If $K$ is a field extension of $F$, then $\alpha \in K$ is algebraic over $F$ if and only if $[F(\alpha):F] < \infty$. Moreover, $K$ is algebraic over $F$ if $[K:F] < \infty$
$\it{Proof}.$
$(\Longrightarrow)$ Suppose that $\alpha \in K$ is algebraic over $F$. Thus, theres exists $\min(F,\alpha)$ and $\deg(\min(F, \alpha)) < \infty$. By the Result 1, $[F(\alpha):F] = \deg(\min(F, \alpha)) < \infty$.
$(\Longleftarrow)$ I couldn't do it. $\square$
Some of the results that I proved:
- $\bf{Result\; 1:}$ Let $K$ is a fiel extension of $F$ and let $\alpha \in K$ be algebraic over $F$. Then
If $n=\deg(\min(F,\alpha))$, then the elements $1, \alpha, …, \alpha^{n-1}$ form a basis for $F(\alpha)$ over $F$, so $[F(\alpha):F] = \deg(\min(F, \alpha)) < \infty$. Moreover, $F(\alpha) = F[\alpha]$.
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$\bf{Result\; 2:}$ If $K$ is a finite extension of $F$, then $K$ is algebraic and finitely generated over $F$.
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$\bf{Result\; 3:}$ Let $F \subseteq L \subseteq K$ be fields. Then
$$[K:F] = [K:L].[L:F].$$ -
$\bf{Result\; 4:}$ Let $K$ be a field extension of $F$. If each $\alpha_{i} \in K$ is algebraic over $F$, then $F[\alpha_{1}, …, \alpha_{n}]$ is a finite dimensional field extension of $F$ with
$$[F[\alpha_{1}, …, \alpha_{n}]:F]\leq \prod\limits_{i=1}^{n}[F(\alpha_{i}):F].$$
I know it's something small I can't see. I liked any hint, no complete solutions. I think it's enough to use Results 2 and 4. But I don't know how.
Best Answer
Hint: if $[F(\alpha) : F] < \infty$, the elements $1, \alpha, \alpha^2, \ldots$ cannot be linearly independent in $F(\alpha)$ viewed as a vector space over $F$.