Does anyone know how to find the Galois group of x^3-3 over the finite field F_13?

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- May 6th 2010, 07:27 AMKSM08Galois group
Does anyone know how to find the Galois group of x^3-3 over the finite field F_13?

- May 6th 2010, 07:53 AMtonio

As $\displaystyle \sqrt[3]{3}\notin \mathbb{F}_{13}\,\,\,and\,\,\,3^3=9^3=1\!\!\!\pmod {13}$ , the roots of $\displaystyle f(x)=x^3-3$ are $\displaystyle w:=\sqrt[3]{3}\,,\,3w\,,\,9w$ , and from here that the spliiting field of

$\displaystyle f(x)$ over $\displaystyle \mathbb{F}_{13}$ is $\displaystyle K:=\mathbb{F}_{13}(w)\Longrightarrow [K:\mathbb{F}_{13}]=3$ ...

Tonio - May 7th 2010, 12:42 AMKSM08
Thanks- I had worked out that the degree of the splitting field is three, but what does this tell me? Since the polynomial is not separable, presumably it's not the case that the size of the Galois group is 3. (But that 3 divides the size?) What automorphisms should I be trying?

- May 7th 2010, 01:01 AMkompikTheorem 9.5.1 in Roman's Field Theory
I think Theorem 9.5.1 here answers your question.

- May 7th 2010, 01:03 AMNonCommAlg
since $\displaystyle x^3-3$ is irreducible over $\displaystyle \mathbb{F}_{13},$ the splitting field of this polynomial is $\displaystyle \frac{\mathbb{F}_{13}[x]}{(x^3-3)} \cong \mathbb{F}_{13^3}=K.$ thus $\displaystyle 3=[K:\mathbb{F}_{13}]=|Gal(K/\mathbb{F}_{13})|$ and so the Galois group is just $\displaystyle \mathbb{Z}/3\mathbb{Z}.$

- May 7th 2010, 03:27 AMtonio