Timeline for Can Schwartz-Zippel be formulated for commutative rings instead of fields?

10 events

when toggle format	what		by	license	comment
Jun 15, 2020 at 7:27	history	edited	CommunityBot		Commonmark migration
S Sep 17, 2015 at 17:07	history	suggested	Anurag		added some more relevant tags
Sep 17, 2015 at 16:50	review	Suggested edits
S Sep 17, 2015 at 17:07
Sep 3, 2015 at 23:54	answer	added	Anurag		timeline score: 5
Nov 8, 2014 at 13:29	history	edited	Thomas Klimpel	CC BY-SA 3.0	Added answer to question 1 based on the observations from the comments
Nov 4, 2014 at 23:02	comment	added	Thomas Klimpel		I now see the next problem: $q\prod_{i=0}^{p-1}(x-i)$ in $\mathbb Z/pq\mathbb Z$ is a non-zero polynomial of total degree $p$, but it evaluates to $0$ for any $x\in\mathbb Z/pq\mathbb Z$. So even Emil Jeřábek's nice version of the lemma/conjecture can't be used directly to probabilistically check $(1+x^n)=1+x^n(\operatorname{mod}n)$, which was one of the motivations for this question in the first place. One can try to go to some extension ring to fix this, but the hoped for/suggested simplicity is lost nevertheless.
Nov 3, 2014 at 18:29	vote	accept	Thomas Klimpel
Nov 3, 2014 at 15:59	answer	added	Emil Jeřábek		timeline score: 10
Nov 3, 2014 at 13:08	comment	added	YCor		For a commutative ring, "subring of a direct product of fields" is equivalent to being reduced (no nonzero nilpotent elements). In a noetherian commutative ring it's also equivalent to "subring of a finite direct product of fields".
Nov 3, 2014 at 8:04	history	asked	Thomas Klimpel	CC BY-SA 3.0