Find the degree of the splitting field of $x^4 + 1$ over $\mathbb{Q}$

I think I first need to find the $4$ roots of this polynomial and then calculate $\mathbb{Q}(\mbox{root }1, \mbox{root }2, \mbox{root }3, \mbox{root }4)$, right?

I know that this polynomail has roots only in the complex field, so I need to find them:

$$x^4 + 1 = (x^2-i)(x^2+i) = (x-\sqrt{i})(x+\sqrt{i})(x-\sqrt{-i})(x+\sqrt{-i})$$

so I need to calculate

$$\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i})$$

What do I need to do in order to calculate the degree of these? I thought about doing:

$$[\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i}):\mathbb{Q}(\sqrt{i}, -\sqrt{i})][\mathbb{Q}(\sqrt{i}, -\sqrt{i}):\mathbb{Q}]$$

is this right?

Then how to calculate $[\mathbb{Q}(\sqrt{i}, -\sqrt{i}):\mathbb{Q}]$? Because it'd be the field $\mathbb{Q}$ with $\pm\sqrt{i}$, but it must contain also its multiplicative inverse $\frac{1}{\sqrt{i}}$. I've discovered that this field must contain at least the elements $a, b\sqrt{i}, c\frac{1}{i}$ for $a,b\in\mathbb{Q},c\in\mathbb{Q}$. But how do I know that $\frac{1}{i}$ can't be formed with $a+b\sqrt{i}$ for example? If I find all the possible elements in the field $[\mathbb{Q}(\sqrt{i}, -\sqrt{i})]$, I can find a basis for it and then take its degree over $\mathbb{Q}$

Then, for the degree $[\mathbb{Q}(\sqrt{i}, -\sqrt{i}, \sqrt{-i}, -\sqrt{-i}):\mathbb{Q}(\sqrt{i}, -\sqrt{i})]$ I should verify if $\sqrt{i}$ and $\sqrt{-i}$ are independent. If we take $w = \sqrt{i}$ then $w^2 = i$ and $w^2$ is still in $\mathbb{Q}$, so $-w^2 = -i$. Is there an element in $\mathbb{Q}(\sqrt{i}, -\sqrt{i})$ such that its square is $-i$?

  • $\begingroup$ You might find it helpful to find a more explicit expression for a square root of $i$. $\endgroup$ – carmichael561 Sep 21 '17 at 20:47

Let $\alpha$ be a root of $x^4+1$. Then $\alpha^3,\,\alpha^5,\,\alpha^7$ are also (distinct!) roots of $x^4+1$. An easy way to see this is to consider 8th roots of unity, i.e. roots of $x^8-1 = (x^4+1)(x^4-1)$. If $\alpha$ is a primitive 8th root of unity, then every odd power $\alpha^{2k+1}$ is a root of $x^4+1$ (just draw a picture). Or you could just check it directly.

This means that $\mathbb Q[\alpha]$ is the splitting field of $x^4+1$. All you have to do now is to prove that $x^4+1$ is irreducible over $\mathbb Q$ to conclude that the degree of the splitting field is $4$.

EDIT: Perhaps a better way to show that $[\mathbb Q[\alpha]:\mathbb Q] = 4$ is to first notice that since $\alpha$ is a root of $x^4+1$ that $[\mathbb Q[\alpha]:\mathbb Q] \leq 4$. Now notice that $\alpha + \alpha^7 = \sqrt 2$, so $\mathbb Q[\sqrt 2]\subseteq \mathbb Q[\alpha]$. But, $\mathbb Q[\sqrt 2]\subseteq\mathbb R$, while $\alpha$ is complex, so $\mathbb Q[\sqrt 2]\neq \mathbb Q[\alpha]$, so it must be that $[\mathbb Q[\alpha]:\mathbb Q] = 4$.

In the bellow graph are shown 8th roots of unity. Red are roots of $x^4+1$ and blue are roots of $x^4-1$.

  • $\begingroup$ I understood, but how do I prove that $x^4+1$ is irreducible over $\mathbb{Q}$? $\endgroup$ – Guerlando OCs Sep 21 '17 at 22:55
  • $\begingroup$ @Guerlando OCs, $x^4+1 = (x^2+x\sqrt 2 + 1)(x^2-x\sqrt 2 + 1)$ is the unique factorization over $\mathbb R$. $\endgroup$ – Ennar Sep 22 '17 at 4:58
  • $\begingroup$ @Guerlando OCs, please see my edit. $\endgroup$ – Ennar Sep 22 '17 at 16:58

$x^4+1=\Phi_8(x)$, i.e. the roots of $x^4+1$ are the primitive eighth roots of unity and $x^4+1$ is the minimal polynomial of $\alpha=\exp\left(\frac{2\pi i}{8}\right)=\frac{1+i}{\sqrt{2}}$. The field extension $\mathbb{Q}\mapsto\mathbb{Q}(\sqrt{2})$ trivially has degree two and the same holds for the field extension $\mathbb{Q}(\sqrt{2})\mapsto\mathbb{Q}(\sqrt{2},1+i)$, since it is a complex extension and the minimal polynomial of $1+i$ over $\mathbb{Q}$ is given by $x^2-2x+2$.
$x^4+1$ has no rational root and neither $(x-\alpha)(x-\alpha^3)$, nor $(x-\alpha)(x-\alpha^5)$ nor $(x-\alpha)(x-\alpha^7)$ belong to $\mathbb{Q}[x]$, hence $x^4+1$ is irreducible over $\mathbb{Q}$.
Long story short, the degree of the splitting field of $\Phi_8(x)$ over $\mathbb{Q}$ is $\varphi(8)=4$.

  • $\begingroup$ I think I might have misunderstood what a splitting field is. I thought it was just a field with the roots adjoined. Does my reasoning and your answer work for the case where the degree to be calculated is not of the splitting field but instead is of the field with roots adjoined? $\endgroup$ – Guerlando OCs Sep 22 '17 at 3:12


$\zeta=\mathrm e^{\tfrac{i\pi}4}$ is one root of $x^4+1$. What are the other roots?

  • $\begingroup$ $e^{\frac{i\pi}{4}}\cdot e^{\frac{n\pi}{2}}$ for $n=1,2,3$, right? $\endgroup$ – Guerlando OCs Sep 21 '17 at 22:15
  • $\begingroup$ Almost. You're missing an $i$ in the second exponential. $\endgroup$ – Bernard Sep 21 '17 at 22:22
  • $\begingroup$ Thanks. Now, how should I proceed? I guess I'd just substitute these roots in the place of the ones in my question and try the same thing I was trying. Am I right? If so, how should I deal with the part about finding the basis of one extension over another? $\endgroup$ – Guerlando OCs Sep 21 '17 at 22:54
  • $\begingroup$ Just observe the roots of unity involved are powers of $\zeta$. $\endgroup$ – Bernard Sep 21 '17 at 22:55
  • $\begingroup$ So if the others are just powers, that means the field of the four roots, is equal to the field of just one root, right? So now how much is the degree of this field of only one root over $\mathbb{Q}$? $\endgroup$ – Guerlando OCs Sep 24 '17 at 0:39

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