Formal Laurent series and field of fractions of Taylor series Let us define the field of formal Laurent series of $n$ variables as $K=k((x_1))((x_2))...((x_n))$. As a subring it contains the ring of formal Taylor series $R=k[[x_1]][[x_2]]...[[x_n]]$. And the third ring, let's call it $R_0$, is defined to be the subring of $K$ consisting of series such that for each variable the degree is bounded from below. I have two questions:
(1) Does the structure of these two rings depend on the order of variables?
(2) What is $Frac(R)$ and how is it different from $K$? And how this subfield is related to $R_0$?
 A: If $R$ is a ring, then $R((t))$ is the localization of $R[[t]]$ with respect to the element $t$, that is every element of $R((t))$ is of form $t^mf(t)$ where $f\in R[[t]]$, $m$ being any element of $\Bbb Z$. In particular, $R((t))$ is most certainly not the fraction field of $R[[t]]$, unless you prove a theorem that it is, as, for example when $R$ is a field. I’m sure that when you write $\Bbb Z((t))$ you don’t mean for the rationals to be contained.
To get to your question, let’s just look at the two-variable case; I presume you mean for $k$ to be any field.
First, you seem to ask whether $k((x))((y))$ and $k((y))((x))$ have the same structure. Of course they are isomorphic as rings, but I don’t think that they are the same set. For instance, consider
$$
\sum_0^\infty\frac{x^n}{y^n}=1+\frac xy+\frac{x^2}{y^2}+ \cdots\,,
$$
which is certainly in $k((y))((x))$, even in $k((y))[[x]]$, but not in $k((x))((y))$, because the $y$-degrees in the denominators are unbounded. We know, from our high-school education, that this is
$$
\frac1{1-\frac yx}\,,
$$
and so is in the fraction-field of $k[x,y]$.
Because $k[[x]][[y]]=k[[y]][[x]]$ as sets (needs verification), it’s conventionally written $k[[x,y]]$; but one would have to be very careful about defining exactly what one meant by $k((x,y))$. Try it, and make sure that multiplication is well defined. And don’t expect the result to be a field.
EDIT (Correction):
Instead of deleting a partially wrong answer and replacing it with one that I hope is now right, I will leave the above and give below what I think the right approach might be.
All thanks to you, @Anna Abasheva, for pointing out that since $K=k((x))((y))$ is a field containing $k[[x]][[y]]$, we must have $K\supset\text{frac}\bigl(k[[x,y]]\bigr)$. I hope the following will answer at least some of your questions.
I now contend that
$$
F(x,y)=\sum_n\frac{y^n}{x^{n^2}}\in k((x))[[y]]
$$
is not a quotient of power series in $k[[x,y]]$.
For this, I will be thinking of the power series ring as $D[[y]]$, where $D$ is the complete discrete valuation ring $k[[x]]$, with fraction-field $k((x))$. On this ring and its fraction field we have the (additive) valuation $v_x$, measuring divisibility by the prime element $x$. And for this ring, we have the Weierstrass Preparation Theorem, according to which every $f\in D[[y]]$ may be written $f(y)=x^rP(y)U(y)$, where $r$ is the minimum valuation of all coefficients of $f$. There will be a smallest degree $d$ for which the coefficient of $y^d$ has $v$-value $r$. Then the Theorem says:
There is a (unique) monic polynomial $P(y)\in D[y]$ of degree $d$ and a (unique) unit power series $U(y)\in D[[y]]^\times$ such that $f(y)=x^rP(y)U(y)$.
Now, $P$ is a polynomial over the fraction field $k((x))$ of $D$, and therefore has at most finitely many roots in $D$.
We now apply Weierstrass Preparation to any element $f/g\in\text{frac}\bigl(k[[x,y]]\bigr)$, that is to any
fraction with numerator and denominator in our $D[[y]]$. We get
$$
\frac{f(x,y)}{g(x,y)}=\frac{x^rP(y)U(y)}{x^{r'}Q(y)U'(y)}\,,
$$
where $P,Q$ are monic $D$-polynomials of some degree, and $U,U'$ are both unit power series, which just means that as elements of $k[[x,y]]$, they have nonzero constant term.
For every element $\varepsilon(x)$ of the maximal ideal of $D$, excepting zero and the finitely many roots in $D$ of the polynomials $P$ and $Q$, then, $y\mapsto\varepsilon(x)$ gives an evaluation of $f/g$ to a nonzero element of $k((x))$.
But there is no evaluation of the Laurent series (power series, actually, even) $F$ above that can be done for any such $\varepsilon(x)$. Essentially, the radius of convergence of this series is zero. Say that $v(\varepsilon)=s\ge1$. Then the monomial $y^n/x^{n^2}$ is evaluated to $\varepsilon^n/x^{n^2}$, for which the $v$-value is $ns-n^2$, doesn’t go to $\infty$, so that the series is not convergent.
I think this exhibits an element of $k((x))((y))$ that isn’t in $\text{frac}\bigl(k[[x,y]]\bigr)$.
A: Edit My previous answer was is still completely wrong, and is now fixed I should leave this question to the experts.


*

*The order of the variables does not matter. You can show $k[[x_1]][[x_2]]\cong k[[x_2]][[x_1]].$

*$\operatorname{Frac}(R)$ is the set of ordered pairs of the form $(f,g)$, where $f,g$ are formal Taylor series and $g\neq 0$. Two such pairs $(f_1,g_1)$ and $(f_2,g_2)$ are considered equivalent when $f_1g_2=g_1f_2$. The interpretation is $(f,g)=\frac{f}g$.Every formal Laurent series is an element of $\operatorname{Frac}(R)$. However, when $n>1$, not every element of Frac$(R)$ is a formal Laurent series. An example is $\frac1{x_1+x_2}$.
