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May 18, 2016 at 6:42 history edited GH from MO
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Apr 18, 2011 at 18:46 vote accept zamanjan
Apr 18, 2011 at 2:11 comment added GH from MO @Gerald: This problem is purely algebraic. Many formal power series don't converge at all (off the center) yet they participate in the problem.
Apr 18, 2011 at 2:00 answer added GH from MO timeline score: 15
Apr 18, 2011 at 1:56 comment added Gerald Edgar $$e^{zw} = \exp(\exp(\log z + \log w))$$ I left out the "rational" part of the question. And consideration of domains...
Apr 18, 2011 at 1:10 comment added KConrad Michael: you want to show $e^{z^2}$ is not a rational function of $z$. Or instead let $w = 2$ instead of $w = z$.
Apr 17, 2011 at 23:56 answer added user12806 timeline score: 4
Apr 17, 2011 at 20:10 comment added Martin Brandenburg 1+, very interesting question. I've also worked on that to find explicit examples that the functor from complex manifolds to complex locally ringed spaces does not preserve products.
Apr 17, 2011 at 19:51 comment added zamanjan Thanks Fedor, indeed my original title was not good.
Apr 17, 2011 at 19:33 comment added Fedor Petrov @Gerald: I've tried to make the title more related to the question
Apr 17, 2011 at 19:32 history edited Fedor Petrov CC BY-SA 3.0
title changed
Apr 17, 2011 at 19:24 comment added Gerald Edgar Presumably the vote to close is because the question in the title is easy as Michael showed. (Homework level for a first course in complex variables.) So the title should be changed!
Apr 17, 2011 at 19:03 comment added user12806 Zamanjan search to prove that $e^{zw} \neq Q(f_1(z),...,f_k(z),g_1(w),...,g_n(w))$ for all rational function $Q$ and all series $f_1(z),...,f_k(z),g_1(w),...,g_n(w)$.
Apr 17, 2011 at 18:53 comment added Michael Lugo For the particular case you're interested in, wouldn't it suffice to let $z = w$ and show that $e^{2z}$ can't be written as a rational function of $z$?
Apr 17, 2011 at 17:29 history asked zamanjan CC BY-SA 3.0