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I want to find the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt{10})$. I saw this link, but I really don't know in what it's conclusive. I don't see how he get the minimal polynomial of $\sqrt{3}+\sqrt 5$. By the way, since $\mathbb Q(\sqrt{10})$ is not a subfield of $\mathbb Q(\sqrt3+\sqrt 5)$, the minimal polynomial a priori doesn't exist, no ? Or maybe there is something I don't get.

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  • $\begingroup$ First of all set $X=\sqrt 3 + \sqrt 5$ and try to find a polynomial vanishing at $X$. Then ask yourself if there could be one of smaller degree. $\endgroup$ – Maffred Apr 20 '17 at 15:32
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It's the same minimum polynomial as over $\Bbb Q$. Otherwise one of these would be the minimum polynomial: $$X-\sqrt3-\sqrt5,$$ $$(X-\sqrt3-\sqrt5)(X-\sqrt3+\sqrt5),$$ $$(X-\sqrt3-\sqrt5)(X+\sqrt3-\sqrt5),$$ $$(X-\sqrt3-\sqrt5)(X+\sqrt3+\sqrt5)$$ would be. But you can check, at your leisure, that none have coefficients in $\Bbb Q(\sqrt{10})$.

It may help to note that $\Bbb Q(\sqrt2,\sqrt3,\sqrt5)$ has degree $8$ over $\Bbb Q$ by Kummer theory.

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  • $\begingroup$ Thanks. But why is it important to note that $\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)/\mathbb Q$ has degree $8$ ? $\endgroup$ – MSE Apr 21 '17 at 11:40
  • $\begingroup$ @MSE It's a power of $2$, so all minimal polynomials must have $2$-power degree. $\endgroup$ – Angina Seng Apr 21 '17 at 11:49
  • $\begingroup$ thank you. But one thing is strange; how can we find the minimal polynomial of $\mathbb Q(\sqrt 3+\sqrt 5)$ over $\mathbb Q(\sqrt{10})$ whereas $\mathbb Q(\sqrt{10})$ is not a subfield of $\mathbb Q(\sqrt 3+\sqrt 5)$ ? $\endgroup$ – MSE Apr 21 '17 at 17:02
  • $\begingroup$ They are all subfields of $\Bbb Q(\sqrt2,\sqrt3,\sqrt5)$. @MSE $\endgroup$ – Angina Seng Apr 21 '17 at 17:03

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