Timeline for Base change of a finite morphism

Current License: CC BY-SA 3.0

13 events

when toggle format	what		by	license	comment
Oct 9, 2017 at 15:55	history	edited	Jason Starr	CC BY-SA 3.0	deleted 2 characters in body
Oct 9, 2017 at 15:23	history	edited	Jason Starr	CC BY-SA 3.0	added 163 characters in body
Oct 9, 2017 at 15:10	history	edited	Jason Starr	CC BY-SA 3.0	added 144 characters in body
Oct 9, 2017 at 14:59	history	edited	Jason Starr	CC BY-SA 3.0	added 2237 characters in body
Oct 9, 2017 at 14:46	history	edited	Jason Starr	CC BY-SA 3.0	added 2237 characters in body
Oct 9, 2017 at 14:41	history	edited	Jason Starr	CC BY-SA 3.0	added 2237 characters in body
Oct 9, 2017 at 12:33	comment	added	Pierre		I wonder why, as stated in the above answer, the closed immersion $\Delta$ failing to be an isomorphism over a every point of $C$, which is non regular locus of ${\mathrm Spec}\,k[u,v]/(u^2 - v^3)$, implies that $X^{\mathrm{nor}} \times_X X^{\mathrm{nor}}$ is not integral. Is this some standard fact in scheme theory?
Oct 9, 2017 at 12:27	comment	added	Pierre		Under the modified assumption ${\cal O}_Y \cap {\cal O}_Z = {\cal O}_X$, there should be NO example.
Oct 9, 2017 at 12:19	comment	added	Pierre		Maybe $4(a^3 - b^3)$ but NOT $2(a^3 - b^3)$.
Oct 9, 2017 at 12:11	comment	added	Jason Starr		For notation's sake, I will denote by $a$ and $b$ the elements $T\otimes 1$ and $1\otimes T.$ Then $R$ equals $k[a,b]/\langle a^2-b^2,a^3-b^3\rangle.$ Thus, the element $n=a-b$ is nilpotent. Indeed $n^3 = (a-b)^3 = a^3-3a^2b+3ab^2-b^3 = a^3 -3(b^2)b + 3a(a^2) - b^3 = 2(a^3-b^3) = 0.$
Oct 9, 2017 at 12:04	comment	added	Pierre		If we write $R \colon= k[T] \otimes_{k[u,v]/(u^2 - v^3)} k[T]$, with the embedding $u \mapsto T^3, v \mapsto T^2$. What is the nilpotent element in R?
S Oct 9, 2017 at 11:01	history	answered	Jason Starr	CC BY-SA 3.0
S Oct 9, 2017 at 11:01	history	made wiki			Post Made Community Wiki by Jason Starr