Field Extension Proof

Asked Oct 19 '17 at 19:37

Active Oct 19 '17 at 19:37

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I want to show that |$\mathbb{Q}(\sqrt a,\sqrt b)|=4$ with a, b and ab not being squares

Is it enough to say that,

$\mathbb{Q}(\sqrt a,\sqrt b)=\mathbb{Q}(\sqrt a)(\sqrt b)$= {p+q$\sqrt b | p,q \in \mathbb{Q}(\sqrt a)$}= {$p+q\sqrt a +r \sqrt b +s \sqrt ab$} which is clearly spanned by $1,\sqrt a, \sqrt b, \sqrt ab$

asked Oct 19 '17 at 19:37

Bradley Hill

1

You have to show $1$, $\sqrt a$, $\sqrt b$, $\sqrt{ab}$ are linearly independent over $\Bbb Q$, which is true, but not entirely trivial. – Angina Seng Oct 19 '17 at 19:43
Take a peek at Bill Dubuque's answer. Variants and special cases have been handled on other occastions also. – Jyrki Lahtonen Oct 19 '17 at 19:52
Assume that $\sqrt{a} =c+d \sqrt{b}$ for some rationals $c,d$ and show a contradiction. This will show $x^2-a$ is irreducible over $\mathbb{Q}(\sqrt{b})$ – reuns Oct 19 '17 at 20:09

Field Extension Proof

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