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Let $\alpha = i\sqrt{2} + \sqrt{3}$ and let $K:=\mathbb{Q}(i\sqrt{2}, \sqrt{3})$.

(i) Find the degree of extension $K:\mathbb{Q}$.

(ii) Prove that the extension $K:\mathbb{Q}$ is simple.

(iii) Find the minimal polynomial of $\alpha$ over $\mathbb{Q}$.

I think at some point I'm meant to square $\alpha$ which is when I get $\alpha^2 = 1+2i\sqrt{6}$ so $\alpha^2 -1 = 2i\sqrt{6}\quad$ so squaring both sides again I get $\alpha^4 -2\alpha^2 +25=0$.

Again, I think I'm meant to do that since I've seen it in similar posts but I'm unsure. Now that I'm doing more research I believe my part (iii) is basically answered in Minimal polynomial with $f(\alpha)=0$ but would still be great to have confirmation/helps in other parts!

whatismath
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1 Answers1

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If I proove part ii), then i) can be derived by ii) and iii).

$\frac1\alpha=\frac{\bar\alpha}{|\alpha|^2}=\frac{\sqrt3-i\sqrt2}5$. So, $\sqrt3=\frac12(\alpha+5\alpha^{-1})$ and $i\sqrt2=\frac12(\alpha-5\alpha^{-1})$.

Therefore, $K=\mathbb Q(\alpha)$. So $K/\mathbb Q$ is simple extension.

Then $[K:\mathbb Q]=\deg(\text{irr}(\alpha,\mathbb Q))=4$.

MH.Lee
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