# Do algebraic completion/topological completion of fields always terminate? If so, are they unique?

Take the field $$\mathbb{Q}$$, If we complete it topologically with respect to the Euclidean norm, we get $$\mathbb{R}$$, then if we complete it algebraically, we get $$\mathbb{C}$$. On the other hand, the algebraic completion of $$\mathbb{Q}$$ is $$\overline{\mathbb{Q}}$$, and if we complete it with respect to the Euclidean norm now, we get $$\mathbb{C}$$. So it indeed terminates and in fact the result is the same field/space.

Now, another example is the topological completion of $$\mathbb{Q}$$ with respect to the p-adic norm, giving us $$\mathbb{Q}_p$$ the p-adic numbers, then the algebraic completion $$\overline{\mathbb{Q}_p}$$, and then once more with topological completion under p-adic norm, we get $$\mathbb{C}_p$$ which is the end.

Maybe this question can be better phrased in terms of ascending chains/homology but if anyone has any insight, that would be great.

• I think the sentence "On the other hand, the algebraic completion of $\mathbb{Q}$ is $\overline{\mathbb{Q}}$ and if we complete it with respect to the Euclidean norm now, we get $\mathbb{C}$" is a bit confusing, because for a finite extension of $\mathbb{Q}$, there is no unique extension of the norm (the Euclidean norm on $\mathbb{Q}[\sqrt{2}]$ is not invariant under the Galois action). – YCor Feb 8 at 18:05
• I think this answer of Laurent Moret-Bailly's at least addresses YCor's important objection (by observing that all extensions of the norm are conjugate) and gives some insight on the question. Does it make a difference if we consider valuations in an ordered abelian group versus norms valued in $\mathbb R$? – Tim Campion Feb 8 at 18:10
• In fact, the problem does not make much sense if we don't restrict it to valued fields, because in general there is no preferred choice of a topology on the algebraic closure (not even up to isomorphism). – Laurent Moret-Bailly Feb 10 at 14:36
• Another general comment: the (topological) completion of a topological field is not always a field. For instance, $\mathbb{Q}$ is dense in $\mathbb{Q}_2\times\mathbb{Q}_3$, which is therefore the completion of $\mathbb{Q}$ for the induced topology. (This topology is generated by the 2-adic and 3-adic ones). – Laurent Moret-Bailly Apr 3 at 9:26
• @FrançoisG.Dorais It is the product ring, with the product topology. Density is just (a special case of) the approximation theorem for independent valuations. – Laurent Moret-Bailly Apr 3 at 18:35