There is a natural homomorphism
$$\phi:F[x]\to F(\alpha)$$
$$\phi(p(x))=p(\alpha)$$
obtained by a mapping which leaves coefficient of polynomial unchanged and sends $x$ to $\alpha$.
Clearly $m_\alpha(x)\in \operatorname{ker}\phi$. Where $\operatorname{ker}\phi=\{p(x)\in F[x]:p(\alpha)=0\}$. So we obtain an induced homomorphism:
$$\phi:F[x]/(m_\alpha(x))\to F(\alpha).$$
$F[x]$ is P.I.D. which is generated by unique monic polynomial $m_\alpha(x)$. which is irreducible since the homomorphism $\phi$ is a surjection onto the field $F(\alpha)$.
Therefore $F[x]/(m_\alpha(x))$ is a field. the map $\phi$ is not the zero map .
Therefore , $\phi$ is injective ring homomorphism. By the definition of $F(\alpha)$ the map $\phi$ must be surjective.
It follows that $\phi$ is isomorphism. Hence $F[x]/(m_\alpha(x))\cong F(\alpha)$.
Is the proof correct?
Best Answer
The homomorphism $\psi$ is defined in a confusing way. Actually, it is defined as follows: for any polynomial $p(X)\in F[X]$, we set $$\psi(p(X)))=p(\alpha)$$ which amounts to defines the image of the indeterminate $X$ as $X\mapsto\alpha$.
Now $\,\ker\psi=\{p(X)\in F[X]\mid p(\alpha)=0\}$, i.e. the ideal of polynomials which have $\alpha$ as a root. As $F[X]$ is a P.I.D., it is generated by a unique minimal polynomial, which is irreducible since the homomorphism $\psi$ is a surjection onto the field $F[\alpha]$.