# Let k⊆K⊆E fields and if E/k is a finite extension, the E/K and K/k are also finite extensions.

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!

• It is correct, but I dont know whether saying "Then, $dim_{k}E=\infty=n < \infty$." is correct. – B.Swan Apr 7 at 19:17
• MathJax works in the title section. Also, please use $\operatorname{dim}$ for $\operatorname{dim}$. – Shaun Apr 8 at 4:41

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.