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edited Aug 1 2010 at 11:58 |
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.)
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edited Jul 31 2010 at 18:24 |
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.
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edited Jul 31 2010 at 14:59 |
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.
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edited Jul 31 2010 at 14:45 |
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 $p/q = f/g * p_0/q_0$, where $p_0/q_0 \in K[[x,y]]$ and $f/g$ is a ratio of homogeneous polynomials. If you like, you can dehomogenize them and write $f/g = y^n f_0/g_0$, where $n \in \mathbb{Z}$ and $f_0, g_0 \in K[x/y]$. So $K((x,y)) = K(y) K(x/y) K[[x,y]]$. 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 $p(x,y)$ \ell(x,y)$ be any Laurent polynomial in $x/y$ which is not a rational function. Examples:
$$p(x,y) $\ell(x,y) = \exp(x/y) = \sum_{n = 0}^\infty \frac{1}{n!} x^n y^{-n}$$
$$p(x,y) $\ell(x,y) = \sum_{n = 0}^\infty b_n x^n y^{-n}$$
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.
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answered Jul 31 2010 at 14:35 |
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 $p/q = f/g * p_0/q_0$, where $p_0/q_0 \in K[[x,y]]$ and $f/g$ is a ratio of homogeneous polynomials. If you like, you can dehomogenize them and write $f/g = y^n f_0/g_0$, where $n \in \mathbb{Z}$ and $f_0, g_0 \in K[x/y]$. So $K((x,y)) = K(x/y) K[[x,y]]$. 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 $p(x,y)$ be any Laurent polynomial in $x/y$ which is not a rational function. Examples:
$$p(x,y) = \exp(x/y) = \sum_{n = 0}^\infty \frac{1}{n!} x^n y^{-n}$$
$$p(x,y) = \sum_{n = 0}^\infty b_n x^n y^{-n}$$
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.
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