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Aug 3, 2020 at 1:03 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Jul 4, 2020 at 0:16 answer added tomasz timeline score: 1
Jun 11, 2016 at 10:46 comment added matthias.p @YCor But in $(\mathbf Z/4\mathbf Z)[t]/(t^2,2t)$, $t$ is not a prime element, because $2\cdot2\in(t)$, but $2\notin(t)$.
Jun 10, 2016 at 16:16 comment added YCor And it's false also in artinian rings (replace $\mathbf{Z}$ with $\mathbf{Z}/4\mathbf{Z}$), and in reduced noetherian rings (replace $\mathbf{Z}[t]/(t^2,2t)$ with $\mathbf{Z}[t]/(t^2-2t)$.
Jun 10, 2016 at 16:15 comment added matthias.p @YCor Thanks! So even in Noetherian rings this is false (while I've found books that state the opposite...)
Jun 10, 2016 at 15:49 comment added YCor $R=\mathbf{Z}[t]/(t^2,2t)$; $k=2$, $a=2$, $b=p=t$.
Jun 10, 2016 at 15:43 review First posts
Jun 10, 2016 at 16:15
Jun 10, 2016 at 15:39 history asked matthias.p CC BY-SA 3.0