The problem is as follows:

Determine the Galois group of the polynomial $f(x)=x^4-2$ over the finite field with $101$ elements, $\mathbb{F}_{101}$.

I am not really sure how to go about this, but here are some things I've thought about:

So, I know that Finite fields have cyclic Galois group. So, if $f(x)$ were irreducible over $\mathbb{F}_{101}$ then it would follow that it's splitting field is $\mathbb{F}_{101^4}$, and hence its Galois group would by isomorphic to $\mathbb{Z}_4.$ Similarly, if it reduced into a linear and a cubic, then the Galois group would be isomorphic to $\mathbb{Z}_3,$ and so on..

But in this case, it doesn't seem like it is the intention of the problem to check how $f(x)$ factors over $\mathbb{F}_{101}$. There seem to be way too many things that could happen. If it were over a field $\mathbb{F}_p$ for a reasonably small prime $p$ then one could check the irreducibility of $f(x)$ over this field.

The roots of $f(x)$ are certainly $\sqrt[4]{2}\zeta_4^k$, for $k=0,1,2,3,$ where $\zeta_4$ is the $4th$ primitive root of unity, so it is a root of the cyclotomic polynoial $\Phi_4(x)$. This in turn is a factor of $x^{101}-x$. So, in particular, it splits completely over $\mathbb{F}_{101}$, since this last one is the splitting field of $x^{101}-x.$

Hence, the splitting field of $f(x)$ over $\mathbb{F}_{101}$ is $\mathbb{F}_{101}(\sqrt[4]{2}).$

So, I guess, then the question is really if the following is true $$x^4\equiv 2\mod 101$$ for any $x$ in $\mathbb{F}_{101}.$

Another approach would be, i think, to try and compute the discriminant of $f(x)$ and it's cubic resolvent, but i am encountering similar issues of whether a polynomial is irreducible over this field. So, this approach seems to be even less worth pursuing.

I'd appreciate any help, or if there is any general strategy how to approach similar problems?!

  • $\begingroup$ Well, in general, Galois group of extensions of finite fields are cyclic, generated by the approprite Frobenious automorphism. This can be shown in general. $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '14 at 4:15

Why don't you see if $2$ is quadratic residue mod $101?$ WHat about $-1?$


If you know the result that $2$ is a square modulo a prime $p$ if and only if $p \equiv \pm 1 \mod 8$, then you can do this with virtually no calculation. Since $101 \equiv 5 \mod 8$ (you do need to do that calculation!), $2$ is a non-square in ${\mathbb F}_{101}$. Since $4$ does not divide $102$ (more calculation!), it follows that the splitting field of $x^4-2$ must be ${\mathbb F}_{101^4}$.

To see this, let $\beta$ be a square root of $2$ in ${\mathbb F}_{101^2}$. Let $w$ be a primitive element of ${\mathbb F}_{101^2}$ and $\beta = w^k$. So $w^{2k}=\beta^2=2 \in {\mathbb F}_{101}$, hence $102 \vert 2k$. So, if $k$ was even, we would have $102 \vert k$, but then $\beta \in {\mathbb F}_{101}$, contradiction. So $k$ is odd, and hence $\beta$ has no square root in ${\mathbb F}_{101^2}$. So the splitting field is not ${\mathbb F}_{101^2}$, and it must be ${\mathbb F}_{101^4}$.

  • $\begingroup$ Dear Derek, There is a typo on the third-last line: $2 = 2^{2k}$ should read $2 = \beta^{2k}$ (I think). Regards, $\endgroup$ – Matt E Jan 10 '14 at 12:16
  • $\begingroup$ Thanks! It should have been $w^{2k}$. It's corrected now. $\endgroup$ – Derek Holt Jan 10 '14 at 12:36
  • $\begingroup$ Ah yes, I confused $\beta$ and $w$. It looks good now! Cheers, $\endgroup$ – Matt E Jan 10 '14 at 16:23
  • $\begingroup$ @DerekHolt Thanks, I like this solution as well!( I didn't know that fact about 2 being a square modulo a prime $p$) Just to make sure i'm not missing something: So, we reduced the problem to saying that the splitting field of $x^4-2$ over $\mathbb{F}_{101}$ is $\mathbb{F}_{101}(\sqrt[4]{2})$. Next, since $2$ is not a square modulo $101$, we know that $\mathbb{F}_{101}(\sqrt{2})=\mathbb{F}_{101^2}.$ Now, we are claiming that $\sqrt{2}$ is also not a square in $\mathbb{F}_{101^2}$. Hence, it follows that $\mathbb{F}_{101^2}(\sqrt[4]{2})=\mathbb{F}_{101^4}$ is the splitting field. $\endgroup$ – V-B Jan 10 '14 at 18:06
  • $\begingroup$ sorry, to dredge this up, but I am not seeing why $102\vert 2k$ is implied? $\endgroup$ – qbert Apr 29 '18 at 2:30

Let $\alpha$ be a root of $X^4-2$ in $\overline{\mathbf{F}_{101}}$, what can you say about $\alpha^{{101}^k}$ for $k = 1,2,3$?

  • $\begingroup$ Well, if $\alpha$ is in the closure of $\mathbb{F}_{101}$ then it would be contained in some $\mathbb{F}_{101^k}$ for some $k$. So then, $\alpha^{101^k}=\alpha.$ If $k=1$,then $\alpha$ would be an element of $\mathbb{F}_{101}$. I don't see how this is helping me though? $\endgroup$ – V-B Jan 10 '14 at 4:31
  • $\begingroup$ Just compute $\alpha^{{101}^k}$ for $k = 1,2,3$, and you will see. $\endgroup$ – fkraiem Jan 10 '14 at 4:35
  • $\begingroup$ So, $\alpha^{101}=2^{25}\alpha$, which yields $\alpha^{100}=2^{25}$ So, it cannot be in $\mathbb{F}_{101}$ unless $2^{25}\equiv 1 \mod 101.$ Similarly, $\alpha^{101^2}=2^{25\cdot 102}\alpha$, which implies $\alpha^{101^2-1}=2^{25\cdot 102}$. So, it cannot be in $\mathbb{F}_{101^2}$ unless $2^{25\cdot 102}=\equiv 1\mod 101.$ We already know that $\alpha$ is in $\mathbb{F}_{101^4}$. So, I don't think we even need to check for $k=3$ right? Since $\mathbb{F}_{101^3}$ is not a subfield of $\mathbb{F}_{101^4}$. $\endgroup$ – V-B Jan 10 '14 at 5:11
  • $\begingroup$ I take it back when i said above that we already know $\alpha$ is in $\mathbb{F}_{101^4}$. So, $\alpha^{101^3}=2^{25\cdot(102\cdot 101+1)}\alpha$, implying that $\alpha^{101^3-1}=2^{25\cdot(102\cdot 101+1)}$. So, then this would imply that $\alpha$ must be in $\mathbb{F}_{101^4}$, and since $\alpha$ was arbitrary, this is actually the splitting field for $x^4-2$, right? $\endgroup$ – V-B Jan 10 '14 at 5:50
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    $\begingroup$ Computing $2^{25}\bmod{101}$ is not too difficult. ;) But yes, that's the idea. You have this element $\alpha$, which is a root of some irreducible factor of $f$. You do not know which one, but you do know that the Frobenius map $x \mapsto x^{101}$ sends it to another root of that same irreducible polynomial. So you repeatedly apply the Frobenius, you get four distinct values, so you know that the irreducible factor has degree $4$, and so must be te whole polynomial. $\endgroup$ – fkraiem Jan 10 '14 at 6:06

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