# Extension Fields

• Apr 22nd 2011, 06:02 PM
mulaosmanovicben
Extension Fields
Prove that
$\mathbb{Q}$( $\sqrt{2}$, $\sqrt{3}$) is isomorphic to $\mathbb{Q}$( $\sqrt{2}$ + $\sqrt{3}$)

ATTEMPT:

Well we have been learning about extension fields. I know that if F is a field and a is a zero of a polynomial p(x) in F[x] then F(a) is isomorphic to F[x]/<p(x)>. So if i can find a polynomial such that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{2}$ + $\sqrt{3}$ are all zero that would be great.

I really have no idea what to do can you guys give me hints?
• Apr 22nd 2011, 07:08 PM
mulaosmanovicben
Quote:

Originally Posted by mulaosmanovicben
Prove that
$\mathbb{Q}$( $\sqrt{2}$, $\sqrt{3}$) is isomorphic to $\mathbb{Q}$( $\sqrt{2}$ + $\sqrt{3}$)

ATTEMPT:

Well we have been learning about extension fields. I know that if F is a field and a is a zero of a polynomial p(x) in F[x] then F(a) is isomorphic to F[x]/<p(x)>. So if i can find a polynomial such that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{2}$ + $\sqrt{3}$ are all zero that would be great.

I really have no idea what to do can you guys give me hints?

I figured it out:

We will show each set is a subset of the other:

elements of $\mathbb{Q}$( $\sqrt{2}$, $\sqrt{3}$) have the form (a+b $\sqrt{2}$) + (c+d $\sqrt{2}$) $\sqrt{3}$

let q be an element of $\mathbb{Q}$( $\sqrt{2}$ + $\sqrt{3}$) then q has the form:

q = a + b( $\sqrt{2}$ + $\sqrt{3}$) = a + b $\sqrt{2}$+b $\sqrt{3}$ which is of the form mentioned above with a=a, b=b, c=0, d=b. Therefore $\mathbb{Q}$( $\sqrt{2}$ + $\sqrt{3}$) is a subset of $\mathbb{Q}$( $\sqrt{2}$, $\sqrt{3}$)

Now consider ( $\sqrt{2}$ + $\sqrt{3}$)^3 = 11 $\sqrt{2}$ + 9 $\sqrt{3}$ ( I did the math)

subtract 9( $\sqrt{2}$+ $\sqrt{3}$) from both sides and you get:

2 $\sqrt{2}$ = ( $\sqrt{2}$+ $\sqrt{3}$)^3 - 9( $\sqrt{2}$+ $\sqrt{3}$) so $\sqrt{2}$ is an element of $\mathbb{Q}$( $\sqrt{2}$ + $\sqrt{3}$) similar procedure can show that $\sqrt{3}$ is also an element of that set and therefore $\mathbb{Q}$( $\sqrt{2}$, $\sqrt{3}$) is a subset of $\mathbb{Q}$( $\sqrt{2}$ + $\sqrt{3}$)

So the sets are the same
• Apr 22nd 2011, 07:41 PM
Deveno