Let $\alpha$ be an algebraic integer which satisfies the polynomial $X^5 + 2x + 2=0$. Then we have $\mathbb{Z}[\alpha]$ is isomorphic to $\frac{\mathbb{Z}[X]}{X^5 +2X +2}$. I then want to work out whether $(5)$ is a prime ideal in this ring.
My attempt: $(5)$ a prime ideal iff $\frac{\mathbb{Z}[\alpha]}{(5)}$ is an integral domain. However, we see that $\frac{\mathbb{Z}[\alpha]}{(5)}$ has only a finite number of elements. Hence if it is an integral domain, it is also a field as finite integral domains are always fields. Thus in this case $(5)$ is a prime ideal iff it is maximal. but we can quickly verify that it is not in fact maximal, e.g. $(2X, 5)$. We see that $(X^5 +2X +2, 5, 2X)$ is not equal to $\mathbb{Z}[X]$ and hence $(2X,5)$ is an ideal of $\frac{\mathbb{Z}[X]}{X^5 + 2X + 2}$
My question is then: (1) does my above proof work? and (2) is there a more elegant or shorter way of solving it if this does indeed work?
To solve this problem, recall an ideal $I$ is prime iff $R/I$ is an integral domain.
we observe: $\mathbb{Z}[\alpha]/(5)$ is isomorphic to $\frac{\mathbb{Z_5}[X]}{(X^5+2X+2)}$. This can seen by using that $\mathbb{Z}[\alpha]$ is isomorphic to $\frac{\mathbb{Z}[X]}{(X^5 +2X+2)}$.
However, $X^5 + 2X + 2$ has a root, $1$, in $\mathbb{Z}_5$. As $X-1$ has leading term which divides that of $X^5 + 2X + 2$, we therefore find that $X-1$ divides $X^5 +2X +2$. Hence $\frac{\mathbb{Z_5}[X]}{(X^5 +2X+2)}$ is not an integral domain, as we found a non zero polynomial $P$ such that $P(X)(X-1)=0$
(directly solving for this polynomial P gives $P(X) = X^4 + X^3 +X^2 + X + 3$. the final term may cause confusion - but $-3=2 \bmod 5$.)
P.S. I wrote this up after seeing Bernard's hint.