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.

  • 1
    $\begingroup$ You just need to prove there's no such $a,\,b,\,c,\,d$. $\endgroup$ – J.G. Mar 31 at 12:07
  • $\begingroup$ Kummer theory${}$? $\endgroup$ – Lord Shark the Unknown Mar 31 at 12:09
  • $\begingroup$ 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 $\endgroup$ – 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:

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


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

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.

  • $\begingroup$ By (4), can I conclude that "$\sqrt{q}$ is in $Q$, which is a contradiction" ? $\endgroup$ – Sophia Apr 2 at 4:40
  • $\begingroup$ @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). $\endgroup$ – CopyPasteIt Apr 2 at 4:52
  • $\begingroup$ Now I notice what I missed. Thanks a lot!! $\endgroup$ – 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.


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