Timeline for Liftings and closed immersions

7 events

when toggle format	what		by	license	comment
May 18, 2018 at 21:05	comment	added	Piotr Achinger		By construction x and y are zero mod p because of the choice of $\bar I=(x,y)$.
May 18, 2018 at 15:41	comment	added	Neil Epstein		I think if you alter $A$ so that $A= \mathbb Z_p[\![x,y]\!] / (xy-p)$, then everything you want holds. This is because now $A$ is local with max ideal $m=(x,y,p)$, and since $I$ doesn't contain $p$, $I \subseteq m$. The map $A \rightarrow B$ is then a local homomorphism, so the image is in $pB$, the maximal ideal of $B$.
May 18, 2018 at 12:31	comment	added	Neil Epstein		Why do $x,y$ have to land in $pB$?
May 18, 2018 at 5:52	history	edited	Piotr Achinger	CC BY-SA 4.0	added 1 character in body
May 18, 2018 at 5:51	comment	added	Piotr Achinger		Thanks! I meant solutions with $x$ and $y$ zero modulo $p$ ;)
May 17, 2018 at 20:49	comment	added	Neil Epstein		There are at least some solutions to $xy=p$ in $(\mathbb Z / p^2 \mathbb Z)^2$, for instance one always has $x=1, y=p$. Or in the special case $p=5$, you have the nontrivial solution $x=3$, $y=10$.
May 17, 2018 at 19:04	history	answered	Piotr Achinger	CC BY-SA 4.0