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Math Help - Galois group

  1. #1
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    Galois group

    Does anyone know how to find the Galois group of x^3-3 over the finite field F_13?
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  2. #2
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    Quote Originally Posted by KSM08 View Post
    Does anyone know how to find the Galois group of x^3-3 over the finite field F_13?


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

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

    Tonio
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  3. #3
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    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?
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  4. #4
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    Theorem 9.5.1 in Roman's Field Theory

    I think Theorem 9.5.1 here answers your question.
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  5. #5
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    since x^3-3 is irreducible over \mathbb{F}_{13}, the splitting field of this polynomial is \frac{\mathbb{F}_{13}[x]}{(x^3-3)} \cong \mathbb{F}_{13^3}=K. thus 3=[K:\mathbb{F}_{13}]=|Gal(K/\mathbb{F}_{13})| and so the Galois group is just \mathbb{Z}/3\mathbb{Z}.
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    Quote Originally Posted by KSM08 View Post
    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,


    Why do you think so?? Of course it is separable, and if you have any doubt just remember the basic theorem: any finite extension of a finite field (or any extension of a characteristic zero field) is separable.
    Furthermore, if you already knew the degree of the splitting field then, this being a prime, doesn't leave you many choices for the Galois group, does it?

    Tonio


    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?
    .
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