Dimension of the vector space $\mathbb{Q}(\sqrt{3}, \sqrt{5}, \sqrt{11})$ over $\mathbb{Q}$

Question

Let $\mathbb{Q}(\sqrt{3}, \sqrt{5}, \sqrt{11})$ be the smallest field that contains all rational numbers, $\sqrt{3}, \sqrt{5}$ and $\sqrt{11}$. Consider this field to be a vector space over $\mathbb{Q}$. Find the dimension of this vector space.

My attempt is to demonstrate $\mathbb{Q}(\sqrt{3}, \sqrt{5}, \sqrt{11})$ as $\mathbb{Q}(\sqrt{3}, \sqrt{5})(\sqrt{11})$ and then apply the formula $[A:B][B:C] = [A:C]$ for fields $A$, $B$ and $C$ sastifying $C \le B \le A$ (the notation $[A;B]$ means the dimension of $A$ over $B$)

I got this

$$[\mathbb{Q}(\sqrt{3}, \sqrt{5})(\sqrt{11}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{3} + \sqrt{5})(\sqrt{11}) : \mathbb{Q}] \\= [\mathbb{Q}(\sqrt{3} + \sqrt{5})(\sqrt{11}) : \mathbb{Q}(\sqrt{3} + \sqrt{5})].[\mathbb{Q}(\sqrt{3} + \sqrt{5}): \mathbb{Q}] $$ The latter of the product appears to be $4$, as $\sqrt{3} + \sqrt{5}$ has a minimal polynomial of degree $4$ on $\mathbb{Q}[x]$. The other, however, I'm not sure how to determine its value.

Currently I'm stuck and have to way to proceed.

Please give me a hint. Thank you.

Answer 1

If you know a little bit of Galois theory, and if you proved already that $[\Bbb Q(\sqrt{3},\sqrt{5}):\Bbb Q]=4$ (which can be done in a similar way):

The automorphisms of $K:=\Bbb Q(\sqrt{3},\sqrt{5})$ are given by $\sqrt{3}\mapsto\pm\sqrt{3}$, $\sqrt{5}\mapsto\pm\sqrt{5}$ - the signs are independent, the group of automorphisms has order $4$. Now if $\sqrt{11}\in K$, it must be mapped by every such automorphism to $\pm\sqrt{11}$. However, the only elements $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{3}\sqrt{5}\in K$ ($a,b,c,d\in\Bbb Q$) that are mapped to $\pm$ themselves are those where at most one of $a,b,c,d$ is non-zero. But $\sqrt{11}$ is certainly not of that form ( just take the square and look at the powers of $11$ in the factorization to primes).

So $\sqrt{11}\notin K$, and thus $[K(\sqrt{11}):K]=2$, and so $[K(\sqrt{11}):\Bbb Q]=8$.

Answer 2

You should expect that if you adjoin the square roots of $n$ essentially different integers to $\Bbb Q$, the field extension degree (i.e. the dimension you speak of) will be $2^n$. For the “essentially different” condition, it should enough that the various numbers $n$ are relatively prime in pairs.

But in your case, you should see whether $1,\sqrt3,\sqrt5,\sqrt{11},\sqrt{15},\sqrt{33},\sqrt{55},$ and $\sqrt{165}$ do the trick.

For a look into your future, there’s a side-branch of Galois Theory that describes all quadratic extensions of a field of characteristic $\ne2$, called Kummer Theory. By using that one can say that since $\{3,5,11\}$ generate a multiplicative subgroup of $\Bbb Q$ that’s free of rank three, the adjunction of their square roots gives an extension of $\Bbb Q$ of degree $8=2^3$.