# Thread: Show Q[i + sqrt(2)] = Q[i][sqrt(2)]

1. ## Show Q[i + sqrt(2)] = Q[i][sqrt(2)]

Show $\displaystyle Q[i + \sqrt{2}] = Q[i][\sqrt{2}]$

Now, when I do this, I find:
Basis of $\displaystyle Q[i + \sqrt{2}] = \{1, i+\sqrt{2}, 1+2i\sqrt{2}, 5i-\sqrt{2}\}$ which is equal to $\displaystyle \{1, i, \sqrt{2}, i\sqrt{2}\}$. Which is the same basis as $\displaystyle Q[i][\sqrt{2}]$.

If the basis of each field is shown to be equal, are the fields equal? It seems like an obvious yes to me, but I would like to someone else to confirm this thought if possible. I cannot find it in my text or any other resource online. Also, if someone feels like double-checking the basis for each, I would appreciate it.

2. Originally Posted by tcRom
Show $\displaystyle Q[i + \sqrt{2}] = Q[i][\sqrt{2}]$

Now, when I do this, I find:
Basis of $\displaystyle Q[i + \sqrt{2}] = \{1, i+\sqrt{2}, 1+2i\sqrt{2}, 5i-\sqrt{2}\}$ which is equal to $\displaystyle \{1, i, \sqrt{2}, i\sqrt{2}\}$. Which is the same basis as $\displaystyle Q[i][\sqrt{2}]$.

If the basis of each field is shown to be equal, are the fields equal? It seems like an obvious yes to me, but I would like to someone else to confirm this thought if possible. I cannot find it in my text or any other resource online. Also, if someone feels like double-checking the basis for each, I would appreciate it.
Yes, that is how you do this sort of problem. You should write each member of the basis of Q[i+/sqrt(2)] as a linear combination of basis elements from Q[i,/sqrt(2)] and vice versa.

3. Great, thank you.

I end up with
let $\displaystyle a,b,c,d \in Q$
$\displaystyle \Rightarrow a + b(i+\sqrt{2}) + c(1 + 2i\sqrt{2}) + d(5i - \sqrt{2}) =$
$\displaystyle (a+c)(1) + (b+5d)(i) + (b-d)(\sqrt{2}) + (2c)(i\sqrt{2})$... a linear combination of the basis of Q[i][sqrt(2)]. Thus, the fields are equal.

Am I able to say that the basis are equal, even though it takes a linear combination to reach the other?

4. Originally Posted by tcRom
Great, thank you.

I end up with
let $\displaystyle a,b,c,d \in Q$
$\displaystyle \Rightarrow a + b(i+\sqrt{2}) + c(1 + 2i\sqrt{2}) + d(5i - \sqrt{2}) =$
$\displaystyle (a+c)(1) + (b+5d)(i) + (b-d)(\sqrt{2}) + (2c)(i\sqrt{2})$... a linear combination of the basis of Q[i][sqrt(2)]. Thus, the fields are equal.

Am I able to say that the basis are equal, even though it takes a linear combination to reach the other?
The fields, and their bases, are equal up to isomorphism. Sometimes we have different criteria for what it means for two things to be equal. But that is semantics. You can say that they are equal (equivalent, isomorphic, etc.)