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Reference request: does anyone have a reference for studying the Galois theory of the extension $\mathbb{Q}_{p}/\mathbb{Q}$? I can't find anything on it. There must be something to say, as these are always non-trivial extensions.

Questions: is this extension Galois for every prime? I guess not for the infinite prime; so is it Galois at all finite primes? Say we fix a prime, $p$; What are the intermediate (finite) Galois extensions $\mathbb{Q}\subseteq E \subseteq \mathbb{Q}_{p}$?

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    $\begingroup$ $\mathbb{Q}_p/\mathbb{Q}$ is not an algebraic extension, so I'm not sure what you're looking for. $\endgroup$ Commented Feb 14, 2018 at 1:06
  • $\begingroup$ Since $\mathbb Q_p$ is not algebraic over $\mathbb Q$, one possible interpretation of this is to ask what algebraic extensions of $\mathbb Q$ imbed to $\mathbb Q_p$, in parallel to asking which extensions imbed to $\mathbb R$ (as opposed to $\mathbb C$). Is this what you might mean? $\endgroup$ Commented Feb 14, 2018 at 1:14
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    $\begingroup$ I suppose one could interpret your question as asking to describe the automorphisms of $\Bbb Q_p$ (all of which naturally leave rational numbers fixed). But like $\Bbb R$, the $p$-adic field $\Bbb Q_p$ has only the identity as automorphism. $\endgroup$
    – Lubin
    Commented Feb 14, 2018 at 5:05
  • $\begingroup$ Yeah, that was silly of me. All I am asking about is the finite algebraic extensions between $\mathbb{Q}_{p}$ and $\mathbb{Q}$. $\endgroup$ Commented Feb 14, 2018 at 9:56
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    $\begingroup$ Related: math.stackexchange.com/questions/449424/… $\endgroup$
    – Watson
    Commented Feb 14, 2018 at 20:17

3 Answers 3

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Let $E/\mathbb{Q}$ be a finite extension and let $\mathfrak{p}|p$ be unramified in the ring of integers of $E$. Then we have the extension $E_{\mathfrak{p}}/\mathbb{Q}_p$ and we can view the group $G(E_{\mathfrak{p}}/\mathbb{Q}_p)$ as a subgroup of $G(E/\mathbb{Q})$, and consists of the elements $\sigma$ so that $\sigma(\mathfrak{p})=\mathfrak{p}$. What we can do is use $G(E_{\mathfrak{p}}/\mathbb{Q}_p)$ as a subgroup of $G(E/\mathbb{Q})$ to, in a way, measure the "$p$-adicness" of the extension $E$. This is done by finding how much of $E$ is contained in $\mathbb{Q}_p$.

Define the field $E(\mathfrak{p})/\mathbb{Q}$ to be the fixed field of $G(E_{\mathfrak{p}}/\mathbb{Q}_p)$ when viewed as a subgroup of $G(E/\mathbb{Q})$. Then $E(\mathfrak{p})$ is the largest field inside $E$ (up to conjugation) that embeds into $\mathbb{Q}_p$ (it is a simple exercise to prove this). This reduces the problem of finding how much $E$ is contained in $\mathbb{Q}_p$ to the question of finding $G(E_{\mathfrak{p}}/\mathbb{Q}_p)$ as a subgroup of $G(E/\mathbb{Q})$. This is usually relegated to finding a Frobenius for the extension $\mathfrak{p}|p$. Particularly, it follows that $E$ itself is contained in $\mathbb{Q}_p$ if and only if $p$ splits completely in $E$.

As a corollary, a quadratic extension $E=\mathbb{Q}\left(\sqrt{(-1)^{(q-1)/2}q}\right)$ is contained in $\mathbb{Q}_p$ if and only if $p$ is a quadratic residue mod $q$. Additionally, for an arbitrary Galois extension, we have $E(\mathfrak{p})=\mathbb{Q}$ if and only if $G(E_{\mathfrak{p}}/\mathbb{Q}_p)=G(E/\mathbb{Q})$. This can only happen if $p$ is completely inert in the extension $E/\mathbb{Q}$. But, this can only happen when $E/\mathbb{Q}$ is cyclic (see Here). Therefore, every noncyclic extension $E/\mathbb{Q}$ will have share some nontrivial field with $\mathbb{Q}_p$ for any prime that does not ramify.

This then allows us to, not only find the extensions contained in $\mathbb{Q}_p$, as MatheinBoulomenos did, but also to extract the maximal subextension of $E/\mathbb{Q}$ that is contained in $\mathbb{Q}_p$.

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While $\mathbb{Q}_p / \mathbb{Q}$ is not algebraic, as mentioned in the comments, the question "What are the intermediate (finite) Galois extensions $\mathbb{Q}\subseteq E \subseteq \mathbb{Q}_{p}$?" still makes sense.

There's no reason to restrict oneself to Galois extensions in the following argument, so I'll describe the finite extensions $\mathbb Q \subseteq E \subseteq \mathbb{Q}_p$.

