Is a normal extension $K \subset L$ neccesarily of finite degree?

First let me state the definition:

A normal extension is an extension field $K \subset L$ s.t $L$ is algebraic and for each $\alpha \in L$ the polynomial $\text {Irr}(\alpha,K)$ splits over $L$.

I know that a splitting field $K(\alpha_1, \dots, \alpha_n)$ of $f \in K[X]$ is neccesarily of finite degree considered as a vector space over $K$.

However, text has made me believe that a normal extension $K \subset L$ must be a finite degree as well. I've not been able to prove this.

I see that a normal extension is some sense a splitting field over many polynomials.

An extension is Galois iff it is normal and separable. So just take a Galois extension which is not finite, such as $\overline{\mathbb Q}/\mathbb Q$ and you get a normal extension of infinite degree!
• Thank you Ferra. First what does $\bar {\mathbb Q}$ denote ? Second, I'm working my way to Galois theory, is there any simpler way of showing this ? – Andreas Lykke Iversen Oct 3 '14 at 9:12
• $\overline{\mathbb Q}$ denotes the algebraic closure of $\mathbb Q$. In general, pick any field $K$. Then its algebraic closure $\overline{K}$ is normal over $K$ by definition, since it's algebraic and every polynomial of $K[x]$ splits in $\overline{K}$. If you want an easier example, pick $K=\mathbb Q(\sqrt{p}\colon p\mbox{ is prime})$. This extension is obviously algebraic and has infinite degree over $\mathbb Q$ since for example the set $\{\sqrt{p}\colon p\mbox{ is prime}\}$ is linearly independent. It might be a good exercise for you to check that the remaining property holds! – Ferra Oct 3 '14 at 9:38