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Find a suitable number $a$ such that $\mathbb Q(\sqrt 3, i)=\mathbb Q(a)$

I'm thinking about $a=\sqrt 3 + i$, but I don't know how to prove it. I need help


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up vote 4 down vote accepted

Yes, your idea seems correct. All you have to prove is, that $a\in\Bbb Q(\sqrt3,i)$ -which is obvious,- and that $\sqrt3,i\in\Bbb Q(a)$. For this,

Hint: $(\sqrt3-i)a=?$, and use $\frac12\in\Bbb Q$.

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How do you know that $\sqrt{3}-i\in \mathbb{Q}(a)$? – Kevin Oct 30 '12 at 16:42
$\mathbb Q(a)$ is a field. If $\sqrt{3}\in \mathbb Q(a)$ and $i\in \mathbb Q(a)$, then $\sqrt{3}-i\in \mathbb Q(a)$. – Reader Oct 30 '12 at 17:15
@Kevin: $\sqrt3-i=4/a$. – Berci Oct 30 '12 at 17:30
Ok, I can prove that $a\in\Bbb Q(\sqrt3,i)$ How can we go further from that? – user42912 Oct 30 '12 at 22:52
You also need $\sqrt 3$ and $i$ in $\Bbb Q(a)$. And, $\sqrt 3-i$ is there, as $4/a$ and thus $\sqrt 3$, too. – Berci Nov 1 '12 at 23:42

This is probably a bit advanced but I would like to propose a solution that doesn't really require calculations:

Note that a basis for $K=\mathbb{Q}(\sqrt{3},i)$ over $F=\mathbb{Q}$is $\{\alpha_{i}\beta_{j}\}_{1\leq i,j,\leq2}$ where $\alpha_{1}=1,\alpha_{2}=\sqrt{3},\beta_{1}=1,\beta_{2}=i$.

We have $4$ maps defined by $\varphi:K\to K$ by $1\to1$ and $\sqrt{3}\to\pm\sqrt{3},i\to\pm i$.

Verify that all $4$ $\varphi$ are automorphisms of $K$ that fix $F$. Since this is the splittinf field of $(x^2-3)(x^2+1)$ over a perfect field we have it that $K/F$ is Galois since the degre of the extension is $4$ we have it that $Gal(K/F)=\{\varphi_{i}\}_{i=1}^{i=4}$ where the $\varphi_{i}'s$ are the ones I defined above.

Note that the only automorphism of $K$ that fix $\sqrt{3}+i$ is $Id_{K}$ hence it is a primitive element of the extension, since otherwise $\mathbb{Q}(\sqrt{3}+i)$ is a proper subfield of $K/F$ hence correspond to a proper subgroup of $K/F$, in contradiction.

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Well, you do need to check that the four maps are automorphisms :-) – Mariano Suárez-Alvarez Oct 30 '12 at 16:40
@MarianoSuárez-Alvarez - hmm, yes. but $2$ of them are for free! :) [see next comment please for a correction] – Belgi Oct 30 '12 at 16:41
@MarianoSuárez-Alvarez - in second thought - the degree of the extension is $4$ and the extension is Galois (see revised argument) so there are $4$ automorphisms. since any automorphism sends $i,\sqrt{3}$ to some conjugate of them and since they generate the extension we have it that all of the maps have to be automorphisms - no calculations required! :-) – Belgi Oct 30 '12 at 16:43
:-) ${}{}{}{}{}$ – Mariano Suárez-Alvarez Oct 30 '12 at 16:46

You are correct. $K=Q(\sqrt{3},i)$ has degree 4 and $Q(a)\subset K$. The minimal polynomial for $a$ has degree 4, it is $x^4-4x^2+16$; so $Q(a)=K$.

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How do you prove that the degree of the minimal polynomial is $4$ ? – Belgi Oct 30 '12 at 16:07
@Belgi Since $F=Q[\sqrt{3}]\subseteq R$, it's impossible for $F$ to contain $i$, so $[F[i]:F]=2$. By the multiplicative rule for extensions, $[F[i]:Q]=[F[i]:F][F:Q]=4$. – rschwieb Oct 30 '12 at 16:17
@Belgi The polynomial $x^4-4x^2+16$ can't be factored over $Q$ and $a$ is a root. – i. m. soloveichik Oct 30 '12 at 16:18
@rschwieb - this only shows that the degree of the extension is $4$. You do not know that $\mathbb{Q}(a)$ is not a proper subfield. – Belgi Oct 30 '12 at 16:19
@Belgi It's obvious that $Q[a]\subseteq Q[\sqrt{3},i]$. On the other hand, $i=(a-4a^{-1})/2$, so $i$ and $a-i=\sqrt{3}$ are in $Q[a]$, thus $Q[a]= Q[\sqrt{3},i]$. – rschwieb Oct 30 '12 at 16:30

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