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

Question

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

Examples of common false beliefs in mathematics

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 $K((x))((y)) \setminus K((x,y))$.

Answer 1

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.

Answer 2

There is a characterization of $K((x,y))$ which makes it clear what sorts of examples to expect. That is, it is a subring of $K(z)((y))$ (where $x = yz$) of series with certain regularity properties.

Let $R \subset K[z][[y]]$ be the "filtered power series" ring, consisting of series $p(z,y) = \sum_{n \geq 0} p_n(z) y^n$ with $\deg p_n \leq n$; equivalently, $p_n(x/y) y^n$ is homogeneous in $x,y$ of degree $n$. Then by definition the substitution $x = yz$ furnishes an isomorphism $R \cong K[[x,y]]$. One verifies that $R$ is indeed a ring and that if $p(z,0) = 1$, then $p$ is a unit in $R$, since $$p(z,y)^{-1} = (1 + \sum_{n \geq 1} p_n(z) y^n)^{-1} = \sum_{m \geq 0} (-1)^m (\sum_{n \geq 1} p_n(z) y^n)^m.$$

Now we consider $\mathrm{Frac}(R)$. If $p(z,y) = \sum_{n \geq a} p_n(z) y^n$ with $p_a \neq 0$, then we can write $$p(z,y) = p_a(z) y^a q(z,y)$$ in $K(z)[[y]]$, where if $q(z,y) = \sum_{n \geq 0} q_n(z) y^n$, then $q_n(x/y) y^n$ is homogeneous of degree $n$ (as a rational function). Then $p^{-1} = p_a^{-1} y^{-a} q^{-1} \in K(z)((y))$ also has this property.

Conversely, if $\ell(z,y) \in K(z)((y))$ is a "filtered Laurent series": $\ell(z,y) = \sum_{n \geq -a} \ell_n(z) y^n$ with $a \in \mathbb{Z}$, $\ell_n(z) \in K(z)$, and $\ell_n(x/y)y^n$ homogeneous of degree $n \in \mathbb{Z}$, then $\ell(x/y,y)$ makes sense in $K((x,y))$ if each $\ell_n$ is additionally of the form $q_{n + a}(z) p_a(z)^{-1}$ for polynomials $q_{n + a}, p_a$ of degrees at most $n + a, a$ respectively, since then $$\ell(x/y,y) = p_a(x/y)^{-1} y^{-a} \sum_{m \geq 0} q_m(x/y) y^m,$$ with $p_a(x/y)^{-1} \in K(x/y) \subset K((x,y))$ and $q_m(x/y) y^m \in K[x,y]$ homogeneous of degree $m$. More concisely, we can say: $K((x,y))$ is isomorphic to the subring of $K(z)((y))$ consisting of series $$\ell(z,y) = \sum_{n \geq -a} \ell_n(z) y^n$$ such that each $\ell_n(x/y)y^n$ is homogeneous of degree $n$ and all the $\ell_n(z)$ together have only finitely many poles in $\bar{K}$.

Thus, the following classes of functions in $K((x))((y))$ are not in $K((x,y))$:

• Series of the form $\ell(x,y) = t(x/y) = \sum_{n \geq 0} t_n x^n y^{-n}$ for which $t(z) \in K[[z]]$ is not a rational function. For example, $$t(z) = \exp(z) = \sum_{n \geq 0} \frac{1}{n!} z^n, \qquad t(z) = \sum_{n \geq 0} b_n z^n,$$ where the first only makes sense over a field of characteristic zero, and where in the second, $b_n \in \{0,1\}$ is a random sequence of bits (this will almost surely not be rational). More generally, series of the form $\ell(x,y) = \sum_{a,b} \ell_{a,b} x^a y^b$ such that some $\ell_n(z) = \sum_{a + b = n} \ell_{a,b} z^a$ is not rational.
• Series $\ell(x,y)$ for which the $\ell_n(z)$ have "infinitely many poles" among them. This is hard to detect on the level of formal series, but one criterion is similar to that of Tony Scholl: the radii of convergence of the $\ell_n(z)$ have infimum zero or infinity. Over $\mathbb{C}$, by the root test this is the same as $\lim_{n \to \infty} \sqrt[n]{|\ell_{a, n - a}|}$ being unbounded in $a$, or their inverses being unbounded. (This also works over any field with a complete real valuation.)

