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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, $F(\alpha) = F[\alpha]$.

  • $\bf{Result\; 2:}$ If $K$ is a finite extension of $F$, then $K$ is algebraic and finitely generated over $F$.

  • $\bf{Result\; 3:}$ 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.

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  • $\begingroup$ If you've proved result 1, then you've proved one direction. The other direction uses more linear algebra than field theory. $\endgroup$ – Robert Wolfe Mar 29 '18 at 22:16
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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 $F$.

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  • $\begingroup$ $[F(\alpha):F] = n$ implies $a_{0} + a_{1}\alpha + ... + a_{n}\alpha^{n} = 0$ with $a_{i} \neq 0$ and $a_{i} \in F$? $\endgroup$ – Lucas Corrêa Mar 29 '18 at 22:35
  • $\begingroup$ That sounds good to me. $\endgroup$ – Rob Arthan Mar 29 '18 at 22:36
  • $\begingroup$ Thanks for the help! $\endgroup$ – Lucas Corrêa Mar 29 '18 at 22:39
  • $\begingroup$ @LucasCorrêa: well, that's mostly correct. You can't assert that $a_{i} \neq 0$ for all $i$, only that not all $a_{i}$ are $0$. (But your main idea is correct.) $\endgroup$ – Alex Wertheim Mar 29 '18 at 22:59

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