Let $\mathbb{F}_{p^n}$ be a finite field of characteristic $p$, then we can define discrete valuation $v_p$ in $\mathbb{F}_{p^n}((x))$ by $$ v_p(f):=\min \{n~|~a_n \neq 0\}, $$ where $f(x)=\sum_{n \geq i} a_n x^n \in \mathbb{F}_{p^n}((x))$.
This may be called a one-dimensional local field known as the field of formal Laurent power series.
I am interested if there are suitable generalisations of $\mathbb F_{p^n}((x))$ to more than one variable. For example, here I see $\mathbb F_{p^n}((x))((y))$ is a two-dimensional local field.
- Is $\mathbb F_{p^n}((x))((y)) \cong \mathbb F_{p^n}((x, y))$?
Consider the field $\mathbb F_{p^n}((x,y))$ of formal Laurent power series in $x$ and $y$. To show $\mathbb F_{p^n}((x,y))$ is a local field, at first, we need to define a discrete valuation. If $f(x,y)=\sum_{m \geq i, \\n \geq j} a_{mn} x^m y^n \in \mathbb F_{p^n}((x,y))$, then I guess such a valuation is $$v_p(f)=\min\{m,n~|~a_{mn} \neq 0\}.$$
We have to show it satisfies the properties of valuation as follows:
- (multiplicativity): Let $f(x,y)=\sum_{m,n} a_{mn}x^my^n \in \mathbb F_{p^n}((x,y))$ and $g(x,y)=\sum_{m,n}b_{mn}x^my^n \in \mathbb F_{p^n}((x,y))$, then $$fg=\sum_{m,n}c_{mn}x^my^n, ~\text{where}~c_{mn}=\sum_{k=0}^{i}a_{kj}b_{i-k,j}.$$ I have checked particular examples, giving $v_p(fg)=v_p(f)+v_p(g)$. I think it is multiplicative.
I have not checked if $v_p(f+g) \geq \min \{v_p(f), v_p(g)\}$ holds.
I would like to get your comments if $\mathbb F_{p^n}((x,y))$ is really a local field. We know $\mathbb F_{p^n}((x))((y))$ is a local field.
As pointed out in Pete L. Clark's answer to the MathOverflow thread "Examples of Common False Beliefs in Mathematics", and comments to it, for any field $k$, there is a proper inclusion of fields
$$k((x,y)) \subsetneq k((x))((y))$$
where the left hand side is, by definition, the quotient field of $k[[x,y]]$, while the right hand side, properly written $\color{red}(k((x))\color{red})((y))$, is the field of formal Laurent series in $y$ over the ("coefficient") field of formal Laurent series in $x$.
In fact, there is a bit of discussion there showing that while e.g. the element
$$\sum_{i\ge 0} x^{-i}y^{i}$$
somewhat surprisingly does lie in the left hand side (it is $=\frac{x}{x-y}$), but the upgraded example
$$\sum_{i\ge 0} x^{-i^2}y^{i}$$
does not. (More abstractly, one can check that actually, $k((x))((y))$ is "not symmetric in $x,y$", while the smaller field obviously is.)
By the way, the first example $\sum_{i\ge 0} x^{-i}y^{i}=\frac{x}{x-y}$ shows that your attempt of a valuation is not well defined because not every element of $k((x,y))$ van be written the way you claim it.
You can of course, e.g., take the valuation from $k((x))((y))$ (with the coefficients from $k((x))$ having valuation $v(k((x))) = 0$), and restrict it to that subfield. Pretty sure it will not be complete with respect to that valuation.