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I'm trying to find the degree of the Field extension $\mathbb{Q}(\sqrt{-1},\sqrt{3}) $ over $ \mathbb{Q}(\sqrt{3}) $, namely $$ \vert\mathbb{Q}(\sqrt{-1},\sqrt{3}): \mathbb{Q}(\sqrt{3}) \vert = n $$

Does one usually approach such a problem with the minimal polynomial? And if so, how?

To find the degree $d$ of $$\vert \mathbb{Q}(\sqrt{3}):\mathbb{Q}\vert = d$$ for example, one would take $\alpha = \sqrt{3} $ and find that its minimal polynomial is $m^{\alpha}(x) = x^2-3$, hence $$ 2 = d = deg(m^\alpha)$$ Any help on how to use this knowledge on the former example would be appreciated.

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The degree is $$ [\mathbb{Q}(i,\sqrt{3}):\mathbb{Q}(\sqrt{3})]=2, $$ because the minimal polynomial of $i$ over $\mathbb{Q}(\sqrt{3})$ is $x^2+1$. By definition, the polynomial is the monic polynomial of minimal degree having $i$ as root. Clearly, a linear polynomial with coefficients in $\mathbb{Q}(\sqrt{3})$ cannot have $i$ as root, so the degree $2$ is minimal.

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  • $\begingroup$ Thank you, that clarifies everything $\endgroup$ – Schief Dec 6 '16 at 19:36

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