Finding the irreducible polynomial corresponding to the element $\sqrt3+\sqrt5$ in the field $\mathbb Q(\sqrt15)$

Question

I had the idea of taking the polynomial $p(x)\in \mathbb Q (\sqrt3,\sqrt5)$

$p(x)= (\sqrt3+\sqrt5)-x$

then computing $p^2(x)$ so that:

$p^2(x)= x^2-2\sqrt3 x-2\sqrt5 x +8 + 2\sqrt15$

Noticing that $\sqrt3+\sqrt5$ is the solution to

$x^2+(8+2\sqrt15 =x(2\sqrt3+\sqrt5)$

and then computing:

$(x^2+(8+2\sqrt15)^2=(x(2\sqrt3+\sqrt5))^2$

which gives (if I´m not wrong):

$x^4-16x^2+124+32\sqrt15=0$

So that the polynomial $q(x)\in \mathbb Q (\sqrt15)$

$q(x)=x^4-16x^2+124+32\sqrt15$

has as a solution the element $\sqrt3+\sqrt5$

But I don´t have a clue on how to prove this is in fact irreducible, or if it even is irreducible. Any hint would be very much appreciated.

Answer 1

It is in fact not irreducible, since $[\Bbb Q(\sqrt 3 + \sqrt 5):\Bbb Q(\sqrt{15}]=2$ (the generator of the big field is a primitive element by the constructive proof of the primitive element theorem).

So what is the minimal polynomial? You can easily verify that is is just $(x-\sqrt 3-\sqrt 5)(x+\sqrt 3 + \sqrt 5) = x^2 -(\sqrt 3+\sqrt 5)^2 = x^2 -(8+2\sqrt 15)$.

If you know something of Galois theory, I figured this out by looking at the Galois conjugates of $\sqrt 3+\sqrt 5$ fixing the element $\sqrt{15}$, which are exactly the identity and the one which maps both of them to their opposites (since then $(-\sqrt 3)(-\sqrt 5) = \sqrt{15}$).

Answer 2

$(\sqrt3+\sqrt5)^2=8+2\sqrt{15}$, so $\sqrt3+\sqrt5$ is a root of the polynomial $x^2-(8+2\sqrt{15})$.