Showing that a ring homomorphism is injective Let $f\in\mathbb{Z}[X]$ be a monic irreducible polynomial, $\alpha$ a root of $f$ and $k\in \mathbb{Z}$. Show that the map
$$\varphi : \mathbb{Z}/f(k)\mathbb{Z} \to \mathbb{Z}[\alpha]/(k - \alpha)\mathbb{Z}[\alpha], \quad z + f(k)\mathbb{Z} \mapsto z + (k - \alpha)\mathbb{Z}[\alpha]$$
is a ring-isomorphism.
I'm looking for a direct proof, using the theory taught in standard classes in abstract algebra.
I was able to show that $\varphi$ is well-defined. The homomorphism property is clear. But now I have problems showing that $\varphi$ is injective, which I want to do by showing that the kernel of $\varphi$ is trivial.
 A: I know you want a direct proof, but I can't help but point out what I think is the clearest way to think of this isomorphism.  Note that $\mathbb{Z}[\alpha] \cong \mathbb{Z}[x] / (f(x))$.  Thus
$$
\mathbb{Z}[\alpha]/(k-\alpha) \cong \mathbb{Z}[x]/(f(x), k - x) \cong \mathbb{Z} / (f(k)).
$$
The first isomorphism comes from reducing $\mathbb{Z}[x]$ by $(f(x))$ first, then quotienting out by $(k - \alpha)$; the second isomorphism comes from reducing $\mathbb{Z}[x]$ by $(k - x)$ first, then quotienting out by $(f(k))$.
As for the direct proof, it is equivalent to show that the (principal) ideal
$$I = \{ n \in \mathbb{Z} : n = c \cdot (k - \alpha) \text{ for some } c \in \mathbb{Z}[\alpha] \}$$
is generated by $f(k)$.  Suppose $n = c \cdot (k - \alpha)$, where $c = p(\alpha)$ for some $p(x) \in \mathbb{Z}[x]$.  Consider the polynomial $g(x) = n - p(x)(k-x)$; since $g(\alpha) = 0$, $f(x)$ divides $g(x)$.  So $n = p(x)(k-x) + b(x)f(x)$ for some $b(x) \in \mathbb{Z}[x]$.  Evaluating at $x = k$ gives $n = b(k)f(k)$, or $f(k)$ divides $n$, so $I = (f(k))$.  If you work through this argument, you'll see it is simply making the isomorphisms mentioned above explicit.
