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Let $k \subseteq K \subseteq E$ fields and if $E/k$ is a finite extension, the $E/K$ and $K/k$ are also finite extensions.

As $E/k$ is finite extension, then $dim_{k}E=n < \infty$. So as, $K \subseteq E$ we have that $K$ is a $k$-vectorial subspace of $E$. And by a well known result in linear algebra we have that $dim_{k}K \leq dim_{k}E=n=\infty$. Proving that $K/k$ is also a finite field extension.

Now I want to prove that; $E/K$ is a finite field extension. By some field theory result it should happen that if $E/K$ is a finite field extension

$$(dim_{k}K) (dim_{K}E)=dim_{k}E.$$

So if I suppose $E/K$ is an infinite field extension, then

$$(dim_{k}K) (dim_{K}E)=(dim_{k}K)(\infty)=dim_{k}E=\infty.$$

Then, $dim_{k}E=\infty=n < \infty$. Proving that $E/K$ is a finite field extension.

Is my proof right? If not what Im suppose to correct? Also If this proof right would appreciate to see another solutions for this problem. Thanks!

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    $\begingroup$ It is correct, but I dont know whether saying "Then, $dim_{k}E=\infty=n < \infty$." is correct. $\endgroup$ – B.Swan Apr 7 at 19:17
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    $\begingroup$ MathJax works in the title section. Also, please use $\operatorname{dim}$ for $\operatorname{dim}$. $\endgroup$ – Shaun Apr 8 at 4:41
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There's some awkwardness, where you say things like $n=\infty$ and $\infty=n<\infty.$

The core of the approach is simply to note that $$\left(\operatorname{dim}_kK\right)\left(\operatorname{dim}_KE\right)=\operatorname{dim}_kE.$$ Since $\operatorname{dim}_kE$ is finite, and since $\operatorname{dim}_kK$ and $\operatorname{dim}_KE$ are either positive integers or infinite, then they must be positive integers, so both $E/K$ and $K/k$ are finite extensions.

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