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There is a nice characterization of $K((x,y))$ that which makes it clear what sort sorts of examples to expect. NamelyThat is, the difference between it and $K[[x,y]]$ is no more than the addition a subring of rational functions in $x/y$. If $p(x,y)$ and $q(x,y)$ are in $K[[x,y]]$, then let K(z)((y))$(where$f(x,y)$and x = yz$) of series with certain regularity properties.
Let $g(x,y)$ R \subset K[z][[y]]$be the homogeneous polynomials "filtered power series" ring, consisting of their terms series$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$of lowest total degree and write$p(x,y) n$. Then by definition the substitution$x = f(x,y) p_0(x,y)$yz$ furnishes an isomorphism $R \cong K[[x,y]]$. One verifies that $R$ is indeed a ring and that if $q(x,y) p(z,0) = g(x,y) q_0(x,y)$1$, so then$p/q p$is a unit in$R$, since $$p(z,y)^{-1} = (f/g)(p_0/q_0). If 1 + \sum_{n \geq 1} p_n(z) y^n)^{-1} = \sum_{m \geq 0} (-1)^m (\sum_{n \geq 1} p_n(z) y^n)^m.$$ Now we dehomogenize consider$f,g$\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 $f/g$p(z,y) = y^n f_0/g_0$, p_a(z) y^a q(z,y)$$in K(z)[[y]], where if n q(z,y) = \in sum_{n \mathbb{Z} and geq 0} q_n(z) y^n, then f_0, g_0 q_n(x/y) y^n is homogeneous of degree n (as a rational function). Then p^{-1} = p_a^{-1} y^{-a} q^{-1} \in K[x/y]K(z)((y)) also has this property.Likewise Conversely, we can write any expression if f(x,y)^{-1} x^k y^l (such as appears \ell(z,y) \in K(z)((y)) is a "filtered Laurent series": p_0) \ell(z,y) = \sum_{n \geq -a} \ell_n(z) y^n with a \in the form \mathbb{Z}, (x/y)^k f_0(x/y)^{-1} y^m\ell_n(z) \in K(z),and the same for the terms \ell_n(x/y)y^n homogeneous of degree q_0 using n \in \mathbb{Z}, then g_0. Note that \ell(x/y,y) makes sense in p_0, q_0 have constant term K((x,y)) if each 1, and are therefore invertible series in \ell_n is additionally of the form y having coefficients in q_{n + a}(z) p_a(z)^{-1} for polynomials K(x/y). Summing upq_{n + a}, we have p_a of degrees at most K((x,y)) n + a, a respectively, since then$$\ell(x/y,y) = K(x/y)((y)p_a(x/y)^{-1} y^{-a} \sum_{m \geq 0} q_m(x/y) y^m,$$with p_a(x/y)^{-1} \cong K(z)((y)). In this sense, your wrong example was quite unlucky, though understandablyin K(x/y) \subset K((x,y)) and q_m(x/y) y^m \in K[x,y] homogeneous of degree m. More concisely, since it we can say: K((x,y)) is hard isomorphic to distinguish rational functions from non-rational ones by their Laurent series. The answer the subring of darij grinberg is basically doing just K(z)((y)) consisting of series$$\ell(z,y) = \sum_{n \geq -a} \ell_n(z) y^n$$such that . An infinitude each \ell_n(x/y)y^n is homogeneous of examples then arises by letting degree \ell(x,y) be any Laurent series n and all the \ell_n(z) together have only finitely many poles in x/y \bar{K}. Thus, the following classes of functions in K((x))((y)) are not in K((x,y)): • Series of the form \ell(x,y) = t(x/y) = \sum_{n \geq 0} t_n x^n y^{-n} for which t(z) \in K[[z]] is not a rational function. Examples:For example,$$\ell(x,y)$t(z) = \exp(x/y) exp(z) = \sum_{n = 0}^\infty \geq 0} \frac{1}{n!