Field Theory : What is wrong with this "homomorphism"?

Question

Let $E/F$ be a field extension and $\alpha \in E$ be algebraic over $F$. Let $m(x) \in F[x]$ be irreducible and such that $m(\alpha) \neq 0$. Define $$ \phi : F(\alpha) \to F[x]/(m) $$ by $\phi(f(\alpha)) := \overline{f}$.

Then we have

1. $\phi(f_1(\alpha) f_2(\alpha)) = \phi((f_1f_2)(\alpha)) = \overline{f_1f_2} = \overline{f_1} \,\overline{f_2} = \phi(f_1(\alpha)) \phi(f_2(\alpha))$
2. $\phi(f_1(\alpha) + f_2(\alpha)) = \phi((f_1+f_2)(\alpha)) = \overline{f_1+f_2} = \overline{f_1}+\overline{f_2} = \phi(f_1(\alpha))+\phi(f_2(\alpha))$
3. $\phi(1) = \overline{1}$

Hence $\phi$ is a field homomorphism. In particular we know $\phi$ must be injective. Also, $\phi(0) = \overline{0}$. However, $\phi(m(\alpha)) = \overline{m} = \overline{0}$ with $m(\alpha) \neq 0$, contradicting the injectivity of $\phi$.

Question: What is wrong?

Answer 1

This homomorphism is not well-defined. Let $g$ be the minimal polynomial of $\alpha$. Then $g(\alpha) = 0$ but $\overline g \ne 0$.