Intersection of fixed fields

Solution 1:

It's hard to write this up precisely because you don't say what $\alpha$ is. Since $\sigma(\alpha) = i \alpha$ is an automorphism, it's reasonable to guess that $\alpha = \sqrt[4]{2}$ or similar. I'll be working under this assumption, but it's not hard to modify this approach for whatever your $\alpha$ happens to be.

As a completely brute-force approach, recall

$$ \mathbb{Q}(\alpha) = \{ a + b \alpha + c \alpha^2 + d \alpha^3 \} $$

$$ \mathbb{Q}(\alpha^2, i) = \{ e + b i + c \alpha^2 + d \alpha^2 i \} $$

where all of the coefficients come from $\mathbb{Q}$.

But then what does it mean for an element to be contained in both of these fields? Well it must be of the form

$$ \{ a + c \alpha^2 \} $$

So we get $\mathbb{Q}(\alpha^2)$ as the intersection. Which agrees with what your teacher said.

I hope this helps ^_^