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Let $p$ and $q$ be different prime numbers.

How to show that $[\mathbb{Q}(\sqrt{p}, \sqrt{q}):\mathbb{Q}]=4$?

I did it this way:

$\textbf{Edit: (Is there something to improve?)}$

Proof:

$\mathbb{Q}(\sqrt{p},\sqrt{q})=\mathbb{Q}(\sqrt{p})(\sqrt{q})$, because the first one is the smallest field that contains $\sqrt{p}, \sqrt{q}$ and $\mathbb{Q}$, and the second one is the smallest field that contains $\sqrt{q}$ and $\mathbb{Q}(\sqrt{p}$).

It's $\sqrt{q} \notin \mathbb{Q}(\sqrt{p})$, since if we suppose that we had $\sqrt{q}\in\Bbb Q(\sqrt{p})$ and since $\Bbb Q(\sqrt{p})$ is quadratic over $\Bbb Q$ that would mean that $\sqrt{q}=a+b\sqrt{p}$, for some $a,b \in \mathbb{Q}$.

Taking squares and isolating $\sqrt{p}$:

$\sqrt{p}=\frac{q-a^2-b^2p}{2ab}\in\Bbb Q$, which is impossible.

So $[\mathbb Q(\sqrt{p},\sqrt q):\mathbb Q(\sqrt p)]=2$, since $x^2-q$ is the minimal polynomial of $\sqrt{q}$ over $\mathbb{Q}(\sqrt{p})$.

The same for $[\mathbb{Q}(\sqrt{p}):\mathbb{Q}]$, since $x^2-p$ is symmetric.

$\Rightarrow [\mathbb{Q}(\sqrt{p},\sqrt{q}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{p},\sqrt{q}):\mathbb{Q}(\sqrt{p})]\cdot[\mathbb{Q}(\sqrt{p}):\mathbb{Q}]= 2 \cdot 2 =4$.

Is there something I can improve?

Tartulop
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2 Answers2

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You have to prove that $$[\mathbb Q(\sqrt{p},\sqrt q):\mathbb Q(\sqrt p)]=2.$$

Hint

$X^2-q$ is the minimal polynomial of $\sqrt{q}$ over $\mathbb Q(\sqrt p)$.

Surb
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Suppose that we had $\sqrt{q}\in\Bbb Q(\sqrt{p})$. Since $\Bbb Q(\sqrt{p})$ is quadratic over $\Bbb Q$ that would mean that $$ \sqrt{q}=a+b\sqrt{p}\qquad\text{for some $a$, $b\in\Bbb Q$.} $$ Taking squares and isolating $\sqrt{p}$ : $$ \sqrt{p}=\frac{q-a^2-b^2p}{2ab}\in\Bbb Q $$ which is impossible.

(In fact one would have to discuss separately the case $ab=0$, but this can be dealt with very easily)

Thus $$ \Bbb Q\subset\Bbb Q(\sqrt{p})\subset\Bbb Q(\sqrt{p},\sqrt{q}) $$ is a tower of quadratic extensions and one concludes by multiplicativity of the degree.

Andrea Mori
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