Irreducible of $\mathbb Z/p\mathbb Z$. What are the irreducible number of $\mathbb Z/p\mathbb Z$ ? It looks strange since in a field it looks complicate to talk about irreducible since all element are invertible. So if my question has no sense, how can I prove using Eisenstein that $X^5+2X+2$ is irreducible over $\mathbb Z/3\mathbb Z$ ?

First of all, $f(X)=X^5+2X+2$ does not have roots in $\mathbb Z/ 3\mathbb Z$ ($f(0)=2$, $f(1)=2$, $f(-1)=2$). Suppose by contradiction that the polynomial can be factored. If the polynomial is reducible, then its factors have degree higher than one; the only possibility is that 
$$  X^5+2X+2=(X^2+aX+b)(X^3+cX^2+dX+e)  = \\ X^5+(a+c)X^4+(d+ac+b)X^3+(e+ad+bc)X^2+(ae+bd)X+eb$$
for some $a,\ldots,e\in \mathbb{Z}/3\mathbb Z$. So
$$ \begin{cases} a+c=0 \\ ac+b+d=0 \\ ad+bc+e=0 \\ ae+bd=2 \\ eb=2 \end{cases} $$
From the last equation, either $e=1,b=2$ or $e=2,b=1$. Suppose the latter:
$$\begin{cases} c=-a \\ -a^2+d+1 =0 \\ ad-a+2=0 \\ 2a+d=2\end{cases} \implies \begin{cases}c=-a \\a(a+2)=0 \\-2a^2+a+2=0 \\d=2-2a \end{cases}$$
but the third equation has no solutions in $\mathbb{Z}/3\mathbb{Z}$. The other case is similar.
