# Find the minimal polynomial of $\sqrt 3+\sqrt 5$ over $\mathbb Q(\sqrt{10})$

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.

• 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. Apr 20, 2017 at 15:32

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.
• Thanks. But why is it important to note that $\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)/\mathbb Q$ has degree $8$ ?
• @MSE It's a power of $2$, so all minimal polynomials must have $2$-power degree. Apr 21, 2017 at 11:49
• 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 $\mathbb Q(\sqrt 3+\sqrt 5)$ ?
• They are all subfields of $\Bbb Q(\sqrt2,\sqrt3,\sqrt5)$. @MSE Apr 21, 2017 at 17:03