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 $\Bbb{F}_3$
Question:
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
Best Answer
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 $C_n$.