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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.

ElementX
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    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 02 '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 02 '18 at 17:07
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    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 02 '18 at 17:19
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    This is a special case of this question. The primitive element is discussed here. – Jyrki Lahtonen Oct 02 '18 at 20:55

2 Answers2

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

user8268
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  • 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 02 '18 at 17:16
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    @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 02 '18 at 17:20
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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 $8=2^3$.

Lubin
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