# Dimension as Vector Spaces of Sums of Field Extensions

Suppose that $\mathbb{Q}(a), \mathbb{Q}(b)$ are different finite field extensions such that each has no proper sub-extension. Then if $\mathbb{Q}(a), \mathbb{Q}(b)$ are respectively $m, n$ dimensional vector spaces over $\mathbb{Q}$, it is pretty clear that their sum as vector spaces is $m+n-1$ dimensional since if there were a dependence, this would mean the field extensions had a nontrivial intersection which would correspond to a proper sub-extension.

What about the case where we have more than 2 such extensions? Does some similar characterization of the dimension hold? If not, is there a relatively simple example where this is not the case?

I don't think there exists a similar formula for the dimension of the sum of three extensions. Consider the case where $\mathbb{Q}(a)$, $\mathbb{Q}(b)$, $\mathbb{Q}(c)$ are all cubic extensions (hence have no intermediate fields). Generically we would expect $$\dim \mathbb{Q}(a)+\mathbb{Q}(b)+\mathbb{Q}(c)=7,$$ as each summand usually adds two to the dimension. This would happen for example with $a=\root3\of2$, $b=\root3\of3$, $c=\root3\of5$ (I think).
On the other hand, if $a,b,c$ are the three roots of $x^3-2$, then the dimension of the sum is five, because we have the relations $a+b+c=0$ and $a^2+b^2+c^2=0$.