Let $S$ be a commutative ring with identity with $\operatorname{char}S=p$, where $p$ is a prime number. I wonder if we can always find a ring $R$ such that $\operatorname{char}R=0$ and $R/(p)\cong S$.

I think above question is equivalent 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.

Moreover, it'll be more preferable if such a $R$ admits a canonical projection $\varphi:R\twoheadrightarrow S$ in the sense that every ring homomorphism from a ring of characteristic $0$ to $S$ can be factored through $\varphi$.

  • 1
    $\begingroup$ I highly doubt there's a universal lift to characteristic zero as you describe. $\endgroup$ Commented Jun 5, 2015 at 14:11
  • $\begingroup$ @DustanLevenstein I'm not sure about it, either. So I impose it as an addition requirement. $\endgroup$
    – Censi LI
    Commented Jun 5, 2015 at 14:13
  • $\begingroup$ Cross-posted: mathoverflow.net/questions/208963/… $\endgroup$
    – user26857
    Commented Jul 7, 2015 at 12:22

1 Answer 1


For any field $k$ of characteristic $0$ the ring $k\times S$ satisfies $(k\times S) / (p) \cong S $. Then the projection factors through $\varphi$, i.e. there a map $k \times S \to R$ and $(p,0)=p \cdot 1$ is not in the kernel. This implies that $k\times \{ 0\} \to R$ is injective. But then $\vert R \vert \ge \vert k \vert$ for any field $k$ which can not be true.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .