This question was featured on a qualifying exam at my university:

What's an example of a commutative local ring $R$ of characteristic zero, with a non-maximal prime ideal $P$ such that the characteristic of $R/P$ is not zero?

Our favorite example of a local ring, $\mathbb{Z}_{(p)}$, won't work because it's a PID (a DVR in fact) and won't have any non-maximal prime ideas. I think that the ring of power series $\mathbb{Z}_{(2)}[\![x]\!]$ might be an example, but I haven't worked out the details yet.

  • $\begingroup$ He means localize away from $(p,x)$ which is maximal and $p$ is any prime. This gives maximal ideal. $(p,x)$ after localization. Reduction by $p$ gives $char=p$. $\endgroup$ – user45765 Sep 21 '18 at 21:45
  • $\begingroup$ Take $\mathbb{Z}[X]$ localized at the ideal generated by $p$ and $X$. $\endgroup$ – xarles Sep 21 '18 at 21:45

​An example will be the ring $\mathbb{Z}[x]$ localized at $(x,2)$, so $\mathbb{Z}[x]_{(x,2)}$. An important fact here that makes this a reasonable example to come up is that the localization of a ring $R$ at a prime ideal $P$ will be a local ring $R_P$, the max ideal being $P_P$, and furthermore the prime ideals of $R_P$ will all be of the form $Q_P$ for some prime ideal $Q$ of $R$ that is contained in $P$. So for our particular example, we're looking at the chain of prime ideals $(0) \hookrightarrow (2) \hookrightarrow (2,x) \hookrightarrow \mathbb{Z}[x]$. The ideal $(2)_{(2,x)}$ will be prime in $\mathbb{Z}[x]_{(x,2)}$, and since $\mathbb{Z}[x]_{(x,2)}$ is still unital, the quotient of $\mathbb{Z}[x]_{(x,2)}$ by $(2)_{(2,x)}$ will have characteristic $2$.

The ring $\mathbb{Z}_{(2)}[\![x]\!]$ mentioned in the question is an example too, for nearly the same reason: the ideal $(x,2)$ is maximal and $(2)$ is prime.

  • $\begingroup$ Completion here is different from localization. Completion does make it local. However the last isomorphism is not correct. You do not have infinite series $Z[x]_{(x,2)}$. $\endgroup$ – user45765 Sep 22 '18 at 12:43
  • $\begingroup$ @user45765 I'm convinced now that they're not isomorphic; $\mathbb{Z}_{(2)}[\![x]\!]$ has "way more stuff". But I think we have a inclusion $\mathbb{Z}[x]_{(x,2)} \hookrightarrow \mathbb{Z}_{(2)}[\![x]\!]$. Certainly $\mathbb{Z}[x] \hookrightarrow \mathbb{Z}_{(2)}[\![x]\!]$. Then in $\mathbb{Z}[x]_{(x,2)}$ we've formally inverted all polynomials $f \in \mathbb{Z}[x]$ with odd constant term. These polynomial's will be units in $\mathbb{Z}_{(2)}[\![x]\!]$. Where $f^{-1}$ is just a formal inverse in $\mathbb{Z}[x]_{(x,2)}$, it'll have a power series representation in $\mathbb{Z}_{(2)}[\![x]\!]$. $\endgroup$ – Mike Pierce Sep 22 '18 at 22:53
  • $\begingroup$ The correct statement should be $Z[[x]]_{(x,2)}$. Then you have no trouble for that part. You need to know that completion and quotient commutes like localization commuting with quotient. $\endgroup$ – user45765 Sep 22 '18 at 22:59

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