Is $\mathbb{Z}[x] / (x - a) \cong \mathbb{Z}$ where $a \in \mathbb{Z}$?
I'm trying to get more comfortable with these questions so please critique my attempt, or if you can, suggest other (better) ways of doing this if you can, thanks.
The two rings are isomorphic. Consider the evaluation map $\phi : \mathbb{Z}[x] \to \mathbb{Z}, x \mapsto a$. Since this is an evaluation map it is a homomorphism and it is surjective since $\phi(n) = n$ for any $n \in \mathbb{Z}$.
Now we show $\ker(\phi) = (x - a)$. If $f \in \ker(\phi)$, then $f(a) = 0$, thus $x - a \mid f$ hence $f \in (x - a)$. If $f \in (x-a)$, then $f(x) = g(x)(x-a)$ hence $f(a) = 0$, so $f \in \ker(\phi)$. Thus by the first isomorphism theorem $\mathbb{Z}[x] / (x-a) \cong \mathbb{Z}$.