I have a question about the some rings and fields. Is $\mathbb{Q}[[v^{-1}]] \cap \mathbb{Q}(v)=\mathbb{Q}[[v^{-1}]]$?
Is $\mathbb{Q}[[v^{-1}]] \cap \mathbb{Q}(v)=\mathbb{Q}[[v^{-1}]]$?
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2No. The power series that come from rational functions have several special properties and it's not hard to come up with one without those properties. – 2011-06-01
2 Answers
To amplify Qiaochu's comment: when you expand a rational function into a power series, you get an eventually periodic sequence of coefficients, basically for the very same reason the decimal expansion of a rational number is eventually periodic. Just use long division. Therefore, there are many series in $\mathbb{Q}[[v^{-1}]]$ that are not in $\mathbb{Q}(v)$.
EDIT: this was a bit too quick and a bit too simplistic. See David Speyer's comment below. (I was thinking about generating functions over finite fields).
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1The claim about periodicity almost works in the sense that for all but finitely many primes we can reduce any rational power series $\bmod p$ and the result must be (eventually?) periodic. On the other hand it is straightforward to find a power series which cannot be reduced modulo any prime. – 2011-06-03
Or how about this: $\mathbb{Q}(v)$ is countable and $\mathbb{Q}[[v^{-1}]]$ isn't.
Or how about this: take $e^{v^{-1}} = \sum_{n \ge 0} \frac{v^{-n}}{n!}$. The derivatives of $p(v) e^{v^{-1}}$ never vanish identically (induction), so it can never be a polynomial. Hence $e^{v^{-1}}$ can't be a rational function.
Or, for a counterexample that works in any characteristic, take $f(v) = \sum_{n \ge 0} v^{-n^2}$. If $p(v)$ is a polynomial of degree $d$, then $p(v) f(v)$ has infinitely many nonzero coefficients because the gaps between nonzero terms in $f$ eventually become longer than $d$, so again $f(v)$ can't be a rational function.