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I am trying to calculate the Galois group of the polynomial $f=X^4-2X^2+2$. $f$ is Eisenstein with $p=2$, so irreducible over $\mathbf{Q}$. I calculated the zeros to be $\alpha_1=\sqrt{1+i},\alpha_2=\sqrt{1-i},\alpha_3=-\alpha_1$ and $\alpha_4=-\alpha_2$. Let $\Omega_f=\mathbf{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=\mathbf{Q}(\alpha_1,\alpha_2)$ be a splitting field of $f$ over $\mathbf{Q}$. Since $\alpha_1\alpha_2=\sqrt{1+i}\sqrt{1-i}=\sqrt{2}$, we have $\Omega_f=\mathbf{Q}(\sqrt{1+i},\sqrt{2})$.

So if we can prove that $[\Omega_f:\mathbf{Q}]=8$, then we have $\#\operatorname{Gal} (f)=8$ and for $\operatorname{Gal}(f)\subset S_4$, we must have that it is isomorphic to the dihedral group $D_4$.

How do I go about proving $[\mathbf{Q}(\sqrt{1+i},\sqrt{2})]=8$?

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  • $\begingroup$ Degree $8$ is proved here. $\endgroup$ Commented Apr 15, 2019 at 18:43
  • $\begingroup$ @DietrichBurde there is no need for a downvote, I was aware of that question but I did not find the explanation clear enough. $\endgroup$
    – rae306
    Commented Apr 15, 2019 at 19:22

2 Answers 2

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You have shown that your splitting field is $K=\mathbf Q(\sqrt {1+i}, \sqrt {1-i})$. The two fields $\mathbf Q(\sqrt {1\pm i})$ are obviously quadratic extensions of $\mathbf Q(i)$, and these are equal iff $(1+i)(1-i)=2$ is a square in $\mathbf Q(i)$, iff $\sqrt 2\in \mathbf Q(i)$: impossible. Hence $K$ is a biquadratic extension of $\mathbf Q(i)$, and $[K:\mathbf Q]=8$.

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Break it into towers.

Look at the two extensions $\mathbb Q ( \sqrt {1+i} )|_{Q(i)} $ and $\mathbb Q ( \sqrt {1-i} )|_{Q(i)} $.

Each has degree $2$ ( since $1+i$ and $1-i$ are primes in $\mathbb Z[i] $ which is a UFD ).

Their compositum is the field you are interested in i.e. $ \mathbb Q (\sqrt {1+i} , \sqrt {1-i} )$

At this point you can have a look at the question Finding degree of a finite field extension

I hope you have seen the similarity.

You have a UFD $\mathbb Z[i]$ , it's field of fractions $\mathbb Q(i)$ and you have adjoined square roots of two distinct primes $1+i , 1-i$.

By a similar argument as in the question above we argue the Galois group is of the form $\mathbb Z_2 ^k$ with $k\leq 2$.

But then you got $3$ distinct degree $2$ subextensions $\mathbb Q(\sqrt {1+i}), \mathbb Q(\sqrt {1-i}) , \mathbb Q(\sqrt {2}) $

This gives you $k\geq 2$ So the upshot is $$\mathbb Q(\sqrt {1+i}, \sqrt 2): \mathbb Q(i)=4$$

And conclusion follows.

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