Timeline for If $A$ is an integer ring such that each $P \in A_L[X]$ has a finite number of zeros in $A$, is $A$ commutative?

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Dec 13, 2015 at 21:31	comment	added	YCor		I would try anyway the case of a group algebra over a free group (as suggested by David Speyer) or over the Heisenberg group. I tend to believe that these are counterexamples to your conjecture.
Dec 13, 2015 at 11:34	comment	added	Stabilo		I am sorry Yuan, I haven't the mathematical background to understand your comment... Should I have a look at Lie algebras?
Dec 12, 2015 at 17:44	comment	added	Qiaochu Yuan		Have you looked at the universal enveloping algebra of a nonabelian Lie algebra?
Dec 12, 2015 at 11:17	comment	added	Stabilo		Of course... I meant not necessarily commutative Integral domains.
Dec 12, 2015 at 11:09	history	edited	YCor		edited tags; edited tags
Dec 12, 2015 at 11:07	comment	added	YCor		but English "integer" is not a translation of French "intègre" in any meaning!
Dec 12, 2015 at 11:03	history	edited	Stabilo	CC BY-SA 3.0	added 132 characters in body
Dec 11, 2015 at 18:44	comment	added	Qiaochu Yuan		The contrapositive of your conjecture is that if $A$ is noncommutative, then there is some polynomial with an infinite number of zeroes. This is clearly false if $A$ is finite, so you'll have to tell us whether "integer ring" excludes that possibility.
Dec 11, 2015 at 18:34	comment	added	David E Speyer		What is an integer ring? In particular, are finite rings integer rings? What about the group algebra $\mathbb{Z}[G]$, where $G$ is a free group?
Dec 11, 2015 at 17:27	history	asked	Stabilo	CC BY-SA 3.0