Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having a bit of difficulty trying to answer the following question:

What is the Galois group of $X^8-1$ over $\mathbb{F}_{11}$?

So far I have factored $X^8-1$ as


I know $X^2+1$ is irreducible over $\mathbb{F}_{11}$ since $10$ is not a square modulo $11$. Also, $X^4+1$ is irreducible over $\mathbb{F}_{11}$. The roots of $X^2+1$ and $X^4+1$ over $\mathbb{Q}$ are $\pm i$ and $\pm \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2} i$, respectively. We also see that $\sqrt{2} \not \in \mathbb{F}_{11}$ since no element squared is equal to $2$. I would then think that $\mathbb{F}_{11}(i, \sqrt{2})$ is a splitting field for $x^8-1$ over $\mathbb{F}_{11}$, which is clearly Galois. If all this were true, I would then venture that the Galois group is $V_4$. I have the feeling, however, that I have made many mistakes in my reasoning. How should one approach a problem like this?

share|cite|improve this question
The Galois group cannot be $V_4$, because extensions of Galois groups are always cyclic: their Galois groups are cyclic. – Arturo Magidin Jun 20 '12 at 13:37
@DonAntonio I wrote it as $X+10$ to emphasize I was working modulo 11 – Holdsworth88 Jun 20 '12 at 14:07
See this question and my answer to it for an explanation as to why $x^4+1$ is reducible over any finite field. TonyK's comments allowed me to write down an algorithm for finding its factors. – Jyrki Lahtonen Jun 20 '12 at 14:30
So in particular, the Galois group of $x^8-1$ over $\mathbb{F}_p$ is either trivial (if $p\equiv1\pmod8$) or cyclic of order two (if $p\not\equiv1\pmod8$). – Jyrki Lahtonen Jun 20 '12 at 15:00
@Holdsworth, indeed. I'll edit it back. – DonAntonio Jun 20 '12 at 15:30
up vote 15 down vote accepted

An extension of finite fields is always cyclic: the Galois group must be cyclic. So the Galois group certainly cannot be $V_4$.

Note that $F_{11}(i)$ does have a square root of $2$: $(3i)^2 = -9\equiv 2\pmod{11}$. So once you adjoint $i$ to $F_{11}$, you also get $\sqrt{2}$. Thus, $F_{11}(i,\sqrt{2}) = F_{11}(i)$.

Likewise, $X^2+1$ is reducible over $F_{11}(\sqrt{2})$, since $(4\sqrt{2})^2 = 32\equiv -1\pmod{11}$. That is, $F_{11}(i) = F_{11}(\sqrt{2})$.

The key to remember is that there is a unique field of order $121=11^2$; so any extension you get from $F_{11}$ by adjoining the square root of a nonquadratic residue is the same. So $F_{11}(i) = F_{11}(\sqrt{2}) = F_{11}(\sqrt{6}) = F_{11}(\sqrt{7}) = F_{11}(\sqrt{8})$.

Your other mistake is that $x^4+1$ is not irreducible over $F_{11}$: it splits as a product of two irreducible quadratics. You can figure this out by replacing $\frac{\sqrt{2}}{2}$ with $7i$ and $\frac{\sqrt{2}}{2}i$ with $-7$ (the values in $F_{121}$), or by solving a system of simple equations. Either way, you get $$x^4 + 1 = (x^2-3x-1)(x^2+3x-1)$$ in $F_{11}$. (Had $x^4+1$ been irreducible, then your extension would have been of degree $4$: you need an extension of degree $4$ to get a root for $x^4+1$, that one already contains the quadratic extension, and so you would get all the roots you need. In that case, the Galois group would have been cyclic of order $4$).

share|cite|improve this answer
+1 for a complete answer. The link I gave in my other comment leads to an alternative way of factoring $x^4+1$. – Jyrki Lahtonen Jun 20 '12 at 14:33

Your mistake was in assuming that $x^4 + 1$ was irreducible over $\Bbb{F}_{11}$. In fact there is something incredible about this polynomial; it is irreducible over $\Bbb{Z}$ but reducible over $\Bbb{F}_p$ for every prime $p$!

The following is the proof given in Dummit and Foote:

If $p = 2$ the polynomial is clearly reducible by the schoolboy binomial theorem. Now if $p$ is odd, notice that

$$p^2 \equiv 1 \mod 8$$

for every prime $p$. This is because mod 8, $p$ is congruent to 1,3,5, or 7 all of which square to 1 mod 8. It follows that we have the following divisibilities:

$$x^4 + 1 | x^8 - 1 | x^{p^2 - 1} - 1| x^{p^2} - x.$$

In particular we conclude that $x^4 + 1 | x^{p^2} - x$ and so all the roots of $x^4 + 1$ are roots of $x^{p^2} - x$. It follows that the extension generated by any root of $x^4 + 1$ is at most a degree 2 extension, and so $x^4 + 1$ cannot be irreducible.

$\hspace{6in} \square$

It follows that the Galois group of $x^8 - 1$ over $\Bbb{F}_{11}$ is isomorphic to the cyclic group of order 2.

share|cite|improve this answer

Let $E$ be a splitting field of $x^8 - 1$ over $F_{11}$. Then $E = F_{11}(\alpha)$, where $\alpha$ is a primitive $8$th root of unity.

It follows that $G = \operatorname{Gal}(E/F)$ is isomorphic to a subgroup of $(\mathbb{Z}/8\mathbb{Z})^*$, which is isomorphic to the Klein four-group. Now $G$ cannot have order $1$ because $x^8 - 1$ does not split over $F_{11}$, it cannot have order $4$ because then it wouldn't be cyclic (see this question). Therefore $G$ must be cyclic of order $2$.

In general, suppose $F$ is a field and $E = F(\alpha)$, where $\alpha$ is a primitive $n$th root of unity. Then you can show that $\operatorname{Gal}(E/F)$ is isomorphic to a subgroup of $(\mathbb{Z}/n\mathbb{Z})^*$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.