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Sep 10, 2014 at 21:47 vote accept Siddharth Venkatesh
Apr 4, 2014 at 8:26 comment added Ricky Ah, yes! Merci beaucoup Jérôme !
Apr 4, 2014 at 8:22 comment added Jérôme Poineau @Ricky: Assume that $F \times G$ is a quotient of some noetherian domain $R$. Since $F$ is finite and $F$ is a quotient of $R$, the result tells you that the cardinality of $R$ is at most the continuum. But then $R$ cannot have a quotient that is as big as $G$.
Apr 4, 2014 at 7:58 comment added Ricky Probably a stupid question, but why this result implies that $F \times G$ is not the quotient of a noetherian domain?
Apr 4, 2014 at 2:10 comment added Siddharth Venkatesh I think Jérôme's example (given the correctness of the Harbarter reference for the Noetherianness, which I haven't gone through yet) answers that with a terrific example.
Apr 3, 2014 at 8:05 comment added abx So, finally, we don't know the answer for $\Bbb{Q}\times \Bbb{C}$ ...
Apr 2, 2014 at 19:49 history edited Joël CC BY-SA 3.0
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Apr 2, 2014 at 19:20 history edited Joël CC BY-SA 3.0
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Apr 2, 2014 at 18:55 comment added Eric Wofsey Very nice! I am reminded of a similar cardinality constraint I found in this answer.
Apr 2, 2014 at 18:46 comment added abx I think so too.
Apr 2, 2014 at 18:12 comment added Joël @abx I think Will is wrong. The proof of Lemma 2.1 is very short and simple, and could be given here for the sake of completeness (if user46855 agrees, I can edit his answer to add that proof). Great answer by the way.
Apr 2, 2014 at 15:25 comment added abx Interesting, this seems to contradict Will's answer (with his notation, take $R_1=\mathbb{F}_p$, $R=R_2=$ a very big field of characteristic $p$). Who is wrong?
Apr 2, 2014 at 14:33 history answered user46855 CC BY-SA 3.0