2 typo; missing left parenthesis added

# Explicit elements of $K((x))(y))K((x))((y)) \setminus K((x,y))$

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

http://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 $K((x))(y)) K((x))((y)) \setminus K((x,y))$, 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 $K((x))(y)) K((x))((y)) \setminus K((x,y))$.

1

# Explicit elements of $K((x))(y)) \setminus K((x,y))$

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

http://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 $K((x))(y)) \setminus K((x,y))$.