Suppose $\mathbb{Q} \subset E$ is a finite extension. Let $\mathfrak{p}_1, \dots , \mathfrak{p}_n$ be the prime ideals above $(p)$ in $\mathcal{O}_E$.
These are in bijection to the extensions of the $p$-adic valuation to $E$. For every $\mathfrak{p}_i$, we can complete $E$ with respect to the valuation associated to $\mathfrak{p}_i$, which gives us a finite extension $E_i$ of $\Bbb Q_p$. One can show that $[E_i:\Bbb Q_p]=e(\mathfrak{p}_i/( p)) f(\mathfrak{p}_i/(p))$ where $f$ and $e$ are ramification index and inertia degree, respectively.

If we fix an algebraic closure $\overline{\Bbb Q_p}$, then the $p$-adic valuation on $\Bbb Q_p$ extends uniquely to $\overline{\Bbb Q_p}$. As mentioned in the previous paragraph, we get embeddings $E \to \overline{\Bbb Q_p}$ for every $\mathfrak{p_i}$ (or equivalently for every extension of the $p$-adic valuation to $E$.) From these embeddings, we can reconstruct the valuation (just restrict the valuation from $\overline{\Bbb Q_p}$), so we get bijections

$$\{ \text{prime ideals in }\mathcal{O}_E \text{ above }(p)\} \leftrightarrow \{ \text{extensions of }v_p\text{ to $E$}\} \leftrightarrow\{\text{embeddings } E \to \overline{\Bbb Q_p}\} $$

And the degree of the extension is given by $(\mathfrak{p}_i/( p)) f(\mathfrak{p}_i/(p))$.

It follows that the number of embeddings $E \to \Bbb{Q}_p$ is equal to the number of prime ideals $\mathfrak{p}$ in $\mathcal{O}_E$ above $(p)$ that satisfy $e(\mathfrak{p}/(p))=f(\mathfrak{p}/(p))=1$

This describes all finite intermediate extensions $\Bbb Q \subset E \subset \Bbb{Q}_p$ and by extension, all intermediate extensions that are algebraic over $\Bbb Q$.

As for intermediate extensions transcendental over $\Bbb Q$, there are a lot, as $\Bbb Q_p$ has uncountable transcendence degree over $\Bbb Q$, but I'm not sure if we can say anything more about them.

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Concerning "Galois theory for $\mathbf Q_p/\mathbf Q$", we must first clarify the definition of a normal extension of fields $M/K$, whether it is algebraic or transcendental. Following E. Artin, let $G=Gal(M/K)$ (or $Aut(M/K))$ be the group of the automorphisms of $M$ which fix every element of $K$. Then $M/K$ is called normal iff $K=M^G:=$ the subfield of $M$ fixed elementwise by all $G$. Let us now determine $G=Gal(\mathbf Q_p/\mathbf Q)$. As pointed out by Lubin, $G= (1)$ because every $s \in G$ is continuous w.r.t. the $p$-adic metric. Proof : it suffices to show that for any $\in \mathbf Z_p , v_p (s(x))=v_p (x)$. But $x$ can be uniquely written $x=up^n$, where $u$ is a $p$-adic unit, and $s(u)$ is a $p$-adic unit because $s$ is an automorphism (this is purely algebraic). Since $G$ fixes $\mathbf Q$, it will fix $\mathbf Q_p$ by continuity, so $G=(1)$, and the extension is not normal. See also the ref. given by Watson.

Concerning the existence, for a given $p$, of finite extensions of $\mathbf Q$ "contained" in $\mathbf Q_p$ (with a slight abuse of language : one should speak of "embedding" instead of "containment"), the answer is "yes": these are exactly the number fields $K$ of degree $n$ such that the prime $p$ splits totally in $K$, i.e. for which $n$ distinct prime ideals $v$ of $K$ lie above $p$ (necessarily with inertia index and ramification index equal to $1$), or equivalently, the completions $K_v$ of $K$ w.r.t. all the places $v$ dividing $p$ are equal to $\mathbf Q_p$. Actually, over a fixed number field $k$, finite Galois extensions $K$ are determined by the splitting of primes in $K/k$ in the following precise sense : let $S$ be a finite set of primes of $k$ and denote by $Spl_S(K/k)$ the set of primes $v\notin S$ s.t. $v$ splits completely in $K$; note that this set has density $[K:k]^{-1}$ by Tchebotarev's theorem. Then, for two Galois extensions $K$ and $L$ of $k$, one has $K=L$ iff $Spl_S(K/k)=Spl_S(L/k)$. See e.g. Cassels-Fröhlich , exercise 6.1.

The problem remains to characterize $Spl_S(K/k)$ in some way. Not much is known in general, but for an abelian $K/k$, the job is done in principle by CFT. Over $k=\mathbf Q$, the Kronecker-Weber theorem is enough. Fix $p$ , let $q$ be any odd prime $\neq p$, and introduce the cyclotomic field $\mathbf Q(\zeta_q)$, which is a cyclic extension of $\mathbf Q$ of degree $(q-1)$; for any divisor $d$ of $(q-1)$, denote by $K_d$ the unique subextension of $\mathbf Q(\zeta_q)/\mathbf Q$ of degree $d$. Then $p$ splits completely in $K_d$ iff $p$ is a $d$-th power mod $q$. NB : in a Galois extension $K/k$, if a prime $v$ of $k$ is inert in $K$, then $Gal(K/k)$ is cyclic. See Marcus, chap. 4, ex. 5 (a), or the ref. given by @CW .

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