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

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.

## marked as duplicate by Jyrki Lahtonen, Leucippus, Community♦Oct 3 '18 at 4:55

• No way of rationally put $\sqrt{11}$ with the two other generators. You do have the most probable $8$ as answer. Do just as you have did at your beginning. – Piquito Oct 2 '18 at 16:52
• Sir, you meant that I should split into 3, instead of 2? Or should I prove that $Q(\sqrt{3},\sqrt{5},\sqrt{11}) = Q(\sqrt{3} + \sqrt{5} + \sqrt{11})$? – ElementX Oct 2 '18 at 17:07
• Exactly ("Primitive element" theorem for french people). You can do also $Q(a\sqrt{3} + b\sqrt{5} +c \sqrt{11})$ with $a,b,c$ non-zero rationals. – Piquito Oct 2 '18 at 17:19
• This is a special case of this question. The primitive element is discussed here. – Jyrki Lahtonen Oct 2 '18 at 20:55

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 $$[K(\sqrt{11}):\Bbb Q]=8$$.

• Thank you, sir, for giving a solution. However, I am a beginner at field theory. My only allowed knowledge to solve this problem is the basics of expanding fields. Is there any way not to use Galois theory to prove what I desire? Thank you – ElementX Oct 2 '18 at 17:16
• @ElementX: You can always try to prove that $11\notin K$ by trying $\sqrt{11}=a+b\sqrt{3}+c\sqrt{5}+d\sqrt{3}\sqrt{5}$ ($a,b,c,d\in\Bbb Q$), taking the square, and seein that it's impossible - it works, just it requires a bit of calculation – user8268 Oct 2 '18 at 17:20

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 $$8=2^3$$.