Finding the Galois group of $\mathbb{Q}(\alpha,i)|\mathbb{Q}$, whereby $\alpha^2 = \frac{3}{2}+\frac{1}{2}i\sqrt7$

Question

I would like to find the Galois group of the finite extension $\mathbb{Q}(\alpha,i)|\mathbb{Q}$, whereby $\alpha^2 = \frac{3}{2}+\frac{1}{2}i\sqrt7$.

Note that $\mathbb{Q}(\alpha,i)$ is the splitting field of $(X^4-3X^2+4, X^2+1)$.

I would know what to do if I would find a primitive element of the extension, but that seems not so easy to find.

I would appreciate any hints.

Answer 1

Since the four roots of $X^4-3X^2+4$ are equal to $\pm(\frac{\sqrt7\pm i}{2})$, we find $\mathbb Q(\alpha,i)=\mathbb Q(\sqrt7,i)$ and the degree of the extension is $4$. As the Galois group has two elements of order $2$ (one sending $i$ to $-i$ and another sending $\sqrt7$ to $-\sqrt7$), it must be $\mathbb Z/(2)\times\mathbb Z/(2)$.

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Hope this helps.