Degree of splitting fields

I'm learning about splitting fields but I'm not sure if I am right. Hopefully I can get some insights on whether I have been learning correctly.

The question asks to find the degree of the splitting fields over $$\mathbb Q$$.

In $$x^4+x^3+x+1$$, we can factorize it to become $$(x+1)^2(x^2-x+1)=(x+1)^2(x-\frac{1+\sqrt 3i}2)(x-\frac {1-\sqrt 3i}{2})$$. Am I right to say that we want to find $$[\mathbb Q(\sqrt 3i):\mathbb Q]$$? Hence the answer is just $$2$$.

In $$x^4+x^2+x+1$$, we can factorize it to become $$(x-\frac 14\sqrt5+\frac 14-\frac14i\sqrt2\sqrt{5+\sqrt5})(x+\frac 14\sqrt5+\frac 14-\frac14i\sqrt2\sqrt{5-\sqrt5})\\(x+\frac 14\sqrt5+\frac 14+\frac14i\sqrt2\sqrt{5-\sqrt5})(x-\frac 14\sqrt5+\frac 14+\frac14i\sqrt2\sqrt{5+\sqrt5})$$.

Hence, we want to find $$\mathbb Q(\sqrt5,i\sqrt2\sqrt{5+\sqrt5},i\sqrt2\sqrt{5-\sqrt5})$$. But how do we go about doing this?

• Yes, the degree of the spliting field of $(X-1)^2(X^2-X+1)$ is of course equal to $2$. Why do you want to involve $\mathbb Q(\sqrt5,i\sqrt2\sqrt{5+\sqrt5},i\sqrt2\sqrt{5-\sqrt5})$? – Dietrich Burde Jan 9 at 12:11

Your field is $$\mathbb{Q}(\sqrt{5})(i\sqrt{10-2\sqrt{5}})$$, because the product of $$i\sqrt{2}\sqrt{5 \pm \sqrt{5}}$$ is $$-2(5-\sqrt{5})\in \mathbb{Q}(\sqrt{5})$$.

So you have a biquadratic extension and the degree is $$4$$.

• How did we get the degree to be 4? Should it be 8? – Icycarus Jan 9 at 12:03
• I edited, now there shouldn’t be any reason why the degree is $8$. – Mindlack Jan 9 at 12:06
• ahh! okay I understand now! Thank you much! Can I ask if my first question is correct as well? – Icycarus Jan 9 at 12:10
• I think your own answer to the first point is correct. – Mindlack Jan 9 at 12:11