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If $(K, v)$ is a valued field, is the completion of $K$ with respect to $v$ (denoted $\hat{K}$) an algebraic or transcendental extension of $K$?

E.g. $\mathbb{Q}_p$ is transcendental over $\mathbb{Q}$ by cardinality reasons, and I know there exist explicit elements of $\mathbb{Q}_p$ that have been shown to be transcendental over $\mathbb{Q}$. (In general, however, I don't think it's true that $|K| < |\hat{K}|$.)

Edit: fixed LaTeX error.

valqyrie
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    The argument for $\mathbb Q$ should generalize to every countable field with nontrivial $v$. On the other hand, if $K$ is already complete, it's a trivial hence algebraic extension. Maybe a question would be if that's the only case where the completion is an algebraic extension. – Torsten Schoeneberg Jul 08 '21 at 14:58

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As Torsten Schoeneberg said in a comment, the extension can be trivial and hence algebraic.

There are also examples of non-trivial but algebraic extensions. For instance, take formal Laurent series $k((t))$ over a field $k$, and consider the dense subfield $k(t)$ of rational functions. Fix a transcendance basis $\beta$ of $k((t)))$ over $k(t)$* and set $K:=k(t)(\beta)$, so $\hat{K}=k((t))$ is algebraic over $K$.

In order to make sure that $\hat{K} / K$ is non-trivial, take a basis containing the square root of an element of $k(t)$, such as $1+t$, which also has a $4$-th root in $k((t))$. That way $(1+t)^{\frac{1}{4}}$ cannot lie in $K$.

You can obtain an extension of finite degree by defining $K$ as the relative algebraic closure of the field gnerated by $k(t)$ and all elements but one of the basis $\beta$.

*i.e. $\beta$ is a family of algebraically independant elements, and $k((t))$ is algebraic over $k(t)(\beta)$. Equivalently $\beta$ is a maximal family of algebraically independant elements, hence its existence in general is by Zorn's lemma.

nombre
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    For the existence of a fourth root of $1+t$ in $k((t))$ you need $char(k) \neq 2$, right? – Torsten Schoeneberg Jul 08 '21 at 16:59
  • @TorstenSchoeneberg Oh yes, I really didn't think about that. I would assume that the extension I described is never trivial but I don't see a way to prove it. – nombre Jul 08 '21 at 17:01
  • The case of a finite extension often is impossible. If $K$ has characteristic $0$ and $(K,v)$ is not complete, then suppose the completion $\widehat{K}$ is a finite extension: $\widehat{K}/K$ is a finite separable extension with degree greater than $1$. The answer to https://math.stackexchange.com/questions/4660292/subfield-of-a-complete-valuation-field-also-complete (note $K$ here is $L$ there and $\widehat{K}$ here is $K$ there), including the remark at the end, shows that if $\widehat{K}$ is not algebraically closed then $K$ is complete: contradiction. – KCd Jul 02 '23 at 10:13