I'm trying to show that $[L:\mathbb{Q}]=8$, where $L=\mathbb{Q}(i, \sqrt{2}, \sqrt{3})$. I tried using the tower law to show this by saying:

$[L:\mathbb{Q}]=[L:\mathbb{Q(\sqrt{2}, \sqrt{3})}]*[\mathbb{Q}(\sqrt{2}, \sqrt{3}):\mathbb{Q}]$. I know that $[\mathbb{Q}(\sqrt{2}, \sqrt{3}):\mathbb{Q}]=4$, so it remains to show that $[L:\mathbb{Q(\sqrt{2}, \sqrt{3})}]=2$, but I'm struggling with this.

Then given that $L/\mathbb{Q}$ is a Galois extension, how would I compute the Galois group for this extension. In general, what's the best way to compute the galois group for a field.

Finally, if I had an intermediate field $F$, i.e $L\subseteq F\subseteq\mathbb{Q}$. Then how would I show that $L/F$ is also Galois, I'm guessing I somehow use the fact that $L/\mathbb{Q}$ is galois.

I'm grateful for anyone input on my question.


$[L:\mathbb{Q}(\sqrt{2}, \sqrt{3})]= 2$; because the minimal polynomial of $i$ is $x^2+1$. And the degree of that polynomial is 2.

And for the Galois group: All the automorphisms send any root of the minimal polynomial to another root. Consider the polynomial $g(x)=(x^2+1)(x^2-2)(x^2-3)$.

$L/F$ is Galois for a simple reason: It's separable, because $L/\mathbb{Q}$ is (Think the minimal polynomials). And it's normal, because $L$ is the splitting field of a polynomial $g(x)\in \mathbb{Q}[x]$ over $F$ also.

  • $\begingroup$ A point that should perhaps be made explicit is that $i$ can't be in $K={\bf Q}(\sqrt2,\sqrt3)$ because $K$ is a subsetof the reals while $i$ is not real. $\endgroup$ – Gerry Myerson Nov 28 '14 at 8:23

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