Alternative proof for $[\mathbb{Q}(\sqrt{p},\sqrt{q}):\mathbb{Q}]=4$

Question

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

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

I know it can be shown with $\mathbb{Q}(\sqrt{p},\sqrt{q})=\mathbb{Q}(\sqrt{p})(\sqrt{q})$, but I tried it differently and I'm looking for an alternative proof, so I tried it like this, but I'm not sure if my proof is correct:

Let $M$ be a field extension with $\mathbb{Q} \subset M \subset \mathbb{Q}(\sqrt{p},\sqrt{q})$, then

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

It's $M=\mathbb{Q}(\sqrt{p}) \subset \mathbb{Q}(\sqrt{p},\sqrt{q})$

Also, $[\mathbb{Q}(\sqrt{p}):\mathbb{Q}]=2$ with basis $\lbrace 1, \sqrt{p} \rbrace$ and $[\mathbb{Q}(\sqrt{q}):\mathbb{Q}]=2$ with basis $\lbrace 1, \sqrt{q} \rbrace$

That's why $2\mid[\mathbb{Q}(\sqrt p,\sqrt q):\mathbb{Q}]\mid 2\cdot 2=4$

The index is $4$ since:

It's $m=n \in \lbrace 1,2 \rbrace$, with $m=[ \mathbb{Q}(\sqrt{p},\sqrt{q}):\mathbb{Q}(\sqrt{p})]$ and $n=[\mathbb{Q}(\sqrt{p},\sqrt{q}):\mathbb{Q}(\sqrt{q})]$

Suppose $m=n=1$:

Then $\sqrt{q} \in \mathbb{Q}(\sqrt{p})$, so $\sqrt{q}=a+b\sqrt{p}\Leftrightarrow q = a^2 + 2ab\sqrt p + b^2p$ with $a,b \in \mathbb{Q}$.

So $q-a^2-b^2p$ is rational and that's why $ab=0$.

It's $b \neq 0$, since rational numbers can't be equal to irrational numbers and $a \neq 0$, since then $\sqrt q=b\sqrt p$ and $b=\sqrt{q/p}$, which is impossible.

So, $m=n=2$ and

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

Is my proof right? I don't know if it can be shown like this.

See also this question. – Dietrich Burde Dec 18 '19 at 14:28 — Dietrich Burde, Dec 18 '19 at 14:28
Please don't duplicate your questions. – Bill Dubuque Dec 18 '19 at 18:47 — Bill Dubuque, Dec 18 '19 at 18:47

score 0 · Accepted Answer · answered Dec 18 '19 at 12:40

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Your proof has the correct idea, but sorta loops around for a bit before coming to the right conclusion.

Clearly, $[\Bbb Q(\sqrt p):\Bbb Q]=2$ and $[\Bbb Q(\sqrt q):\Bbb Q]=2$. Therefore, $2\mid[\Bbb Q(\sqrt p,\sqrt q):\Bbb Q]\mid 2\cdot 2=4$. So show that the index is 4, you need to rule out the possibility that the index is 2, or $\sqrt q\in\Bbb Q(\sqrt p)$.

If that is the case, then we have $\sqrt q=a+b\sqrt p$, so $q = a^2 + 2ab\sqrt p + b^2p$. Subtracting the rationals to the left hand side, we see that $ab=0$, so either $a=0$ or $b=0$. The latter is impossible since a rational cannot equal an irrational. If the former is true, then $\sqrt q=b\sqrt p$, so $b=\sqrt{q/p}$, which is also impossible. $\square$

answered Dec 18 '19 at 12:40

Alex

2,449

I don't see how you conclude $ab=0$ from $q = a^2 + 2ab\sqrt p + b^2p$. – Tartulop Dec 18 '19 at 14:21
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$q-a^2-b^2p$ is rational, hence $2ab\sqrt{p}$ is rational, i.e., $ab=0$. – Dietrich Burde Dec 18 '19 at 14:27
I corrected it, is the argumentation right now? – Tartulop Dec 18 '19 at 14:50
@Tartulop Yes . – Alex Dec 18 '19 at 14:56

Alternative proof for $[\mathbb{Q}(\sqrt{p},\sqrt{q}):\mathbb{Q}]=4$

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