Let $F$ be a field, and $K$ a finite extension of $F$. I want to show that any element algebraic over $K$ is algebraic over $F$, and conversely.

Well, if $a$ is algebraic over $F$, we can write $c_0+c_1a+\ldots+c_na^n=0$ for some $c_0,\ldots,c_n\in F$. But $c_0,\ldots,c_n\in K$, so $a$ is algebraic over $K$.

Now suppose $a$ is algebraic over $K$, so we can write $c_0+c_1a+\ldots+c_na^n=0$ for some $c_0,\ldots,c_n\in K$. What to do then?

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    $\begingroup$ @GitGud $\sqrt{2}$ is algebraic over $\mathbb{Q}$... $\endgroup$ – Kunal Jan 18 '14 at 2:14
  • $\begingroup$ My bad, sorry and thanks. $\endgroup$ – Git Gud Jan 18 '14 at 2:17

Hint: $$F(a) \subset K(a)$$

And $[K(a):F]=[K(a):K] [K:F] < \infty$. From here you should be able to get that

$$[F(a):F]< \infty \,.$$

  • $\begingroup$ The last one using $[K(a):F]=[K(a):F(a)][F(a):F]$, right? $\endgroup$ – Kunal Jan 18 '14 at 2:15
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    $\begingroup$ @Kunal Or even simpler, any subspace of a finite dimensional $F$-vector space is finite... Which in this case, is the same as saying that a subextension of a finite extension must be finite ... $\endgroup$ – N. S. Jan 18 '14 at 2:18

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