There is an intuitive sense in which one would like to say that the field of formal Laurent series $k((z))$ over a field $k$ has "transcendence degree $1$". Of course, it doesn't have transcendence degree $1$ over $k$ in the usual sense (in fact I think the transcendence degree is countably infinite, by a cardinality argument). However, $k((z))$ is the completion of $k(z)$ for the $z$-adic absolute value, so it would be nice to say that it has transcendence degree one in some sense. Is there a way to make this precise, or is my intuition overly ambitious?
Remark: I think that it is true that $k((x,y)):=k((x))((y)) \cong k((z))$ as $k$-algebras, simply because both sides have the same (infinite) transcendence degree over $k$, in the usual sense. This doesn't bode well for my question, since one would certainly like to say that $k((x,y))$ has "formal" transcendence degree $2$ over $k$. However, there are natural topologies on both of these $k$-algebras (there is the $z$-adic topology on $k((z))$, and there are a few candidates on $k((x,y))$, for instance the topology for which two formal series are close if their difference is divisible by a high power of $x$ or $y$). If one takes these topologies into account, then I don't think they are isomorphic as topological $k$-algebras, so maybe my question can be rescued.