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

In summary, finding the splitting field of x^4-2 in Q[x] over Q is standard enough. However, if we wanted to find the splitting field over F_5, it would just be F_5(2^(1/4)) since once we have one fourth root of 2, the others would just be multiples of it. However, 2^(1/4) would not be a real number in this case, but rather an element of some algebraic extension of F_5.
  • #1
hypermonkey2
102
0
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
 
Physics news on Phys.org
  • #2
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
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!
 
  • #4
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).
 

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

1. What is a splitting field?

A splitting field is a field extension that contains all the roots of a given polynomial. In other words, it is the smallest field in which a polynomial can be factored into linear factors.

2. How do you find the splitting field of a polynomial?

To find the splitting field of a polynomial, you need to factor the polynomial into irreducible factors and then adjoin all the roots of the factors to the base field. This will result in a field extension that contains all the roots and is the splitting field of the polynomial.

3. What is the difference between Q[x] and F5?

Q[x] refers to the field of rational polynomials, while F5 refers to the finite field with 5 elements. Q[x] contains all rational numbers and their polynomial expressions, while F5 contains only the elements 0, 1, 2, 3, and 4.

4. Can the splitting field of x^4-2 be found in both Q[x] and F5?

Yes, the splitting field of x^4-2 can be found in both Q[x] and F5. However, the splitting fields in Q[x] and F5 will be different due to the different nature of the two fields.

5. Which field is a better choice for finding the splitting field of x^4-2?

It depends on the context and the intended use of the splitting field. If you are looking for a field extension that contains all the roots of x^4-2, then Q[x] would be a better choice as it is a larger field than F5. However, if you are working with finite fields, then F5 would be a more suitable choice.

Similar threads

I Splitting Fields - Example 3 - D&F Section 13.4, pages 537 -
Replies
2
Views
2K
Need help understanding splitting fields
Replies
4
Views
2K
MHB Splitting Fields - Example 3 - D&F Section 13.4, pages 537 - 538
Replies
2
Views
1K
MHB Splitting Fields - Beachy and Blair - Example 6.4.2
Replies
5
Views
2K
MHB How Does Polynomial Splitting Occur in Finite Field Extensions?
Replies
1
Views
1K
Constructing a splitting field for polynomial over F_5
Replies
5
Views
2K
MHB Automorphisms, Galois Groups & Splitting Fields
Replies
3
Views
2K
I Need clarification on a theorem about field extensions/isomorphisms
Replies
14
Views
2K
I Splitting ring of polynomials - why is this result unfindable?
Replies
9
Views
1K
MHB Splitting Fields. Prove Q(alpha^4)=Q(alpha)
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top