Is $(\mathbb{Z}_{(2)}[X])/(X^2+1)$ local? I am looking for an example of a discrete valuation ring $R$ and a polynomial $f$ such that $f$ is irreducible over the field of fractions of $R$ and $R[X]/(f)$ is not local.
I was thinking of $R=\mathbb{Z}_{(2)}$ and $f=X^2+1$, where $\mathbb{Z}_{(2)}$ is the localization of the integers under the prime ideal $(2)$. Clearly $R$ is a DVR and $f$ is irreducible. Also, $R[X]/(f)$ is isomorphic to $\mathbb{Z}_{(2)}[i]$. 
I think that this quotient ring is not local, since $\mathbb{Z}[i]$ is not. However, I do not know how to prove this. Can anyone help me to do so?
 A: The prime $(2)\subseteq \mathbb{Z}$ is ramified in $\mathbb{Z}[i]$, because we can write it as
$$ 2\mathbb{Z}[i]=(1+i)^{2} \subseteq \mathbb{Z}[i] $$
This means that there is only one prime above $(2)$ in $\mathbb{Z}[i]$, namely $(1+i)$. If we take any other non-zero prime $\mathfrak{p}\subseteq \mathbb{Z}[i]$, its intersection with $\mathbb{Z}$ will be some odd prime $(p)$ which is generated by a unit in $\mathbb{Z}_{(2)}$. Hence any other non-zero $\mathfrak{p}\subseteq \mathbb{Z}_{(2)}[i]$ contains a unit and is therefore the whole ring $\mathbb{Z}_{(2)}[i]$. We deduce from this that $\mathbb{Z}_{(2)}[i]$ is a local ring with maximal ideal $(1+i)$.
Added: On the other hand, consider the extension $\mathbb{Z}_{(5)}[i]$ of the local ring $\mathbb{Z}_{(5)}$. The prime $(5)\subseteq \mathbb{Z}$ is split in $\mathbb{Z}[i]$, because we can write it as
$$ 5\mathbb{Z}[i]=(5,i+2)(5,i-2) \subseteq \mathbb{Z}[i] $$
[Side note: if you are wondering how to find these expressions, take a look at Kummer's theorem on ramification of primes in certain extensions of Dedekind domains.]
This means that there are only two primes in $\mathbb{Z}[i]$ above $(5)$, namely $(5,i+2)$ and $(5,i-2)$. These two ideals are different non-zero prime ideals (=maximal ideals, as we are in a Dedekind domain). So $\mathbb{Z}_{(5)}[i]$ is not a local ring.
We can also say as before (although we don't need it anymore) that any other non-zero prime ideal in $\mathbb{Z}[i]$ will be above some prime number $p\in\mathbb{Z}$ different from $5$, and therefore it will become trivial in $\mathbb{Z}_{(5)}[i]$. So the only non-zero primes in $\mathbb{Z}_{(5)}[i]$ were actually $(5,i+2)$ and $(5,i-2)$.
Remark: There are a bunch of general results that we have implicitly used in those arguments. I will mention them for the sake of completeness:


*

*Correspondence between primes in the localization and primes disjoint with the corresponding multiplicative subset.

*Localization of a Dedekind domain is still a Dedekind domain.

*Every non-zero prime ideal of a Dedekind domain is maximal.

*If $A$ is a Dedekind domain with fraction field $K$, $L/K$ is a finite separable field extension and $B$ is the integral closure of $A$ in $L$, then $B$ is also a Dedekind domain. Edit: the separability assumption isn't actually necessary due the to Krull-Akizuki theorem. Thanks to Georges for the remark (see comments below).

*Integral closure commutes with localization.
