Timeline for When does prime elements remain prime in certain integral extension

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May 21, 2018 at 4:27	history	edited	user111492	CC BY-SA 4.0	added 207 characters in body
Jan 28, 2018 at 23:49	comment	added	Raymond Cheng		@R.vanDobbendeBruyn: Great, yes: $R/p$ still has the class represented by $t(t^2 - 1)$. Thanks (:
Jan 28, 2018 at 23:27	comment	added	R. van Dobben de Bruyn		@RaymondCheng: your element $p$ is not prime (but it is irreducible). The ring is $k[x,y]/(y^2-x^3-x^2)$, so if $p = x$ then the quotient is $k[y]/(y^2)$, not $k$.
Jan 28, 2018 at 23:04	comment	added	Raymond Cheng		No, even if $R$ is assumed to be Noetherian. For an example, let $k$ be a field and set $R = k[t^2 - 1, t(t^2 - 1)]$. Then $p = t^2 - 1$ is a prime element since $R/pR = k$ is an integral domain. However, the normalization $\overline R \cong k[t]$ and $p = (t - 1)(t + 1)$ there. This example comes from geometry: $R$ here is the coordinate ring of a nodal cubic, and the prime element $p$ corresponds to the node at the origin. When you normalize, the node splits off into two points, and so the pullback of this prime element factors into two smaller things.
Jan 28, 2018 at 21:29	answer	added	R. van Dobben de Bruyn		timeline score: 7
Jan 28, 2018 at 9:48	history	edited	user111492	CC BY-SA 3.0	added 85 characters in body; edited tags
Jan 28, 2018 at 7:22	history	edited	user111492	CC BY-SA 3.0	added 304 characters in body
Jan 28, 2018 at 7:06	history	asked	user111492	CC BY-SA 3.0