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

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Discussion Overview

The discussion revolves around finding the splitting field of the polynomial x^4 - 2, first over the field of rational numbers Q and then over the finite field F_5. Participants explore the implications of these different fields on the nature of the roots and the structure of the splitting fields.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that the splitting field over F_5 could be F_5(2^(1/4), i), drawing an analogy to the case over Q.
  • Another participant argues that the splitting field would simply be F_5(2^(1/4)), noting that once one fourth root of 2 is found, the others can be derived from it, considering the properties of F_5.
  • A later reply questions the meaning of 2^(1/4) in this context, expressing uncertainty about its representation as a real number and the existence of an element in F_5 that satisfies x^4 = 2.
  • Another participant clarifies that 2^(1/4) would not be a real number but an element of an algebraic extension of F_5, mentioning that the polynomial x^4 - 2 is irreducible over F_5 and suggesting the use of the quotient ring F_5[X]/(X^4 - 2) for the extension.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the splitting field over F_5, with some proposing F_5(2^(1/4)) and others suggesting the inclusion of an imaginary unit i. The discussion remains unresolved regarding the exact structure of the splitting field and the interpretation of roots in this finite field context.

Contextual Notes

There is uncertainty regarding the representation of 2^(1/4) in F_5 and the implications of irreducibility of the polynomial x^4 - 2 over F_5. The discussion highlights the need for clarification on the nature of algebraic extensions in finite fields.

hypermonkey2
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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
 
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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).
 
Citan Uzuki said:
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!
 
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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