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

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

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