Let $A$ be a flat $\mathbf{Z}_p$-algebra, $\overline{I}\subset A/p$ an ideal.
Can we find an ideal $I\subset A$ such that
- $I$ mod $p$ = $\overline{I}$
- $I$ does not contain $p$.
It's harder than it looks. Am I missing something?
1 Answer
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Consider $A=\mathbf{Z}_p[x,y]/(xy-p)$ and $\bar I = (x,y)$. Suppose $I$ lifts $\bar I$ and does not contain $p$; let $I'=(I,p^2)$, which has the same property as $I$. Then the $\mathbf{Z}/p^2$-algebra $B=A/I'$ satisfies $B/p\cong \mathbf{F}_p$, $p^2 B = 0$ and $pB\neq 0$, so we have $B\cong \mathbf{Z}/p^2\mathbf{Z}$. The map $A\to B=\mathbf{Z}/p^2\mathbf{Z}$ furnishes a solution $(x,y)\in (p\mathbf{Z}/p^2\mathbf{Z})^2$ of the equation $xy=p$, but there are none. Contradiction.
answered May 17, 2018 at 19:04
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1$\begingroup$ 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 $x=3$, $y=10$. $\endgroup$ Commented May 17, 2018 at 20:49
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$\begingroup$ Thanks! I meant solutions with $x$ and $y$ zero modulo $p$ ;) $\endgroup$ Commented May 18, 2018 at 5:51
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$\begingroup$ Why do $x,y$ have to land in $pB$? $\endgroup$ Commented May 18, 2018 at 12:31
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$\begingroup$ 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 $B$. $\endgroup$ Commented May 18, 2018 at 15:41
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$\begingroup$ By construction x and y are zero mod p because of the choice of $\bar I=(x,y)$. $\endgroup$ Commented May 18, 2018 at 21:05