# Let $f(x)=(x^2+x+1)(x^2+x-1)$. Find $(E:\Bbb Q)$

I would like some feedback or verification if the steps I used below to calculate $$(E:\Bbb Q)$$ are correct. Thank you for your time.

Let $$f(x)=(x^2+x+1)(x^2+x-1)$$. If $$E$$ is the minimal splitting field of $$f$$ over $$\Bbb Q$$, I need to determine $$(E:\Bbb Q)$$.

First, I found the roots of $$x^2+x+1$$ to be $$-\frac{1}{2}\pm\frac{\sqrt3}{2}i$$, and the roots of $$x^2+x-1$$ to be $$-\frac{1}{2}\pm\frac{\sqrt5}{2}$$. I then concluded that both $$x^2+x+1$$ and $$x^2+x-1$$ are irreducible over $$\Bbb Q$$, and so $$\Bbb Q(\sqrt3i,\sqrt5)$$ is the minimal splitting field.

$$\Bbb Q(\sqrt3i,\sqrt5)=\frac{\Bbb Q(\sqrt5)[x]}{(x^2+3)}$$ and $$\Bbb Q(\sqrt5)=\frac{\Bbb Q[x]}{(x^2-5)}$$, so $$(\Bbb Q(\sqrt3i,\sqrt5):\Bbb Q(\sqrt5))=2$$ and $$(\Bbb Q(\sqrt5):\Bbb Q)=2.$$

Hence, $$(\Bbb Q(\sqrt3i,\sqrt5):\Bbb Q)=(\Bbb Q(\sqrt3i,\sqrt5):\Bbb Q(\sqrt5))\cdot(\Bbb Q(\sqrt5):\Bbb Q)=2\cdot2=4$$

• To be complete, you just have to explain why $x^2+3$ is irreducible over $\mathbf Q(\sqrt 5)$. – Bernard Mar 28 at 22:47
• Would that involve showing that $\sqrt3i$ cannot be of the form $a+b\sqrt5$ for $a,b\in \Bbb Q$? – KronZ Mar 28 at 22:50
• It ensures a proof that $[\mathbf Q(\sqrt3i,\sqrt5):\mathbf Q(\sqrt5) ]=2$. – Bernard Mar 28 at 22:55