# Does $\mathbb{Q}_p \cap \overline{\mathbb{Q}}$ depend on the embedding of $\overline{\mathbb{Q}}$ into $\overline{\mathbb{Q}_p}$?

I think the title pretty much says it all. I'm getting confused in the subtle parts of a proof, and would appreciate some help.

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Dear Shannon, this is the set of $p$-adic numbers that are algebraic over $\mathbb{Q}$ (i.e. satisfy a polynomial equation with coefficients in $\mathbb{Q}$). Consequently, the particular embedding one chooses is irrelevant. – Akhil Mathew Feb 5 '11 at 18:27
This is similar to how we talk about "the" algebraic closure and "an" algebraic closure. Any two algebraic closures of a field are isomorphic. – PEV Feb 5 '11 at 19:12
@PEV: that is not the issue here. Any two algebraic closures of a field are isomorphic, but when you construct things using two or more fields, your constructions may depend on the choice of isomorphism or embedding. For example, the intersection of two number fields is not well-defined because it depends on their embedding into an algebraic closure. – Qiaochu Yuan Feb 5 '11 at 19:13
@Qiaochu Yuan: I guess this is analogous (in your blog post) to equipping $\mathbb{R}^n$ with an inner product and orientation and looking at $SO(n)$? If we equipped $\mathbb{R}^n$ with something different, then $SO(n)$ wouldn't make sense (or would mean something different)? – PEV Feb 6 '11 at 4:11
@PEV: yes. If you equip R^n with a different inner product, you get a conjugate copy of SO(n) which sits inside GL_n differently (but isomorphically). – Qiaochu Yuan Feb 6 '11 at 12:17

Once you fix an algebraic closure $\overline{\mathbb{Q}_p}$, the subfield $\overline{\mathbb{Q}}$ is uniquely determined as the smallest algebraically closed subfield containing the prime field, so all embeddings have the same image.

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There's a subtlety here.

$\overline{\mathbb{Q}}\cap \mathbb{Q}_p$ is well defined as a subfield of $\mathbb{Q}_p$ since it is simply the set of elements which are algebraic over $\mathbb{Q}$.

However, as a subfield of $\overline{\mathbb{Q}}$, this is not canonical. The choice of embedding $\phi: \overline{\mathbb{Q}}\rightarrow \overline{\mathbb{Q}_p}$ amounts to choosing a place of $\overline{\mathbb{Q}}$ lying over $p$. The induced subfield $\phi^{-1}(\mathbb{Q}_p)\subset \overline{\mathbb{Q}}$ will be the maximal subfield in which this place splits. This depends very much on the place we've chosen! This field is not Galois over $\mathbb{Q}$, and its Galois conjugates reflect the other choices of embeddings.

Here's another way of saying it. The embedding $\phi$ gives a choice of decomposition group

$D_p\subset$ Gal$(\overline{\mathbb{Q}}/\mathbb{Q})$. Then

$\phi^{-1}(\mathbb{Q}_p)=\mathbb{\overline{Q}}^{D_p}$.

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