Let $I = \{p(x) ∈ \Bbb Z[x]: 5\mid p(0)\}$. Prove that $I$ is an ideal of $\Bbb Z[x]$ by finding a ring morphism from $\Bbb Z[x]$ to $\Bbb Z_5$ with kernel $I$. Prove that $I$ is not a principal ideal.
I'm unsure of how to go about finding a ring morphism for this problem. Any help would be great, thank you in advance!
Consider $f:\mathbb{Z}[X]\rightarrow\mathbb{Z}/5$ defined by $f(P)=P(0)$, $\operatorname{Ker}f =I$.
Suppose that $I$ is principal, it is generated by $P_0$ of degree $n$, we have $5=QP_0$ implies that $n=0$ this implies that $P_0$ is a constant divided by $5$. $X+5=RP_0$ impossible since $5$ divides the biggest coefficient of $RP_0$ since $5$ divides the constant number $P_0$.