# How to show that $\sqrt{2}$ is not in $\mathbb Q(\sqrt{3},\sqrt{5})$?

How to show that $$\sqrt{2}$$ is not in $$\mathbb Q(\sqrt{3},\sqrt{5})$$?

First I tried to use the theorem that if $$b$$ is in $$F(a)$$, then $$\deg(b,F)$$ divides $$\deg(a,F)$$. But the theorem can not be applied to this problem. Next I tried to show that $$\sqrt{2}+\sqrt{3}+\sqrt{5}$$ is not in $$\Bbb Q\left[\sqrt{3},\sqrt{5}\right]$$. But it also failed.

Finally I tried this way. Since $$\Big\{1, \sqrt{3}, \sqrt{5}, \sqrt{15}\Big\}$$ is a basis for $$\Bbb Q\Big[\sqrt{3},\sqrt{5}\Big]$$, let $$\sqrt{2}=a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$$ for $$a, b, c, d \in \Bbb Q$$. Then, since there is no such $$a, b, c, d$$, $$\sqrt{2}$$ is not in $$\Bbb Q\Big[\sqrt{3},\sqrt{5}\Big ]$$.

Am I right? If you have better ideas, help would be appreciated. Thanks very much.

• You just need to prove there's no such $a,\,b,\,c,\,d$. – J.G. Mar 31 at 12:07
• Kummer theory${}$? – Lord Shark the Unknown Mar 31 at 12:09
• You can also use that $\Bbb{Q}(\sqrt p_1, \sqrt p_2)=\Bbb{Q}(\sqrt p_1+ \sqrt p_2)$ for $p_i$ prime – B.Swan Mar 31 at 12:17

Let $$p$$, $$q$$, $$r$$ be three distinct prime integers.

We assume the following two statements are known to be true:

$$\tag 1 \mathbb Q\,[\sqrt p] \ne \mathbb Q\,[\sqrt p][\sqrt q]$$

and

$$\tag 2 \mathbb Q\,[\sqrt p] \ne \mathbb Q\,[\sqrt p][\sqrt r]$$

We want to prove that $$\sqrt r \notin \mathbb Q\,[\sqrt p][\sqrt q]$$.

To get a contradiction, assume that $$\sqrt r \in \mathbb Q\,[\sqrt p][\sqrt q]$$. Then there exists $$a,b \in \mathbb Q\,[\sqrt p]$$ such that

$$\tag 3 \sqrt r = a + b \sqrt q$$

If $$b = 0$$ then $$\sqrt r = a \in Q\,[\sqrt p]$$, and by (2) that is not possible.

If $$a = 0$$ then $$\sqrt r = b \sqrt q$$, and so $$\sqrt r = (s + t \sqrt p) \sqrt q$$ for $$s, t \in \mathbb Q$$. Using the prime factorization theorem and elementary 'odd/even logic', we must have that both $$s$$ and $$t$$ are nonzero. But then squaring both sides,

$$\quad r = (s^2 + 2st\sqrt p + pt^2)q$$

and solving, we can rewrite $$\sqrt p$$ as a rational number, which is absurd.

So starting with (3), we must also state that both $$a$$ and $$b$$ are nonzero. Squaring both sides we get

$$\tag 4 2ab \sqrt q = r - a^2 - b^2 q$$

Solving for $$\sqrt q$$, it necessarily follows that $$\sqrt q \in \mathbb Q\,[\sqrt p]$$, contradicting (1).

So indeed, $$\sqrt r \notin \mathbb Q\,[\sqrt p][\sqrt q]$$.

Note: This proof was constructed by adapting to the logic found in

Proof that $$[\Bbb{Q}(\sqrt{q_1},\dots,\sqrt{q_r}):\Bbb{Q}]=2^r$$

and applying it to the OP's question.

• By (4), can I conclude that "$\sqrt{q}$ is in $Q$, which is a contradiction" ? – Sophia Apr 2 at 4:40
• @Sophia $a$ and $b$ are not taken to be rational numbers, but since they are in $\mathbb Q\,[\sqrt p]$ you can still solve for $\sqrt q$ in that field, contradicting (1). – CopyPasteIt Apr 2 at 4:52
• Now I notice what I missed. Thanks a lot!! – Sophia Apr 2 at 5:07

$$\mathbb Q(\sqrt 3, \sqrt 5): \mathbb Q$$ is Galois with Galois group $$\mathbb Z_2 × \mathbb Z_2$$ hence there are exactly $$3$$ subfields of order $$2$$. These are precisely $$\mathbb Q(\sqrt d)$$ where $$d=3,5,15$$. So $$\mathbb Q(\sqrt 2)$$ is not in $$\mathbb Q(\sqrt 3, \sqrt 5)$$

We have $$[\Bbb Q(\sqrt{2},\sqrt{3},\sqrt{5}):\Bbb Q]=8$$ and $$[\Bbb Q(\sqrt{3},\sqrt{5}):\Bbb Q]=4$$, see this duplicate:

Proof that $[\Bbb{Q}(\sqrt{q_1},\dots,\sqrt{q_r}):\Bbb{Q}]=2^r$

Hence $$\sqrt{2}\in \Bbb Q(\sqrt{3},\sqrt{5})$$ would give $$8=4$$, a contradiction.