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.

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.