# Example of a characteristic zero local ring with a quotient of positive characteristic

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.

• 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 $char=p$. – user45765 Sep 21 '18 at 21:45
• Take $\mathbb{Z}[X]$ localized at the ideal generated by $p$ and $X$. – 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.
• Completion here is different from localization. Completion does make it local. However the last isomorphism is not correct. You do not have infinite series $Z[x]_{(x,2)}$. – user45765 Sep 22 '18 at 12:43
• @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 $\mathbb{Z}_{(2)}[\![x]\!]$. – Mike Pierce Sep 22 '18 at 22:53
• 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. – user45765 Sep 22 '18 at 22:59