Finding Splitting Field of x^4-2: Q[x] vs F_5

[hypermonkey2]

I was thinking about this,
finding the splitting field of x^4-2 in Q[x] over Q is standard enough... but would much be different is i wanted the splitting field over F_5? (field with 5 elements)
would it just be F_5(2^(1/4), i) analogously to the Q case? or do any of the arguments break down?

Any thoughts are appreciated,
cheers

 

[Citan Uzuki]

I think it would be just F_5(2^(1/4)), since once you have one fourth root of 2, the others would just be 2*2^(1/4), 4*2^(1/4), and 3*2^(1/4) (since 2^4 = 1 in F_5).

 

[hypermonkey2]

I think it would be just F_5(2^(1/4)), since once you have one fourth root of 2, the others would just be 2*2^(1/4), 4*2^(1/4), and 3*2^(1/4) (since 2^4 = 1 in F_5).

Very true! However, what does 2^(1/4) mean exactly in this case? i don't think it can be a real number since i don't believe there is an extension from F_5 to R...
And since there is no element x in F_5 such that x^4=2...

perhaps i am confused?

Thanks for you reply!

 

[Citan Uzuki]

No, it wouldn't be an element of R. It would be an element of some algebraic extension of F_5. In this case, since the polynomial x^4 - 2 is irreducible over F_5, we can take that extension to be the quotient ring F_5[X]/(X^4 - 2).