3
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.