distinct and square free (and not equal to 1). A rational prime p cannot be totally inert in K. Recall from the previous problem set that a prime p splits in ℚ(\sqrt{n}) if and only if x^{2} − n splits modulo p, in other words, if and only if n is a square modulo p. However, K has exactly three quadratic subfields—namely ℚ(\sqrt{n}), ℚ(\sqrt{m}), ℚ(\sqrt{mn})— and if m, n are non-squared modulo p then mn is a square modulo p (similarly, if m, mn are non-squares, then m is a square modulo p and so forth). This is elementary. It follows that if p is inert in any two of the quadratic subfields of K then it splits (or ramifies) in the third. Thus no rational prime can be totally inert in K since it must already split in a subfield. Observe that K/ℚ is a Galois extension with Galois group ℤ/(2)×ℤ/(2). We thus have the fundamental identity

efd = 4,

where e = e(℘|p), f = f(℘|p), and d is the number of primes lying over p (here ℘|(p) in O_{K}). Moreover, how p splits in the three quadratic subfields of K completely determines how p splits in K. We have seen that a rational prime p cannot be totally inert in K. However, our argument also shows that p either splits in exactly one of the three quadratic subfields or it splits in all three (in terms of the Legendre symbol, the options are −1 · −1 = 1 or 1 · 1 = 1 or −1 · 1 = −1; in any case there are either one or three 1’s). Moreover, a prime p ≠ 2 ramifies in ℚ(\sqrt{n}) if and only if p|n. Since p divides one of m, n if and only if p|mn, we see that a prime p ramifies in any quadratic subfield, it ramifies in two of them. Moreover, since m, n are square free, p cannot ramify in all three subextensions since if m, n contain p as a factor then mn contains p as a square factor and hence p does not divide the discriminant of ℚ(\sqrt{mn}) (similarly if m, mn have a square-free factor of p then p |̸n etc.). Finally, if p ≠ 2 ramifies in any extension, it ramifies in two, and hence must split in the third. If p = 2, the behavior is different: for example, it is possible for p to ramify in all three extensions. Similar criteria can be worked out for p = 2, but this is routine so we omit it here.

Let K_{1}, K_{2}, K_{3} be the three quadratic subfields of K. We thus have the following table of data and examples (up to permutation of the labelling of K_{i}):

K_{1}  K_{2}  K_{3}  K  m n p
splits  inert inert  p = 𝔭_{1}𝔭_{2}  2 -1 5
ramifies ramifies splits  p = 𝔭_{1}^{2}𝔭_{2}^{2}  5 -1 5
splits  splits splits p = 𝔭_{1}𝔭_{2}𝔭_{3}𝔭_{4} 3 -1 13

11. We claim that it is possible to find f ∈ ℤ[x] that is irreducible in ℚ[x] but reducible in ℚ_{p}[x] for every p. First observe that problem 10 gives us an easy way to construct an f ∈ ℤ[x] that is irreducible over ℚ but reducible over 𝔽_{p} for every p. In particular, let f be the field polynomial of the extension ℚ(\sqrt{m}, \sqrt{n}). Certainly f is irreducible of degree 4 over ℚ (since m, n are distinct and square free, not equal to 1). However, as we showed in problem 10, no rational prime p is inert in ℚ(\sqrt{m}, \sqrt{n}), from which it follows that f is reducible mod p for every p (since

ℤ[X]/(f, p) = 𝔽_{p}[x]/(f)

is not an integral domain since p is reducible in ℤ[x]/f, so that f is reducible in 𝔽_{p}[x].)

To construct f ∈ ℤ[x] that is irreducible over ℚ but reducible over ℚ_{p} for every p we proceed as follows. Recall that for p odd, ℚ_{p} has three distinct quadratic extensions, while ℚ_{2} has seven. In particular, the seven quadratic extensions of ℚ_{2} are precisely the seven quadratic subfields of

K_{2} := ℚ_{2}(\sqrt{−1}, \sqrt{2}, \sqrt{5}).

It follows that \sqrt{3} ∈ K_{2} and hence that K_{2}(\sqrt{3})/ℚ_{2} has degree 8. Moreover, since ℚ_{p} for p > 3 has only 3 distinct quadratic extensions, the degree of

ℚ_{p}(\sqrt{−1}, \sqrt{2}, \sqrt{3}, \sqrt{5})/ℚ_{p}

is at most 4.

Now Let f be the field polynomial of ℚ(\sqrt{−1}, \sqrt{2}, \sqrt{3}, \sqrt{5}). A short computation with PARI tells us that

f(x) = x^{16} − 72x^{14} + 1932x^{12} − 22552x^{10} + 154038x^{8} − 582456x^{6} + 1440748x^{4} − 1486824x^{2} + 3721041.

Since f is the field polynomial of a degree 16 extension (and of degree 16), it is certainly irreducible over ℚ. However, if f were irreducible over ℚ_{p} for some p then the degree of the field ℚ_{p}(α)/ℚ_{p} would be 16; but we have just seen that all the roots of f are contained in a field extension of ℚ_{p} of degree at most 8. Thus, f(x) is reducible in ℚ_{p}[x] for every p but irreducible in ℚ[x].