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Notes on class field theory

Kiran S. Kedlaya

Section 1.3 The local Kronecker-Weber theorem

Reference.

[59], Chapter 14.
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We now prove the local Kronecker-Weber theorem (Theorem 1.1.5), modulo some steps which will be left as exercises. As shown previously, this will imply the original Kronecker-Weber theorem (Theorem 1.1.2).
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Subsection Extensions of local fields

We first recall the following facts from the theory of local fields (e.g., see [38] II.7).
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Definition 1.3.1.

Let \(L/K\) be an extension of finite extensions of \(\QQ_p\text{.}\) Let \(\gotho_K, \gotho_L\) be the integral closures of \(\ZZ_p\) in \(K, L\text{.}\) We say that \(L/K\) is unramified if the maximal ideal of \(\gotho_K\) generates the maximal ideal of \(\gotho_L\text{.}\) In other words, any element \(\pi\) of \(K\) which generates the maximal ideal of \(\gotho_K\) (i.e., any uniformizer of \(K\)) is also a uniformizer of \(L\text{.}\) In still other words, the condition is that the ramification index \(e(L/K)\) is equal to \(1\text{.}\)
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In general, there is a maximal subextension of \(L/K\) which is unramified. If this is \(K\) itself, we say that \(L/K\) is totally ramified.
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Let \(U\) be the maximal unramified subextension of \(L/K\text{.}\) We say that \(L/K\) is tamely ramified if the degree \([L:U]\) is not divisible by \(p\text{.}\) In other words, the condition is that \(e(L/K)\) is not divisible by \(p\text{.}\)
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Proof.

Choose \(u \in \gotho_L\) generating the residue field of \(L\) over the residue field of \(K\text{,}\) and let \(P(x)\) be the minimal polynomial of \(u\) over \(K\text{.}\) Then over the residue field of \(K\text{,}\) \(P(x)\) divides the \((q-1)\)-st cyclotomic polynomial, so by Hensel’s lemma it splits over \(K(\zeta_{q-1})\text{.}\) Hence \(L \subseteq K(\zeta_{q-1})\text{,}\) and equality follows by comparing degrees.
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Proof.

Let \(\pi_L\) be a uniformizer of \(L\text{.}\) Then \(\pi_L^e\) can be written as a product of a uniformizer \(\pi\) of \(K\) times an element \(u\) of \(\gotho_L\) congruent to 1 modulo \(\pi_L\text{.}\) By Hensel’s lemma, \(u\) has an \(e\)-th root in \(L\text{,}\) as then does \(\pi\text{.}\)
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Subsection Proof of local Kronecker-Weber

We now proceed to the proof of Theorem 1.3.4, modulo some lemmas which we fill in later.
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Proof.

Since \(\Gal(K/\QQ_p)\) decomposes into a product of cyclic groups of prime-power order, by the structure theorem for finite abelian groups we may write \(K\) as the compositum of extensions of \(\QQ_p\) whose Galois groups are cyclic of prime-power order. In other words, it suffices to prove local Kronecker-Weber under the assumption that \(\Gal(K/\QQ_p) \cong \ZZ/q^r \ZZ\) for some prime \(q\) and some positive integer \(r\text{.}\) We split this discussion into three cases; see Lemma 1.3.5, Lemma 1.3.6, and Lemma 1.3.7.
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Proof.

To begin, note that \(\QQ_p(\zeta_p)/\QQ_p\) is totally tamely ramified of degree \(p-1\text{,}\) so by Lemma 1.3.3 it has the form \(\QQ_p(c^{1/(p-1)})\) for some \(c \in p \ZZ_p^*\text{.}\) (The value of \(c\) won’t be critical here, but see Lemma 1.3.8 for later reference.)
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Let \(L\) be the maximal unramified subextension of \(K\text{.}\) By Lemma 1.3.2, \(L = \QQ_p(\zeta_n)\) for some \(n\text{.}\) Let \(e := [K:L]\text{.}\) Since \(e\) is a power of \(q\text{,}\) \(e\) is not divisible by \(p\text{,}\) so \(K\) is totally and tamely ramified over \(L\text{.}\) Thus by Lemma 1.3.3, there exists \(\pi \in L\) generating the maximal ideal of \(\gotho_L\) such that \(K = L(\pi^{1/e})\text{.}\) Since \(L/\QQ_p\) is unramified, \(p\) also generates the maximal ideal of \(\gotho_L\text{,}\) so we can write \(\pi = cu\) for some unit \(u \in \gotho_L^*\text{.}\) Now \(L(u^{1/e})/L\) is unramified since \(e\) is prime to \(p\) and \(u\) is a unit. In particular, \(L(u^{1/e})/\QQ_p\) is unramified, hence abelian. Then \(K(u^{1/e})/\QQ_p\) is the compositum of the two abelian extensions \(K/\QQ_p\) and \(L(u^{1/e})/\QQ_p\text{,}\) so it’s also abelian. Hence any subextension is abelian, in particular \(\QQ_p(c^{1/e})/\QQ_p\text{.}\)
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For \(\QQ_p(c^{1/e})/\QQ_p\) to be Galois, it must contain the \(e\)-th roots of unity (since it must contain all of the \(e\)-th roots of \(-p\text{,}\) and we can divide one by another to get an \(e\)-th root of unity). But \(\QQ_p(c^{1/e})/\QQ_p\) is totally ramified, whereas \(\QQ_p(\zeta_e)/\QQ_p\) is unramified. This is a contradiction unless \(\QQ_p(\zeta_e)\) is actually equal to \(\QQ_p\text{,}\) which only happens if \(e|(p-1)\) (since the residue field \(\FF_p\) of \(\QQ_p\) contains only \((p-1)\)-st roots of unity).
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Now \(K \subseteq L(c^{1/e}, u^{1/e})\) as noted above. But on one hand, \(L(u^{1/e})\) is unramified over \(L\text{,}\) so \(L(u^{1/e}) = L(\zeta_m)\) for some \(m\text{;}\) on the other hand, because \(e|(p-1)\text{,}\) we have \(\QQ_p(c^{1/e}) \subseteq \QQ_p(c^{1/(p-1)}) = \QQ_p(\zeta_p)\text{.}\) Putting it all together,
\begin{equation*} K \subseteq L(c^{1/e}, u^{1/e}) \subseteq \QQ_p(\zeta_n, \zeta_p, \zeta_m) \subseteq \QQ_p(\zeta_{mnp})\text{.} \end{equation*}
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Proof.

