I have a Galois theory exercise to prove that $$[\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]=8$$ I understand the proof up until the bit where If I prove $$\sqrt5 \notin \mathbb{Q}(\sqrt2,\sqrt3)$$ then $$[\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb Q(\sqrt2,\sqrt3)]=2$$ How does proving that $\sqrt5\notin \mathbb{Q}(\sqrt2,\sqrt3)$ imply that $[\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb Q(\sqrt2,\sqrt3)]=2$??
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You showed that $\sqrt5 \notin \mathbb{Q}(\sqrt2,\sqrt3)$, which means that the minimum polynomial $f$ of $\sqrt5$ over $\mathbb{Q}(\sqrt2,\sqrt3)$ has $\deg(f)>1$. But $f(x)=x^2-5$ is the polynomial over $\mathbb{Q}(\sqrt2,\sqrt3)$ with minimal degree such that $f(\sqrt5)=0$. So $[\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb Q(\sqrt2,\sqrt3)]=\deg(f)=2$
Dan
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Ah so simple! Thanks! – J.Menton Jan 04 '18 at 16:57
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Out of interest, what would change if $\sqrt5 \in \mathbb{Q}(\sqrt2,\sqrt3)$ – J.Menton Jan 05 '18 at 13:06
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In that case, the minimal polynomial of $\sqrt5$ over $\mathbb{Q}(\sqrt2,\sqrt3)$ would just be $f(x)=x-\sqrt5$ of degree $1$, so $[\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb Q(\sqrt2,\sqrt3)]=\deg(f)=1$. – Dan Jan 05 '18 at 13:09
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Got it! thanks again – J.Menton Jan 05 '18 at 13:10