# Thread: How can I prove these two field extensions are equal?

1. ## How can I prove these two field extensions are equal?

How can I prove these two field extensions are equal?

Q(√3, -√3, i, -i) = Q(√3+i) where Q is the field of the rational numbers.

I got Q(√3+i) ⊆ Q(√3, -√3, i, -i), that direction is easy.

How can I prove Q(√3, -√3, i, -i) ⊆ Q(√3+i)?

Let m ∈ Q(√3, -√3, i, -i)

Therefore m = q + a√3 + bi + c(-√3) + d(-i) where q,a,b,c,d ∈ Q

Let p = a-c and r = b-d. Therefore, p,r ∈ Q

Therefore m = q + p√3 + ri

But I have to prove m = q + p√3 + ri = q + (SOMETHING IN Q)* (√3+i)

Any hints?

2. Originally Posted by Seda1
How can I prove these two field extensions are equal?

Q(√3, -√3, i, -i) = Q(√3+i) where Q is the field of the rational numbers.

I got Q(√3+i) ⊆ Q(√3, -√3, i, -i), that direction is easy.

How can I prove Q(√3, -√3, i, -i) ⊆ Q(√3+i)?

Let b ∈ Q(√3, -√3, i, -i)

Therefore b = q + p√3 + ri where q,p,r ∈ Q

The fact that Q is a field and has additive inverses takes care of the -√3 and -i.

But I have to prove b = q + p√3 + ri = q + (SOMETHING IN Q)* (√3+i)

Any hints?
Let a = "something in Q. Then $b = q + p\sqrt{3} + ri = q + a(\sqrt{3}+i) = q + a \sqrt{3} + ai \in Q(\sqrt{3},~-\sqrt{3}, ~i, ~-i)$

-Dan

3. Originally Posted by topsquark
Let a = "something in Q. Then $b = q + p\sqrt{3} + ri = q + a(\sqrt{3}+i) = q + a \sqrt{3} + ai \in Q(\sqrt{3},~-\sqrt{3}, ~i, ~-i)$

-Dan
Doesn't this assume p=r? I don't get it.

4. Originally Posted by Seda1
Doesn't this assume p=r? I don't get it.
My bad. I thought you were trying to show that one was a subset of the other, not equal.

-Dan

5. I'm trying to prove they're equal by showing each is a subset of the other.

I've shown Q(√3+i) ⊆ Q(√3, -√3, i, -i). Very quickly: m from Q(√3+i) can be written m = q +r(√3+i) = q + r√3 + ri, and let r = a-b,
m= q + a(√3) +a(i) + b(-√3) + b(-i), so m is in Q(√3, -√3, i, -i). Therefore Q(√3+i) ⊆ Q(√3, -√3, i, -i)

I'm trying to prove Q(√3, -√3, i, -i) ⊆ Q(√3+i).

Note: Cleaned up the question a bit.