Suppose \(\Gal(K/\QQ_p) \cong \ZZ/p^r\ZZ\text{.}\) We can use roots of unity to construct two other extensions of \(\QQ_p\) with this Galois group. Namely, \(\QQ_{p}(\zeta_{p^{p^r}-1})/\QQ_p\) is unramified of degree \(p^r\text{,}\) and automatically has cyclic Galois group; meanwhile, the index \(p-1\) subfield of \(\QQ_p(\zeta_{p^{r+1}})\) is totally ramified with Galois group \(\ZZ/p^r\ZZ\text{.}\) By assumption, \(K\) is not contained in the compositum of these two fields, so for some \(s>0\text{,}\)
\begin{equation*} \Gal(K(\zeta_{p^{p^r}-1}, \zeta_{p^{r+1}})/\QQ_p) \cong (\ZZ/p^r\ZZ)^2 \times \ZZ/p^s \ZZ \times \ZZ/(p-1)\ZZ\text{.} \end{equation*}
This group admits \((\ZZ/p\ZZ)^3\) as a quotient, so we have an extension of \(\QQ_p\) with Galois group \((\ZZ/p\ZZ)^3\text{.}\) We rule this out using Lemma 1.3.9.
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Proof.

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Subsection Filling in the details

We now return to the lemmas that we skipped over in the proof of Theorem 1.3.4. At this point, we make heavy use of Kummer theory.
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Proof.

Proof.

