For any proper ideal $I\subseteq R = \mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$, $R/I$ has at least $4$ elements. Given the ring $R = \mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$, how can one (begin to) prove that for any proper ideal $I$ of $R$, $R/I$ has at least 4 elements? 
I checked that the polynomial $f$ over $\mathbb{Z}$ given by $f(x) = x^2 - x + 5$ has no roots in either $\mathbb{F}_2$ or $\mathbb{F}_3$. This was given as a hint, but I don't know how to use it.
This question arose as I was given an outline to prove eventually that $R$ is not a Euclidean domain. 
 A: Note that $R/I$ is always a finite ring because every prime ideal is a maximal rank $\Bbb Z-$sub-module, but then if $I\cap \Bbb Z$ is an ideal of $\Bbb Z$, we see that the cardinality of $R/I$ is at least that of $\Bbb Z/(I\cap\Bbb Z)$. So it is enough to note that if $I\ne R$, we have it is contained in some maximal ideal, which then will intersect $\Bbb Z$ in a maximal ideal, which is to say an ideal generated by a prime number $p\in\Bbb Z$. So it is sufficient to show this for maximal ideals, as $I\subseteq J\implies |R/J|<|R/I|$. Now

$$R\cong \Bbb Z[x]/(x^2-x+5)$$

And if $I\subseteq M$ is contained in the maximal ideal $M$ which has the property that $M\cap \Bbb Z=(p)$ we see that

$$R/M\cong \Bbb Z[x]/(x^2-x+5,p)$$

which, as a module over $\Bbb F_p$ has a rank, either $1$ or $2$ depending on whether or not there is an element in $\Bbb F_p$ which satisfies $x^2-x+5=0$. Since this is false for $p=2,3$, we see that the module has rank $2$, and therefore cardinality $p^2\ge 4$. For $p>3$, the cardinality is at least $p>5$, proving the result.
A: A proper ideal of norm $ < 4 $ would have norm $ 2 $ or $ 3 $, so it would be a prime ideal lying above $ 2 $ and $ 3 $. However, these primes are inert in $ \mathbf Q(\sqrt{-19}) $, as can be directly checked by Dedekind's factorization criterion, so there are no such ideals.
