6
$\begingroup$

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 $[\mathbf{Q}(\sqrt{1+i},\sqrt{2})]=8$?

$\endgroup$
  • $\begingroup$ Degree $8$ is proved here. $\endgroup$ – Dietrich Burde Apr 15 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 Apr 15 at 19:22
2
$\begingroup$

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 $[K:\mathbf Q]=8$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.