In a worked exam the following three quotients rings: \begin{equation} R_1=\frac{\mathbb{Z}_5[x]}{(x^2)},\space R_2=\frac{\mathbb{Z}_5[x]}{(x^2+1)}, \space R_3=\frac{\mathbb{Z}_5[x]}{(x^2+2)} \end{equation} are given. The examples determines which of these rings are fields?
In the solution it says that $x^2$ and $x^2+1$ are not irreducible in $\mathbb{Z}_5[x]$ (since that the latter has $2$ as a root in $\mathbb{Z}_5$), while $x^2 + 2$ is irreducible (has no roots in $\mathbb{Z}_5$)
I am having a hard time grasping the concept of an irreducible polynomial. For example in the example above how does it determine that $x^2$ and $x^2+1$ are reducible? I also do not understand how $2$ is a root of $x^2+1$ as $2^2+1=5 \neq 0 (?)$
I know that if $R$ is a Principal Ideal Domain then $R/(a)$ is a field iff $a \in R$ is irreducible so I understand the method it is following, but I get lost when trying to do it on my own.
A degree 2 polynomial in $\mathbb{Z}_5[X]$ is reducible iff it has a root. We know by Root Theorem that if it does have a root it is reducible. If it is reducible then it can be expressed as $aX^2 + bX + c = (a_1X + b_1)(a_2X + b_2).$ Clearly the product has a root.
So because $x^2, x^2 + 1$ have roots (0, $\pm2$ respectively), we can conclude those are reducible. In fact, we factor them into $x^2 = x \cdot x$ and $x^2 + 1 = (x + 2)(x - 2).$