Is the completion of a valued field an algebraic or transcendental extension, in general? 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.
 A: 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.
