Prove that $F(\alpha)\cong F[x]/(m_\alpha(x))$.

Question

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?

Answer 1

First of all we prove a lemma then we prove the main theorem

Lemma: Let $L/K$ is a field extension. Let $\alpha\in L$ is algebraic over $K$. Then the minimal polynomial in $\alpha$ over $K$ is unique and irreducible.

Proof: let $m_\alpha(x)$ be a minimal polynomial with degree $n$. By the uniqueness of minimal polynomial we have that $m_\alpha(x)$ is unique minimal polynomial. Let us assume that leading coefficient of $m_\alpha(x)$ is $1$ (because $K$ is a field). Assuming the contrary. Suppose $m_\alpha(x)$ is reducible. Then there exist non-constant polynomial $g(x),h(x)\in K[x]$ such that: $m_\alpha(x)=g(x)h(x).$ $\Rightarrow$ $g(\alpha)h(\alpha)=0$. Then either $h(\alpha)=0$ or $g(\alpha)=0$(Since $L$ is an integral domain). contradict the minimality of $m_\alpha(x)$.

Proof of the main theorem

Proof: 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:

$$\phi:F[x]/(m_\alpha(x))\to F(\alpha).$$ $\alpha\in F(\alpha)$ is algebraic over $F$. Then the minimal polynomial $m_\alpha(x)$ over $F$ is unique and irreducible polynomial(By above Lemma). Therefore $F[x]/(m_\alpha(x))$ is a field. the map $\phi$ is not the zero map .  Therefore , $\phi$ is injective ring homomorphism. Since $\operatorname{Im}\phi$ is isomorphic to its image which is a subfield of $F(\alpha)$(By $1^{st}$ isomorphosm theorem). By the definition of $F(\alpha)$ the map $\phi$ must be surjective.  It follows that $\phi$ is isomorphism. Hence $F[x]/(m_\alpha(x))\cong F(\alpha)$

Answer 2

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 $F[\alpha]$.