How do i find the minimal polynomial of Q[root(2)+root(3)] over Q[root(6)]? I know how to find it over Q but?

Question

i tend to get confused with anything more complicated! Thanks!

Answer 1

I think you don't want the Q in front of (root2+root3), am I right? That is, you are looking for the minimal poly of the number (root2+root3) over the field Q[root6]? If this is indeed what you are looking for, then here's how to do it:

Notice that (root2+root3)^2=5+2*root6, and this lies in the field Q[root6], so that the min poly of (root2+root3) will be

x^2-(5+2*root6)

Notice that this is quadratic, whereas over Q by itself this is a fourth degree extension.

Answer 2

if y = root(2)+root(3), then

y^2 = 5 + 2 root(6)

That shows that Q[root(2)+root(3)]=Q[y] includes Q[root(6)].
Let us show y is not in Q[root(6)]
If by absurd y is in Q[root(6)] then the degree of y over Q will be 2.

But y satisfies:
(y^2-5)^2 = 24

y^4 -10y^2 + 49 = 0
and this is an irreducible polynomial( by rational root theorem)
Contradiction.

Therefore the minimal polynomial of y is

y^2-5-2 root(6)=0.

We obtain two extensions of degree 2:
Q[y] > Q[root(6) ] > Q,

Grumpy is right: it's the minimal polynomial of root(2)+root(3), not as you wrote "minimal polynomial of Q[root(2)+root(3)] ".

Answer 3

take the square root of 7 and dived it by three then watch emerdale for an hour