I am trying to prove the following:
If there exists a ring isomorphism $f:\mathbb{Q}(\sqrt{2})\longrightarrow\mathbb{Q}(i)$, then there exists a ring isomorphism $g:\mathbb{Q}(\sqrt{2})\longrightarrow\mathbb{Q}(i)$ such that $g(\sqrt{2})=i$.
My approach to the problem was trying to find a ring isomorphism $h:\mathbb{Q}(\sqrt{2})\longrightarrow\mathbb{Q}(\sqrt{2})$ such that $h(\sqrt{2})=a$, where $a\in \mathbb{Q}(\sqrt{2})$ is such that $f(a)=i$. Then I would define $g=f\circ h$. However, I have not been able to find a ring isomorphism $h$ that works.
It would be very helpful if anyone could suggest either a different approach to tackle the problem or maybe the ring isomorphism $h$ I was looking for.
Thanks in advance.
Well, there is no ring isomorphism $f : \Bbb Q(\sqrt{2}) \to \Bbb Q(i)$.
Indeed, if $f$ were such an isomorphism, then $f$ would have to map $1$ to $1$ and by extension, $f(n) = n$ for all $n \in \Bbb Z$.
Now, note that $\sqrt{2}^2 = 2$. Applying $f$ on both sides and using the fact that it is a homomorphism along with the above observation gives $$f(\sqrt{2})^2 = 2.$$ However, $f(\sqrt{2}) \in \Bbb Q(i)$ but there is no element in $\Bbb Q(i)$ whose square is $2$.
Thus, the given statement is true vacuously.