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 $a^2+b^2+c^2=0$.

The scenery may again change, if we exclude cases like this by requiring that no participating field is contained in the sum of the other two. Such a requirement is so natural that I'm not entirely sure that I would call this an answer.

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    $\begingroup$ For some strange reason I first thought that the dimension of the sum would be six = the dimension of the splitting field. IMHO it would be more interesting if the some of two subfields would intersect the third in a subspace of an intermediate dimension. $\endgroup$ – Jyrki Lahtonen Jun 20 '13 at 18:55
  • $\begingroup$ Thanks! As you note, there's definitely more to think about here, but this is a good example illustrating that one doesn't always have linear independence. $\endgroup$ – Alexander Jun 20 '13 at 19:44

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