# Isomorphism of algebraic closure of p-adics with their completion

Consider the following fields:

1) $\mathbb{C}$ the complex numbers

2) $\overline{\mathbb{Q}}_p$

3) $\mathbb{C}_p : = \hat{\overline{\mathbb{Q}}_p}$

They are all the same cardinality, algebraically closed, and of characteristic 0. Therefore they are all isomorphic as fields. However: $\overline{\mathbb{Q}}_p \to \mathbb{C}_p$, since it is the topological completion, and is not surjective. Is there something wrong with this argument or not?

Edit (To make my question more clear): Is $\overline{\mathbb{Q}}_p \to \mathbb{C}_p$ surjective? Or can there exist embeddings of fields into isomorphic fields which are not surjective?

## 1 Answer

Nothing is wrong with the argument. These objects are isomorphic in the category of fields and not in the category of topologial fields (or even the category of topological spaces).

• What about the injection above which is not surjective? – Eins Null Feb 24 '16 at 0:21
• @EinsNull I don't understand your question. That particular map exists and isn't an isomorphism; it doesn't mean the fields aren't isomorphic. To say two fields are isomorphic is to say that there is an isomorphism between them, not that every map between them is an isomorphism. – hunter Feb 24 '16 at 10:01
• @EinsNull now I see your edit. No, that map is certainly not surjective. For a concrete example of something not in the image, choose a sequence $\zeta_1, \zeta_2, \ldots$ where $\zeta_i$ is a primitive $p^i$th root of unity. Then $1 + \zeta_1p + \zeta_2p^2 + \ldots$ is in $\mathbb{C}_p$ but not $\overline{\mathbb{Q}}_p$. – hunter Feb 24 '16 at 10:03
• Thanks! It just seems weird to me that a field can be embedded into itself as a field without surjecting. – Eins Null Feb 24 '16 at 14:33
• @EinsNull maybe an easier example to grasp: $\mathbb{Q}(x_1, x_2, \ldots)$ embeds into itself by the "Hilbert hotel map" $x_i \mapsto x_{i+1}$ – hunter Feb 24 '16 at 14:37