Let $\alpha$ be integral, $f=\min(\alpha,\mathbb{Q})$, and $a,p\in \mathbb{Z}$ be such that $p$ is prime and $f(a)\equiv 0\mod p$. I would like to show that the map $$\phi: \mathbb{Z}[\alpha]\to \mathbb{Z}/p$$ given by $z\mapsto \bar{z}$ and $\alpha\mapsto \bar{a}$ is a ring homomorphism.
My attempt: Let $d=\deg(f)$. I do know $\mathbb{Z}[\alpha]\cong \mathbb{Z}[x]/\langle f(x)\rangle$ and that each element of $\mathbb{Z}[\alpha]$ can be uniquely written as $$\sum_{i=0}^{d-1}z_i\alpha^i$$ for some $z_i\in \mathbb{Z}$. Using this fact makes it very easy to show $\phi$ respects addition. I am having trouble showing it respects multiplication. Any help is appreciated.
If you know that the ring structure stays preserved under quotients by ideals, you know that this map preserves the ring structure, so it is sufficient to show that it is well defined. If you do not know this, you need to prove this.
If you are allowed to use the isomorphism in your attempt, then we can define this isomorphism as the map $$ \psi : \Bbb Z[x]/(f(x)) \to \Bbb Z[\alpha],\; x \mapsto \alpha. $$ Then $$ \phi \circ \psi : \Bbb Z[x]/(f(x)) \to \Bbb Z/(p),\; x \mapsto \bar a $$ This is the map that evaluates a polynomial $g \in \Bbb Z[x]/(f(x))$ at $x = \bar a$. Now it is clear that if this polynomial $g$ is inside the ideal $(f(x))$ then $f(x)$ divides it, so $g(\bar a) \equiv 0 \mod p$. So $(f(x)) \subset \ker \phi \circ \psi$, which shows the map is well defined.