Answer 3

I think that $\displaystyle\sum\limits_{n\geq 0}\frac{y^{2^n}}{x^{2^n}}$ is such a case. In fact, would it lie in $K\left(\left(x,y\right)\right)$, we would have two nonzero power series $\sum\limits_{u,v\geq 0}a_{u,v}x^uy^v$ and $\sum\limits_{u,v\geq 0}b_{u,v}x^uy^v$ satisfying $\displaystyle\sum\limits_{n\geq 0}\frac{y^{2^n}}{x^{2^n}}\cdot \sum\limits_{u,v\geq 0}a_{u,v}x^uy^v=\sum\limits_{u,v\geq 0}b_{u,v}x^uy^v$. This rewrites coefficientwise as

(1) $\displaystyle b_{u,v}=\sum\limits_{n\geq 0}a_{u+2^n,v-2^n}$

for all $u$ and $v$ (maybe the $+$ and the $-$ signs are the other way round here, but I hope it is clear what to do then), where $a_{p,q}$ is defined to be $0$ if $p<0$ or $q<0$. Now, take some $\left(p,q\right)$ satisfying $a_{p,q}\neq 0$, and take some natural $N$ which is way bigger than $p$ and $q$. Apply (1) to $u=p-2^N$ and $v=q+2^N$ and obtain $b_{p-2^N,q+2^N}=a_{p,q}$ (because in the sum on the right hand side of (1), the summands for $n < N$ have $u+2^n<0$ and thus are zero, while the summands for $n > N$ have $v-2^n<0$ and thus are zero as well, so only the $n=N$ summand remains), contradicting $p-2^N<0$ (which is because $N$ is so huge).

Answer 4

Addendum: actually there is a chain of subfields between $K((x,y))$ and $K((x))((y))$ with cardinality c. For instance, for any $\lambda>1$ we may consider the subset $R_\lambda$ of $K((x))((y))$ of all Laurent series $\sum_k c_k(x)y^k$ with $c_k\in K((x))$ satisfying $$\inf_k \lambda ^{-k} \mathrm{ord}(c_k) > -\infty.$$ It's easy to check that it's a subfield of $K((x))((y))$, containing $K[[x,y]]$.

Moreover, since in place of $\lambda^k$ we can use functions $\mathbb{N}\to\mathbb{N}$ with arbitrarily large growth, one can also show that the subfields between $K((x,y))$ and $K((x))((y))$ have uncountable cofinality also.

Answer 5

This really should be a comment to Tony Scholl's nice counterexample above, but I dont have the necessary reputation. (!!!)

A transparent way to see that any $\sum_{n\ge 0} \dfrac{y^n}{x^{r(n)}}$ with $\dfrac{r(n)}{n}$ tending to $\infty$ will serve as a counterexample (without dividing by $b_0$, etc.) is as follows: More generally, suppose $\sum_{k \ge 0} a_k(x) y^k$ is to be a counterexample, with $a_k(x)$ coming from $K((x))$ ($K$ will be the base field), and $v_x(a_k(x)):= -r(k) < 0)$ ($v_x$ being the $x$-adic valuation on $K((x))$ corresponding to the lowest degree term in $x$.) We wish to show that it is impossible to write it as the quotient $\sum_{i \ge 0} p_i(x) y^i$ divided by $\sum_{j \ge 0} q_j(x) y^j$, where, importantly, each $v_x(p_i(x))$ and $v_x(q_j(x))$ are non-negative, and $q_0(x)\neq 0$. (Note that we are embedding $K[[x,y]]$ in $K((x))((y))$ by writing elements of $K[[x,y]]$ as a power series in $y$ with coeffs in $K[[x]]$, see the discussions on this above!) Then cross-multiplying and comparing coeffs of $y^k$ we get $$\sum_{j=0}^k a_{k-j}(x)q_{j}(x) = p_k(x).$$ If $v_x(q_j(x)) := d_j \ge 0$, then the nonzero summands on the left have $v_x$ (i.e., exponent of lowest $x$-terms) $-r(k) + d_0$, $-r(k-1)+d_1$, etc. The lowest $x$-term on the right is of non-negative exponent. To make this an impossibility (so that we get a counterexample), it is sufficient to choose the function $r$ to satisfy $-r(k) + d_0 < -r(k-1)$ for any choice of $d_0\ge 0$ and for all $k > k_0(d_0)$. (Recall that the $r(k)$ are negative integers while the d_i are non-negative integers. This condition makes the $r(k)$ increase monotonically, and dominate all $r(j)$ after a while.) We write this as $r(k) - r(k-1) > d_0$ for any $d_0\ge 0$, for all $k > k_0(d_0)$. Writing this as $r(n) > r(k_0) +(n-k_0)d_0$ for $n > k_0(d_0)$ and dividing by $n$, we find that $\dfrac{r(n)}{n}$ goes to $\infty$. Pietro Majer's comment also follows from this.