# Thread: Extension Fields

1. ## Extension Fields

Prove that
$\displaystyle \mathbb{Q}$($\displaystyle \sqrt{2}$,$\displaystyle \sqrt{3}$) is isomorphic to $\displaystyle \mathbb{Q}$($\displaystyle \sqrt{2}$ + $\displaystyle \sqrt{3}$)

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 $\displaystyle \sqrt{2}$,$\displaystyle \sqrt{3}$ and $\displaystyle \sqrt{2}$ + $\displaystyle \sqrt{3}$ are all zero that would be great.

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

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

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

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

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

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

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

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

So the sets are the same

3. your conclusion is correct, but your reasoning is slightly off.

q has the form a + b√2 + c√3 + d√6, so √2 is the vector (0,1,0,0) and √3 is the vector (0,0,1,0) relative to the basis {1,√2,√3,√6}.

the implication the other way is fine.

a more in-depth look at Q(√p,√q) for p,q distinct primes was posted here: http://www.mathhelpforum.com/math-he...ml?pagenumber=