Choosing splitting field of $x^4+x^2+1$ over $\mathbb{Q}$

Question

I choose my title "choosing" since I found two splitting field (I know splitting fields are unique!) Here is my solution to the question:

First, observe that $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ and I found that the set of solution is: $$\lbrace{\dfrac{-1-i\sqrt{3}}{2},\;\dfrac{-1+i\sqrt{3}}{2},\;\dfrac{1-i\sqrt{3}}{2},\;\dfrac{1+i\sqrt{3}}{2}\rbrace}$$

So I conclude that the splitting field over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt3,i)$.

But then I saw a solution saying splitting field over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{-3})$. At first I thought that they are equal but they aren't since degree of their minimal polynomial are 4 and 2, respectively. So, which one is true? Thanks in advance!

Answer 1

The splitting field of a polynomial is the smallest field such that the polynomial splits into linear factors.

As you can see that over $\Bbb{Q}(i\sqrt{3})$ the polynomial splits into it's linear factors. So if $E$ is the splitting field then $E\subseteq \Bbb{Q}(i\sqrt{3})$

Conversely the $\dfrac{-1+i\sqrt{3}}{2}-\dfrac{-1-i\sqrt{3}}{2}=i\sqrt{3}\,\in E$.

So $\Bbb{Q}(i\sqrt{3})\subseteq E$ as $\Bbb{Q}(i\sqrt{3})$ is the smallest field containing $\Bbb{Q}$ and $i\sqrt{3}$.

This means $\Bbb{Q}(i\sqrt{3})=E$

Now it is obvious that $[\Bbb{Q}(i\sqrt{3}):\Bbb{Q}]=2$ and $[\Bbb{Q}(i,\sqrt{3}),\Bbb{Q}]=4$ and hence $\Bbb{Q}(i\sqrt{3})\subsetneq \Bbb(i,\sqrt{3})$.

It is also obvious from the fact that $i\notin \Bbb{Q}(i\sqrt{3})$ and neither does $\sqrt{3}\notin\Bbb{Q}(i\sqrt{3})$ .

If $i=a+bi\sqrt{3}\,,a,b\in\Bbb{Q}$ then $a=0$ and $b=\frac{1}{\sqrt{3}}$ which is not possible .

And similarly $\sqrt{3}=a+ib\sqrt{3}\implies a=\sqrt{3}\,b=0$ which is not possible as $a\in\Bbb{Q}$.

So although over $\Bbb{Q}(i,\sqrt{3})$ the polynomial splits into linear factors, it is not the smallest extension of $\Bbb{Q}$ such that the "splitting" occurs. Namely it splits over a smaller subfield $\Bbb{Q}(i\sqrt{3})\subsetneq \Bbb{Q}(i,\sqrt{3})$ and we have shown above that it is indeed the splitting field.

Answer 2

Here’s a way of understanding your polynomial $f(x)=x^4+x^2+1$. Its roots are clearly the square roots of the roots of $g(x)=x^2+x+1$. That is, $f(x)=g(x^2)$.

Now I leave it to you to recognize that the roots of $g$ are the primitive cube roots of unity, $\frac{-1\pm\sqrt{-3}}2$. Their square roots all are sixth roots of unity, though not all of these are primitive. Indeed, as you have recognized, the roots of $f$ fall into two classes: roots of $x^2+x+1$ (primitive cube roots of unity) and roots of $x^2-x+1$ (primitive sixth roots of unity).

I think that you see that by adjoining the roots of $x^2-x+1$, you automatically get the roots of the other factor as a bonus. So your splitting field is quadratic, $\Bbb Q(\sqrt{-3\,}$ .