C2 as Galois group of an irreducible cubic

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Homework Statement



If f(x) is an irreducible cubic polynomial over a field F, is it ever possible that C_2 may occur as the \operatorname{Aut}(K/F) 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 \operatorname{Aut}(K/F) would be trivial.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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