# Prove that $[\mathbf{Q}(\sqrt{1+i},\sqrt{2}):\mathbf{Q}]=8$.

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

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

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

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.