We know that if $\Bbb{F}_p$ is the field of integer modulo $p$ a prime. If $L/\Bbb{F}_p$ is a finite extension with degree $[L:\Bbb{F}_p]=n$. Then $Gal(L:\Bbb{F}_p)$ is a cyclic group of order $n$, generated by
$\phi:L \to L$ where $\phi(\alpha)=\alpha^p$, for $\alpha \in L$.
I have to find the galois group of the following:
i) $x^3 + x +1$ over $\Bbb{F}_5$
ii) $x^4-1$ over $\Bbb{F}_7$
iii) $x^4+x-1$ over $\Bbb{F}_3$

Can I say that from the fact that the extension will have some kind cyclic Galois group, generated by the above homomorphism; to obtain a splitting field from an irreducible polynomial we only have to adjoin $1$ root?? And hence we only need to find the degree, say $n$, of the irreducible polynomial in question and then $Gal(f)=C_n$?? This would give the following:

So for i) $f(x)=x^3+x+1$ has no root so is irreducible and so $Gal(F) = C_3$ - We already knew this anyway, we can check the discriminant and see that it is square in $\Bbb{F}_5$, so we have galois group $A_3 \cong C_3$ by well known result.

ii) $f(x) = x^4-1 = (x-1)(x+1)(x^2+1)$ and as $(x^2+1)$ is irreducible over $\Bbb{F}_7$ all we have to do is adjoin roots of this. $Gal(f) = C_2$?

iii) $f(x) = x^4+x-1$ which I've shown is irreducible in $\Bbb{F}_3$ so $Gal(f) = C_4$.

Is this correct- I think it is but it seems a little, simple. Thanks in advance

  • $\begingroup$ For example in iii) we have that the splitting field is $F_3(a_1, a_2, a_3, a_4)$ whose degree over $F_3$ is 4. So we have $[F_3(a_1, a_2, a_3, a_4) : F_3(a_i)][F_3(a_i) : F_3] = 4.$ That is $F_3(a_1, a_2, a_3, a_4) = F_3(a_i).$ Not sure if this is what you're looking for... $\endgroup$
    – green frog
    May 25 '17 at 2:38
  • $\begingroup$ Yeah that is what I'm asking; that due to the cyclic nature of the roots, that the degree of the extension is just the degree of the (irreducible part of the) polynomial, since we only have to add in one root for the splitting field? My question, is exactly that. As I said, I think it's true, but I'm not sure. By my reasoning this will always be the case for these finite fields $\endgroup$
    – SEWillB
    May 25 '17 at 2:42
  • $\begingroup$ Dear @ntntnt, how did you manage to argue a priori that $\mathbb F_3 (a_1, a_2, a_3, a_4)$ must have degree $4$ over $\mathbb F_3$? I'm afraid I don't see it, but it might just be one of those days... $\endgroup$
    – Kenny Wong
    May 25 '17 at 2:52
  • $\begingroup$ The crux of my argument was just to show that for any splitting field of the form $F(a_1, ..., a_n)$ we have the subfield $F(a_i)$ which we know is also the splitting field of $f.$ Hence, $F(a_i) = F(a_1, ..., a_n).$ Might have badly worded or omitted these exact words but this was what I was thinking. As for your question I don't see how you can show apriori that the extension has degree 4...I thought OP wanted to assume this fact. $\endgroup$
    – green frog
    May 25 '17 at 3:06
  • $\begingroup$ Show that if $f \in \mathbb{F}_p[x], f(\alpha) = 0$ then $f(\alpha^p) = f(\alpha^{p^m}) = 0$ (using the Frobenius automorphism $\alpha \mapsto \alpha^p$) $\endgroup$
    – reuns
    May 25 '17 at 3:12

Yes, I believe the result is true. In fact, to argue that the splitting field of a degree-$n$ irreducible polynomial $f(X) \in \mathbb F_{p}[X]$ is generated by a single root $\alpha$, all we need to know is that $\mathbb F_{p}(\alpha) \cong \mathbb F_{p^n}$ is a Galois extension of $\mathbb F_p$ (being the splitting field of $X^{p^n} - X$). Since $f$ is irreducible with at least one root $\alpha$ in the extension field $\mathbb F_p (\alpha)$, and since the extension is Galois, it immediately follows that $f$ splits completely in $\mathbb F_p(\alpha)$. Having established this, it then follows that the Galois group of $f$ is $C_n$.

  • $\begingroup$ Wait can we also just argue that $F_p(\alpha)$ is always a subfield of the splitting field $F_{p^n}$ of the same degree extension so that it too is always Galois? $\endgroup$
    – green frog
    May 25 '17 at 2:50
  • $\begingroup$ Well, $\mathbb F_p(\alpha)$ is actually equal to $\mathbb F_{p^n}$, and since $\mathbb F_{p^n} : \mathbb F_{p}$ is Galois, this means that $\mathbb F_p(\alpha) : \mathbb F_p$ is Galois. That is my argument. $\endgroup$
    – Kenny Wong
    May 25 '17 at 2:53
  • $\begingroup$ Is my observation not true that since the automorphisms (which are $\phi:\alpha \to \alpha^p$) must map roots of the irreducible polynomial back to the roots of the polynomial we have the property I want. In any case this has cleared everything up. To run through the argument as a check: $\Bbb{F}_p(\alpha)$ is a degree $n$ extension and so has $p^n$ elements so the multiplicative group has $p^n-1$ elements and so for any $a \in \Bbb{F}_p$ we have that $a^{p^n-1} = 1$ so all elements are roots of $X^{p^n}-X$ as you said, so this is the splitting field. $\endgroup$
    – SEWillB
    May 25 '17 at 2:56
  • $\begingroup$ And then it's separable as formal derivative of $X^{p^n}-X$ is $-1 \ne 0$. So we're Galois and we're done $\endgroup$
    – SEWillB
    May 25 '17 at 2:57
  • $\begingroup$ @SEWillB Yes, you understood my argument. The separability isn't necessary here, because the statement "$f$ irreducible with one root in extension field $\implies$ $f$ splits completely in extension field" only uses the fact that the field extension is normal. $\endgroup$
    – Kenny Wong
    May 25 '17 at 2:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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