
#1
Feb1809, 08:31 PM

P: 102

I was thinking about this,
finding the splitting field of x^42 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 



#2
Feb1809, 10:21 PM

PF Gold
P: 274

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).




#3
Feb1809, 10:36 PM

P: 102

Very true! However, what does 2^(1/4) mean exactly in this case? i dont think it can be a real number since i dont 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! 



#4
Feb1809, 11:11 PM

PF Gold
P: 274

Splitting Fields
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).



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