Proposition. properties of $\alpha$ being algebraic over $F$ [1403]

Let $K / F$ be a field extension and $\alpha \in K$. The following are equivalent:

$\alpha \in K$ is algebraic over $F$,
the ring homomorphism $\varphi : F[x] \to K, \varphi \left( a(x) \right) = a(\alpha)$ is not injective,
$[F(\alpha) : F] < \infty$,
$[F[\alpha] : F] < \infty$,
$F(\alpha) = F[\alpha]$.

Proof. sketch [1403A]

(1) $\iff$ (2) is trivial.

(5) $\implies$ (1): Since $F(\alpha) = F[\alpha]$, there exists $c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \in F[\alpha]$ with at least one nonzero coefficient such that it equals $\alpha ^{-1} \in F(\alpha)$. Hence $$ \begin{align*} &\alpha ^{-1} = c_{n}\alpha^{n} + \cdots + c_{1}\alpha + c_{0} \\ \iff &1 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha \\ \iff &0 = c_{n}\alpha^{n+1} + \cdots + c_{1}\alpha^{2} + c_{0}\alpha - 1 \in F[\alpha] \end{align*} $$ Therefore, $\alpha$ is algebraic over $F$.

(2) $\implies$ (3)(4)(5): Since $\varphi$ is not injective, $\ker \varphi$ is non-trivial. By the First Isomorphism Theorem, $F[x] / \ker\varphi \cong \mathrm{Im} \varphi$. Here, $\mathrm{Im}\varphi$ is a subring of $K$ containing $1$. Now $K$ is a field $\implies$ $\mathrm{Im} \varphi$ is an ID $\implies$ $F[x] / \ker \varphi$ is an ID $\implies$ $\ker\varphi$ is a PID, i.e., $\ker \varphi = \left( p(x) \right)$ for some $p(x) \in F[x]$ $\implies$ $p(x)$ is prime in $F[x]$ $\implies$ $p(x)$ is irreducible in $F[x]$. By Theorem 1206, $F[x] / \ker \varphi$ is a field and $$ F[x] / \ker\varphi = F + F\overline{x} + \cdots + F\overline{x}^{n-1}. $$ With $x \mapsto \alpha$ and $p(\alpha) = 0$ for $p(x) \in \ker\varphi$, we obtain $$ \mathrm{Im}\varphi = F + F\alpha + \cdots + F\alpha^{n-1}. $$ So $\mathrm{Im}\varphi$ is field extension of $F$ and $[\mathrm{Im}\varphi : F] < \infty$. By definition of $\varphi$, we have $F[\alpha] \subseteq \mathrm{Im}\varphi$. The above implies $\mathrm{Im}\varphi \subseteq F[\alpha]$. So $\mathrm{Im}\varphi = F[\alpha]$. Now $\mathrm{Im}\varphi$ is a field $\implies$ $F[\alpha] = F(\alpha)$. Therefore, $[F(\alpha) : F] < \infty$ and $[F[\alpha] : F] < \infty$.

(4) $\implies$ (1): $[F[\alpha] : F] < \infty$ implies that $F[\alpha] = F + F\alpha + \dots + F \alpha^{n-1}$ for some $n \in \mathbb{Z}_{\ge 0}$. Hence $1 \in F[\alpha]$ can be written as $1 = c_{0} + c_{1}\alpha + \cdots + c_{n-1}\alpha^{n-1}$, i.e., $c_{0} - 1 + c_{1}\alpha + \dots + c_{n-1}\alpha^{n-1} = 0$. So $\alpha$ is algebraic over $F$.

(3) $\implies$ (4): $F(\alpha) \subseteq F[\alpha] \subseteq F$. Then use Theorem 1207.