} x^n y^{-n}\ell(x,y) z^n, \qquad t(z) = \sum_{n = 0}^\infty \geq 0} b_n x^n y^{-n}$$where the first only makes sense over a field of characteristic zero, and where in the secondexample, b_n refers to b_n \in {0,1} is a random sequence of bits . Almost surely, (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.Unlike • Series \ell(x,y) for which the first one\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 actually works when is the same as K has positive characteristic\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.) • 4 deleted 3 characters in body There is a nice characterization of K((x,y)) that makes it clear what sort of examples to expect. Namely, the difference between it and K[[x,y]] is no more than the addition of rational functions in x/y. Namely, if If p(x,y) and q(x,y) are in K[[x,y]], then let f(x,y) and g(x,y) be the homogeneous polynomials consisting of their lowest-degree terms of lowest total degree and write p(x,y) = f(x,y) p_0(x,y) and q(x,y) = g(x,y) q_0(x,y), so p/q = (f/g)(p_0/q_0). If we dehomogenize f,g we can write f/g = y^n f_0/g_0, where n \in \mathbb{Z} and f_0, g_0 \in K[x/y]. Likewise, we can write any expression f(x,y)^{-1} x^k y^l (such as appears in p_0) in the form (x/y)^k f_0(x/y)^{-1} y^m, and the same for the terms of q_0 using g_0. Note that p_0, q_0 have constant term 1, and are therefore invertible series in y having coefficients in K(x/y). Summing up, we have K((x,y)) = K(x/y)((y)) \cong K(z)((y)). In this sense, your wrong example was quite unlucky, though understandably, since it is hard to distinguish rational functions from non-rational ones by their laurent Laurent series. The answer of darij grinberg is basically doing just that. An infinitude of examples then arises by letting \ell(x,y) be any Laurent polynomial series in x/y which is not a rational function. Examples:$$\ell(x,y) = \exp(x/y) = \sum_{n = 0}^\infty \frac{1}{n!} x^n y^{-n}\ell(x,y) = \sum_{n = 0}^\infty b_n x^n y^{-n}$$where in the second example, b_n refers to a random sequence of bits. Almost surely, this will not be rational. Unlike the first one, this actually works when K has positive characteristic. 3 Make correct. There is a nice characterization of K((x,y)) that makes it clear what sort of examples to expect. Namely, the difference between it and K[[x,y]] is no more than the addition of rational functions in x/y. Namely, if p(x,y) and q(x,y) are in K[[x,y]], then let f(x,y) and g(x,y) be the homogeneous polynomials consisting of their lowest-degree terms and write p(x,y) = f(x,y) p_0(x,y) and q(x,y) = g(x,y) q_0(x,y); here p_0 and q_0 have 1 for a lowest degree term and are therefore invertible. So , so p/q = f/g * p_0/q_0, where p_0/q_0 \in K[[x,y]] and f/g is a ratio of homogeneous polynomials. (f/g)(p_0/q_0). If you like, you can we dehomogenize them and f,g we can write f/g = y^n f_0/g_0, where n \in \mathbb{Z} and f_0, g_0 \in K[x/y]. So Likewise, we can write any expression f(x,y)^{-1} x^k y^l (such as appears in p_0) in the form (x/y)^k f_0(x/y)^{-1} y^m, and the same for the terms of q_0 using g_0. Note that p_0, q_0 have constant term 1, and are therefore invertible series in y having coefficients in K(x/y). Summing up, we have K((x,y)) = K(y) K(x/yK(x/y)((y)) K[[x,y]]\cong K(z)((y)). In this sense, your wrong example was quite unlucky, though understandably, since it is hard to distinguish rational functions from non-rational ones by their laurent series. The answer of darij grinberg is basically doing just that. An infinitude of examples then arises by letting \ell(x,y) be any Laurent polynomial in x/y which is not a rational function. Examples:$$\ell(x,y) = \exp(x/y) = \sum_{n = 0}^\infty \frac{1}{n!} x^n y^{-n}\ell(x,y) = \sum_{n = 0}^\infty b_n x^n y^{-n} where in the second example, $b_n$ refers to a random sequence of bits. Almost surely, this will not be rational. Unlike the first one, this actually works when $K$ has positive characteristic.