For convenience, put \(\pi := \zeta_p - 1\text{.}\) Then \(\pi\) is a uniformizer of \(\QQ_p(\zeta_p)\text{.}\)
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If \(\Gal(K/\QQ_p) \cong (\ZZ/p\ZZ)^3\text{,}\) then \(\Gal(K(\zeta_p)/\QQ_p(\zeta_p)) \cong (\ZZ/p\ZZ)^3\) as well, and \(K(\zeta_p)\) is abelian over \(\QQ_p\) with Galois group \((\ZZ/p\ZZ)^* \times (\ZZ/p\ZZ)^3\text{.}\) Applying Kummer theory to \(K(\zeta_p)/\QQ_p(\zeta_p)\) produces a subgroup \(B \subseteq \QQ_p(\zeta_p)^*/(\QQ_p(\zeta_p)^*)^p\) isomorphic to \((\ZZ/p\ZZ)^3\) such that \(K(\zeta_p) = \QQ_p(\zeta_p, B^{1/p})\text{.}\) Let \(\omega\colon \Gal(\QQ_p(\zeta_p)/\QQ_p) \to (\ZZ/p\ZZ)^*\) be the canonical map; since \(\QQ_p(\zeta_p, b^{1/p}) \subseteq K(\zeta_p)\) is also abelian over \(\QQ_p\text{,}\) by Lemma 1.2.11,
\begin{equation*} b^g/b^{\omega(g)} \in (\QQ_p(\zeta_p)^*)^p \qquad (\forall b \in B, g \in \Gal(\QQ_p(\zeta_p)/\QQ_p))\text{.} \end{equation*}
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Recall the structure of \(\QQ_p(\zeta_p)^*\text{:}\) the maximal ideal of \(\ZZ_p[\zeta_p]\) is generated by \(\pi\text{,}\) while each unit of \(\ZZ_p[\zeta_p]\) is congruent to a \((p-1)\)-st root of unity modulo \(\pi\text{,}\) and so
\begin{equation*} \QQ_p(\zeta_p)^* = \pi^\ZZ \times (\zeta_{p-1})^\ZZ \times U_1\text{,} \end{equation*}
where \(U_1\) denotes the set of units of \(\ZZ_p[\zeta_p]\) congruent to 1 modulo \(\pi\text{.}\) Correspondingly,
\begin{equation*} (\QQ_p(\zeta_p)^*)^p = \pi^{p\ZZ} \times (\zeta_{p-1})^{p\ZZ} \times U_1^p\text{.} \end{equation*}
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Now choose a representative \(a \in L^*\) of some nonzero element of \(B\text{;}\) without loss of generality, we may assume \(a = \pi^m u\) for some \(m \in \ZZ\) and \(u \in U_1\text{.}\) Then
\begin{equation*} \frac{a^g}{a^{\omega(g)}} = \frac{(\zeta_p^{\omega(g)}-1)^m}{\pi^{m\omega(g)}} \frac{u^g}{u^{\omega(g)}}\text{;} \end{equation*}
but \(v_\pi(\pi) = v_\pi(\zeta_p^{\omega(g)}-1) = 1\text{.}\) Thus the valuation of the right hand side is \(m(1-\omega(g))\text{,}\) which can only be a multiple of \(p\) for all \(g\) if \(m \equiv 0 \pmod{p}\text{.}\) (Notice we just used that \(p\) is odd!) That is, we could have taken \(m=0\) and \(a = u \in U_1\text{.}\)
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As for \(u^g/u^{\omega(g)}\text{,}\) note that \(U_1^p\) is precisely the set of units congruent to 1 modulo \(\pi^{p+1}\) (see Exercise 2). Since \(\zeta_p = 1 + \pi + O(\pi^2)\text{,}\) we can write \(u = \zeta_p^b(1 + c\pi^d + O(\pi^{d+1}))\text{,}\) with \(c \in \ZZ\) and \(d \geq 2\text{.}\) Since \(\pi^g/\pi \equiv \omega(g) \pmod{\pi}\text{,}\) we get
\begin{align*} u^g &= \zeta_p^{b\omega(g)} (1 + c \omega(g)^d \pi^d + O(\pi^{d+1})),\\ u^{\omega(g)} &= \zeta_p^{b\omega(g)} (1 + c \omega(g) \pi^d + O(\pi^{d+1}))\text{.} \end{align*}
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But these two have to be congruent modulo \(\pi^{p+1}\text{.}\) Thus either \(d \geq p+1\) or \(d \equiv 1 \pmod{p-1}\text{,}\) the latter only occurring for \(d=p\text{.}\)
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What this means is that the set of possible \(u\) is generated by \(\zeta_p\) and by \(1 + \pi^p\text{.}\) But these only generate a subgroup of \(U_1/U_1^p\) isomorphic to \((\ZZ/p\ZZ)^2\text{,}\) whereas \(B \cong (\ZZ/p\ZZ)^3\text{.}\) Contradiction.
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Exercises Exercises

1.

Hint.
Prove that \((\zeta_p-1)^{p-1}/p - 1\) belongs to the maximal ideal of \(\ZZ_p[\zeta_p]\text{.}\)
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2.

Prove that (in the notation of Lemma 1.3.9) \(U_1^p\) is the set of units congruent to 1 modulo \(\pi^{p+1}\text{.}\)
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Hint.
In one direction, write \(u \in U_1\) as a power of \(\zeta_p\) times a unit congruent to 1 modulo \(\pi^2\text{.}\) In the other direction, use the binomial series for \((1+x)^{1/p}\text{.}\) (See Exercise 1 for a generalization of this result.)
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3.

Prove that for any \(r>0\text{,}\) there is an extension of \(\QQ_2\) with Galois group \(\ZZ/2\ZZ \times (\ZZ/2^r\ZZ)^2\) contained in \(\QQ_2(\zeta_n)\) for some \(n>0\text{.}\)
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Hint.
Consider \(L = \mathbb{Q}_2(\zeta_{2^{r+1}}, \zeta_{2^r-1})\text{.}\)
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4.

Suppose that \(K/\QQ_2\) is a \(\ZZ/2^r\ZZ\)-extension not contained in \(\QQ_2(\zeta_n)\) for any \(n>0\text{.}\) Prove that there exists an extension of \(\QQ_2\) with Galois group \((\ZZ/2\ZZ)^4\) or \((\ZZ/4\ZZ)^3\text{.}\)
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Hint.
Compare \(K\) with its compositum with some field \(L\) as in Exercise 3. Use the structure of finite abelian groups to show that if \(LK \neq L\text{,}\) then \(\Gal(LK/\mathbb{Q}_2)\) is forced to have a quotient of the specified form.
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5.

Prove that there is no extension of \(\QQ_2\) with Galois group \((\ZZ/2\ZZ)^4\text{.}\)
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Hint.
Use Kummer theory to show that every quadratic extension of \(\QQ_2\) is contained in \(\QQ_2(\zeta_{24})\text{.}\)
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6.

Prove that there is no extension of \(\QQ_2\) with Galois group \((\ZZ/4\ZZ)^3\text{.}\)
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Hint.
Reduce to showing that there exists no extension of \(\QQ_2\) containing \(\QQ_2(\sqrt{-1})\) with Galois group \(\ZZ/4\ZZ\text{.}\) Prove this by following the argument of Lemma 1.3.9.
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