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Abstract algebra

  • Thread starter buzzmath
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  • #1
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I'm having trouble understanding splitting fields. Some of the problems are find the degree of the splitting field of x^4 + 1 over the rational numbers and if p is a prime prove that the splitting field over the rationals of the polynomial x^p - 1 is of degree p-1. I'm really confused with these type of problems. Can anyone give some helpful advice?

Thanks
 

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  • #2
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could you say that x^4 + 1=(x^2 + i)(x^2 - i). Since this is factored into two polynomials of degree 2 the degree of the splitting field is 2*2=4?
 
  • #3
matt grime
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i is not a rational number. What are the roots of x^4_1? What is the smallest extension of Q that contains them?
 
  • #4
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can't the roots be in an extension field? isnt' the splitting field an extension field so i would be in the extension field? The roots would be x = (-1)^(1/4) but this isn't a rational number. I think because of this the extension would have to be the complex field because no real number satisfies this. How do you know if it's the smalllest extension?
 
  • #5
HallsofIvy
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can't the roots be in an extension field? isnt' the splitting field an extension field so i would be in the extension field? The roots would be x = (-1)^(1/4) but this isn't a rational number. I think because of this the extension would have to be the complex field because no real number satisfies this. How do you know if it's the smalllest extension?
Yes, that's the whole point. The splitting field is the smallest extension of Q that contains all of its roots. No, that splitting field is NOT the set of all complex numbers. What are the four roots of x4/sup]= -1?
 
  • #6
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Can you write it like w = cos(360/4)+isin(360/4) then w^4 =1 so (2)^(1/2)/2+or-(2)^(1/2)/2i are the four roots so the field is Q(2^(1/2),i) but then how do you tell the degree of this field?
 
  • #7
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my experience with degrees is with polynomials not an entire field
 
  • #8
HallsofIvy
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Good. Yes, the splitting field is [itex]Q(\sqrt{2},i)[/itex]. The smallest field containing all rational numbers, [itex]\sqrt{2}[/itex], and i. In particular, that means it must include all numbers of the form [itex]a+ bi+ c\sqrt{2}[/itex] where a, b, c are rational numbers. Since it must be closed under multiplication, it must also include [itex]\isqrt{2}[/itex]. Since that cannot be written in the above form, it must, in fact, include numbers of the form [itex]a+ bi+ c\sqrt{2}+ di\sqrt{2}[/itex]. One can show that any number in this extension field can be written in that form. We can think of that as a vector space over the rational numbers with basis {1, i, [itex]\sqrt{2}[/itex],[itex]i\sqrt{2}[/itex]}: i.e. the vector space has dimension 4 over the rational numbers. THAT is the "degree" of the extension field: its dimension as a vector space over the rational numbers. In this case we could also have seen that by noting that [itex]\sqrt{2}[/itex] and i are both "algebraic of order 2" and that they are "algebraically independent".
 

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