Extension field on $\mathbb Q$

Question

Pick the correct statements

1. $\mathbb Q (\sqrt 2)$ and $\mathbb Q(i)$ are isomorphic as $\mathbb Q$ vector space.

2. $\mathbb Q (\sqrt 2)$ and $\mathbb Q(i)$ are isomorphic as fielods.

3. $Gal_{\mathbb Q}(\mathbb Q(\sqrt 2)/\mathbb Q) $ and $Gal_{\mathbb Q}(\mathbb Q(i)/\mathbb Q) $ are isomorphic.

4. $\mathbb Q(\sqrt 2)$ and $\mathbb Q (i)$ are both Galois extension of $\mathbb Q$.

For (1) define $T(a+b\sqrt 2) = a + bi$ for all a,b $\in \mathbb Q$. This is a linear transformation. So $\mathbb Q (\sqrt 2)$ and $\mathbb Q(i)$ are isomorphic as $\mathbb Q$ vector space.

For (4) Every extension of dimension 2 is normal extension. So $\mathbb Q(\sqrt 2)$ and $\mathbb Q (i)$ are both normal and seprable extension . So $\mathbb Q(\sqrt 2)$ and $\mathbb Q (i)$ are both Galois extension of $\mathbb Q$.

For (3) We know that $|\mathbb Q(\sqrt 2)|= 2$ and $|\mathbb Q (i)| = 2$, So both are cyclic groups . So they are isomorphic.

Please tellme about (2).

Thank you

Answer 1

Hint: In $\mathbb Q(i)$ we have $0 = i^2 + 1$. If $\phi\colon \mathbb Q(i) \to \mathbb Q(\sqrt 2)$ were a field isomorphism, we would have $0 = \phi(i)^2 + 1$. (1), (3), (4) look fine.

Answer 2

Suppose $\eta :\mathbb{Q}(i)\rightarrow \mathbb{Q}(\sqrt{2})$ is an isomorphism...

With all due respect $\eta (i)=a+b\sqrt{2}$ for some $a,b\in \mathbb{Q}$

i.e., $\eta(i)\cdot \eta(i)=(a+b\sqrt{2})^2\Rightarrow ****\Rightarrow -1=(a^2+2b^2)+2ab\sqrt{2}$

Now do you see something is wrong?