C2 as Galois group of an irreducible cubic

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SUMMARY

The discussion centers on the possibility of the cyclic group C2 occurring as the Galois group \operatorname{Aut}(K/F) for an irreducible cubic polynomial f(x) over a field F. It is established that C2 can arise if f(x) is an inseparable polynomial with exactly two distinct roots, particularly in characteristic 3. In contrast, for separable irreducible cubics, the Galois group will be either A3 or S3, ruling out C2 in characteristic 0. The conclusion emphasizes that in cases where f(x) is purely inseparable, \operatorname{Aut}(K/F) will be trivial.

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  • Understanding of Galois theory and Galois groups
  • Knowledge of irreducible cubic polynomials
  • Familiarity with field characteristics, particularly characteristic 0 and characteristic 3
  • Concept of separable and inseparable polynomials
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  • Study the properties of Galois groups, specifically A3 and S3
  • Explore inseparable polynomials in characteristic p, focusing on characteristic 3
  • Learn about splitting fields and their relationship to Galois groups
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Mathematicians, algebraists, and students studying Galois theory, particularly those interested in the properties of cubic polynomials and field characteristics.

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Homework Statement



If f(x) is an irreducible cubic polynomial over a field F, is it ever possible that [itex]C_2[/itex] may occur as the [itex]\operatorname{Aut}(K/F)[/itex] where K is the splitting field of f?

The Attempt at a Solution


It seems that this should be theoretically possible. In particular, if f is an inseparable polynomial which split into precisely two distinct roots, this would be the case. However, I'm trying to think of an example and am having trouble coming up with one. In particular, it feels as though this won't be possible in characteristic 0, but may be possible in char p. Can anybody shed some light on this?
 
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If your irreducible cubic f(x) is separable, then its Galois group will either be A_3 or S_3. If f(x) isn't separable, then necessarily charF=3, and f(x)=x^3-a for some noncube a. Can C_2 be the splitting field of such an f(x)?
 
Ah yes. For some reason it didn't occur to me that the field would necessarily have to have characteristic 3, but that makes sense. In that case, the polynomial is purely inseparable and so [itex]\operatorname{Aut}(K/F)[/itex] would be trivial.
 

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