In an answer to the popular question on common false beliefs in mathematics


[Examples of common false beliefs in mathematics](https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23568#23568)


I mentioned that many people conflate the two different kinds of formal Laurent series field in two variables. Let $K$ be any field. Then we have the **joint Laurent series field**, the field of fractions


$K((x,y)) = \operatorname{Frac}(K[[x,y]])$ 


and also the **iterated Laurent series field** 


$K((x))((y)) = \left(K((x))\right)((y))$. 


These fields are not the same, roughly because when we write out an arbitrary element of the iterated field as a formal Laurent series in $y$, for each (say non-negative) $n$, the coefficient of $y^n$ is allowed to be an arbitrary formal Laurent series in $x$. In particular, as $n$ varies, arbitrarily large negative powers of $x$ may appear. 


However, this is rather far from a convincing argument. Indeed, I gave the following explicit (and fallacious!) example: $\sum\_{n=0}^{\infty} x^{-n} y^n$. But in a comment to my answer, user AS points out that this element is equal (in the iterated field, say) to $\frac{1}{1-\frac{y}{x}}$ and therefore it must lie in the fraction field of $K[[x,y]]$. Evidently the fallacy here is that the fraction field of $K[[x,y]]$ is the field of all formal Laurent series which are finite-tailed in both $x$ and $y$. But as this example shows, the latter isn't even a field, unlike the one-variable case.


[At least when $K = \mathbb{C}$, by less explicit means one can see that these two fields are very different: e.g., the joint field is Hilbertian so has nonabelian Galois group, whereas the iterated field has Galois group $\widehat{\mathbb{Z}}^2$.]


AS offered to write down, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$, so I decided to post a question asking for such a guy. Of course, there is more than one such element -- or better put, more than one type of construction of such elements -- so I would be interested to see multiple answers to:



> 
> Please exhibit, with proof, an explicit element of $K((x))((y)) \setminus K((x,y))$.
> 
> 
> 



Explicit elements of $K((x))((y)) \setminus K((x,y))$
Suppose $R$ is a domain with field of fractions $F$. Let $f\in F[[y]]$ and suppose that $f\in Frac(R[[y]])$. Then $f=h/g$ with $g,h\in R[[y]]]$ and we may assume that $g=b\_0+b\_1y+\dots$ with $b\_0\ne0$. Therefore 
$$
b\_0g^{-1}=(1+(b\_1/b\_0)y+\dots)^{-1}\in R[[y/b\_0]]
$$
and so $f\in b\_0^{-1}R[[y/b\_0]]$. 


So when $R=k[[x]]$, any $f=\sum y^n/x^{r(n)}$ with $r(n)/n \to \infty$ will work.


An analytic paraphrase of this argument (which is relevant to Makhalan's comment) is that (regarding $F=k((x))$ as a valued field) any element of $Frac(k[[x,y]])$ is a ratio of power series converging on the open unit disc in $F$. So it is enough to write down a power series in $y$ with zero radius of convergence.


