All rings here are associative, commutative and unital. By a ring of characteristic zero (resp. of characteristic $p$, for prime $p$) I mean a ring $A$ such that the canonical homomorphism $\mathbb Z\to A$ is injective (resp. factors through $\mathbb{Z}/p\mathbb{Z}$).

A simple observation (from the comments) is that whenever $S$ is a ring of characteristic $p$ then there exists a ring of characteristic zero $R$ and an isomorphism $R/(p)\cong S$: for instance we can simply take $R=S\times\mathbb{Q}$.

Is there a "universal" such construction, in the following sense

Given a ring $S$ of characteristic $p$, does there exist a ring $R$ of characteristic zero $R$ and a fixed isomorphism $R/(p)\cong S$, such that for every ring characteristic zero $R'$ of characteristic zero and homomorphism from a ring of characteristic zero into $S$ can be lifted to a homomorphism $R\to R'$ (i.e., so that the homomorphism $R'\to S$ equals the composite map $R'\to R\twoheadrightarrow R/(p)\cong S$)


Note that such $R\twoheadrightarrow S$ does not have to be minimal (as homomorphisms above are not required to be surjective).

I think that the above question is equivalent (or related) to: if for every $\mathbb Z_{p}$-polynomial algebra $A$ and ideal $I$ of $A$ containing $p$, there exists an ideal $J$ not containing nonzero constants such that $I=(p)+J$. But I'm not sure if the latter simplifies the former.

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    $\begingroup$ This is a cross-post of math.stackexchange.com/questions/1313245/… $\endgroup$ – Censi LI Jun 11 '15 at 7:06
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    $\begingroup$ There's certainly no reason to expect a universal such lift. Already there's no universal lift when $S = \mathbb{F}_4$; here any algebra of the form $\mathbb{Z}[x]/(x^2 - ax - b)$ where $a, b \in \mathbb{Z}$ are odd is a lift. $\endgroup$ – Qiaochu Yuan Jun 11 '15 at 7:16
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    $\begingroup$ For the existence, take $R=S\times\mathbb{Q}$. $\endgroup$ – Laurent Moret-Bailly Jun 11 '15 at 7:21
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    $\begingroup$ What do you mean by characteristic zero ring? $\endgroup$ – Vinteuil Jun 11 '15 at 7:30
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    $\begingroup$ It seems obvious to me that (1) is the correct definition, at least in the context of this question. I don't really understand the votes to close here. $\endgroup$ – Eric Wofsey Jun 11 '15 at 8:27

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