How do I prove this? $$\Bbb Z[x]/(p)\cong \Bbb F_p[x]$$
This implies that $(p)$ must be equal to $p\Bbb Z$ where $p$ is prime. Why is the only prime element that makes this true $p$ where $p$ is prime? Why is $p\Bbb Z$ generated by $(p)$?
I know that $\Bbb Z/ p\Bbb Z\cong \Bbb F_p$ when $p$ is prime. I don't see why $p\Bbb Z\cong (p)$.
Consider the homomorphism of rings $\phi: \mathbf{Z}[x] \to \mathbf{F}_{p}[x]$ given by $\phi(f(x))=f(x)$ mod $p$ (taking the residues mod $p$ of the coefficients). $\phi$ is clearly surjective. An element is in the kernel if and only if it gets mapped to the zero polynomial in $\mathbf{F}_{p}[x]$ hence if and only if all its coefficients are multiples of $p$, so the kernel is the ideal $(p)=p\mathbf{Z}[x]$. The rest follos from the first isomorphism theorem. (In this case, the ideal is $p\mathbf{Z}[x]$ and not $p \mathbf{Z}$)