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If $k$ is a finite field, then $k((x))$ is a local field, and we can define a discrete valuation on $k((x))$ with respect to which it is complete. It is sometimes called a 1-dimensional local field.

I am interested in knowing if there are proper generalisations of $k((x))$ in several variables preserving the localness.

For example, I found here and here, there is a 2-dimensional local field $k((x))((t))$, which is the field of Laurent series in the variable $t$ over the 1-dimensional local field $k((x))$.

  • If it is a local field, then what is the valuation? Is it the same valuation as $k((x))$?

More generally, consider the field $k((x,t))$ of the Laurent power series in $x,t$. There are no results claiming it is a local field. To be a local field, there must be a discrete valuation on $k((x,t))$.

I need to understand if $k((x,t))$ can not be a local field, or, we just don't know yet if there is a discrete valuation making $k((x,t))$ a local field. Can someone clarify it please?

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    $\begingroup$ I think that you are confused (understandably) by the fact that "2-dimensional local fields" are not local fields. $\endgroup$
    – Tom Goodwillie
    Commented Mar 25 at 12:00
  • $\begingroup$ @TomGoodwillie, ok, so there is no discrete valuation on $k((x))((t))$. Right? same on $k((x,t))$ $\endgroup$
    – MAS
    Commented Mar 25 at 12:18
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    $\begingroup$ Of course there are discrete valuations, namely there are discrete valuations on $L((t))$ for every field $L$. The point is that for $L=k((x))$, this makes $k((x))$ a bounded subfield. $\endgroup$
    – YCor
    Commented Mar 25 at 12:28
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    $\begingroup$ @MAS a valuation defines a distance $d(x,y)=\exp(-v(x-y))$. Bounded means bounded in this metric; for a subring it just means that the valuation takes values $\ge 0$. For every field $L$, $L((x))$ is naturally a complete field for the discrete valuation wrt $x$. $\endgroup$
    – YCor
    Commented Mar 25 at 13:25
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    $\begingroup$ @MAS I think that "local field" sometimes assume that the topology is locally compact (which for $k((t))$ is true iff $k$ is finite). Of course these fields are of interest even when $k$ is infinite. However the topology for which $k((x))$ is bounded is or is not useful, depending on the context. Note that $k((x))((t))$ is not symmetric in $x,t$: for instance $\sum_{n\ge 0} x^{-n}t^n$ makes sense in it, while it is ill-defined in $k((t))((x))$. So the notation $k((x,t))$ can be confusing, or referring to something else. $\endgroup$
    – YCor
    Commented Mar 25 at 13:35

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