I am learning to solve problems like these. I think I should follow the following steps:

  1. Determine if $x^3-2$ is reducible in $\mathbb{F}_7$. By simply plugging in the equivalence classes I see that none of them will give me the equivalence class containing 0. Hence the polynomial is irreducible over the field.

  2. Now, I need to find the roots and adjoin them to the field. From examples given in the book dummit and foote I know that $\sqrt[3]{2}, \zeta_3 \sqrt[3]{2}, \zeta_3^2 \sqrt[3]{2}$ are roots for $x^3-2$. So, I think I am looking at the field $\mathbb{F}_7(\sqrt[3]{2}, \zeta_3 \sqrt[3]{2}, \zeta_3^2 \sqrt[3]{2})$, since no multiple roots, this means the the Automorphism group of the field is the Galois group?

  3. Now, I compute the Galois group by using the fact automorphism takes generators to generators. So, I just work with $\sqrt[3]{2}, \zeta_3 \sqrt[3]{2}, \zeta_3^2 \sqrt[3]{2}$ and look at maps onto themselves and then determine which one is or is not an isomorphism. There are total 9 possible such maps , since I know that the degree of the splitting field is at most $($deg $x^3-2)!= 3!$ which is also the cardinality of the Galois group. So I must have 3 maps that arent isomorphisms.

Where $\zeta_3$ is the primitive third root of unity. Then I arrived at the same Galois group as $\mathbb{Q}[x]/(x^3-2)$, $S_3$. Is this correct? And is that generally how one should solve such type of problem? Thanks

  • $\begingroup$ I'm not sure what you mean by "total 9 possible such maps". Any onto map of a finite set to itself must also be injective. So the automorphisms permute the generators, and some additional facts are known about the permutation group they form. $\endgroup$
    – hardmath
    Commented May 13 at 19:23
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    $\begingroup$ No, this is not correct. You have a few steps in the process reasonably correct, but there are also meaningful errors. You need to be careful about what you mean when you say $\zeta_{3}$, which I assume you are using to denote a primitive third root of unity. It is also not quite precise to say that "automorphism of the splitting field takes generators to generators". Finally, you are begging the question when you say that $3!$ "is also the cardinality of the Galois group", since you do not know how big the Galois group is a priori and haven't proven anything about it yet. $\endgroup$ Commented May 13 at 19:25
  • $\begingroup$ In fact, the Galois group of this extension is cyclic of degree three, not $S_{3}$. Let's start with the first big clarifying point: what did you mean in your post when you wrote $\zeta_{3}$? What is your definition of primitive 3rd root of unity? $\endgroup$ Commented May 13 at 19:26
  • $\begingroup$ @AlexWertheim Thanks! and yes it is primitive third root of unity. Thank you for pointing out everything. I basically just trying to mimic the steps I see in Dummit and Foote when it solves for Galois group of $x^3-2$ over $\mathbb{Q}$. $\endgroup$
    – Remu X
    Commented May 13 at 19:56
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    $\begingroup$ You missed the fact that $2^3=1$ in $\Bbb{F}_7$. Therefore $2$ and $2^2=4$ play the roles of $\zeta_3$ and $\zeta_3^2$. Because $\zeta_3\in\Bbb{F}_7$ you don't need to adjoin it, and you only get a degree three extension. A more general fact is that any finite extension $L/K$ of finite fields is normal, implying that any irreducible (over $K$) polynomial with at least a single zero in $L$, will split completely. $\endgroup$ Commented May 14 at 4:43

1 Answer 1


The order of the Galois group is the degree of the extension.

Since $2$ has order $3$ in $\Bbb F_7,$ a root $\alpha $ of $x^3-2$ will have order $9.$

So if we call the degree of the extension $n,$ we need $9\mid (7^n-1).$

The smallest such $n$ is $n=3.$

Thus the Galois group has order $3,$ and must be cyclic.

Or, note that $x^3-2$ is monic irreducible, so the minimal polynomial of each of its roots.

We only need to attach $1$ root then, because we're going into a Galois field.

Thus the extension is degree $3.$

And there's only one group of order $3.$

As a slight variant, you can also look for the smallest iterate of the Frobenius ($x\to x^7$) for which $\alpha $ is fixed.

  • $\begingroup$ Thanks! I just want to make sure I understand everything here. It looks I made the mistake by forgetting that we are working in $\mathbb{F}_7$. So, after I made sure that $x^3-2$ has no root in $\mathbb{F}_7$. I should let $\alpha$ be any root of it and work with that instead of adjoining real roots. I see that $2$ has order $3$ so it serves as primitive roots of unity here. But how do I know that $\alpha$ has order 9? I must be missing some basic facts. $\endgroup$
    – Remu X
    Commented May 13 at 20:35
  • $\begingroup$ Ohh I see $\alpha^3=2$ and $2$ has order 3 hence $\alpha ^9 = 1$ $\endgroup$
    – Remu X
    Commented May 13 at 20:37
  • $\begingroup$ But now could you help me understand why your third line holds? Why we must have $9| (7^n-1)$? $\endgroup$
    – Remu X
    Commented May 13 at 20:39
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    $\begingroup$ Yes. Since a root $\alpha $ has order $9,$ the multiplicative group of the extension nust have order a multiple of $9.$ Meanwhile that group necessarily has order $7^n-1,$ where $n$ is the degree of the extension. Good question. $\endgroup$
    – i can try
    Commented May 13 at 20:52

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