These are very basic questions but I need a little help and clarity getting started.
Find the splitting field for the following over GF(3)
a.) x^4+2
b.) x^4-2
For a.) I figured x^4+2=x^4-1 in GF(3). But in GF(5), x^4-1=x^5-x since x^5=x mod p by Fermat's Corollary and so GF(5) is the splitting field of the polynomial x^4-1.
Another way I naively approached it was to split x^4-1=(x^2+1)(x-1)(x+1) and saw that in GF(5), 1=-4 so then x^2+1=x^2-4=(x+2)(x-2). Are any of these for part a.) correct? If not, please explain. Thank you
Find the splitting field for the following over GF(3)
a.) x^4+2
b.) x^4-2
For a.) I figured x^4+2=x^4-1 in GF(3). But in GF(5), x^4-1=x^5-x since x^5=x mod p by Fermat's Corollary and so GF(5) is the splitting field of the polynomial x^4-1.
Another way I naively approached it was to split x^4-1=(x^2+1)(x-1)(x+1) and saw that in GF(5), 1=-4 so then x^2+1=x^2-4=(x+2)(x-2). Are any of these for part a.) correct? If not, please explain. Thank you
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First of all, the splitting field must be a field which contains the original field. GF(5) does not contain GF(3), so GF(5) cannot be the answer. Since GF(3) has characteristic 3, so does any field containing it. Therefore the splitting fields must also have characteristic 3.
a) Let K = GF(3). In GF(3), x^4 + 2 = x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x+1)(x-1)(x^2 + 1). Therefore the splitting field of x^4 + 2 is the splitting field of x^2 + 1. Note that x^2 + 1 does not split in K. [Since x^2 + 1 has degree 2, this is equivalent to the statement that x^2 + 1 has no roots in K, which you can check by brute force.] Let i denote a root of x^2 + 1. Then the roots of x^2 + 1 are i and -i, so the splitting field is K(i). Note also that K(i) = GF(9). [Since i is a root of an irreducible quadratic polynomial, {1, i} is a basis for K(i) over K. Since K has order 3, it follows that K(i) has order 3^2 = 9.]
b) Let s denote a fourth root of 2, so that s is a root of x^4 - 2. In K(s), x^4 - 2 factors as
x^4 - 2 = (x^2 - s^2)(x^2 + s^2)
= (x^2 - s^2)(x^2 - 2s^2)
= (x^2 - s^2)(x^2 - s^6)
= (x + s)(x - s)(x + s^3)(x - s^3),
This proves that K(s) is the splitting field. [The splitting field must contain K(s), since s is in the splitting field. But K(s) contains all the roots of x^4 - 2, so K(s) contains the splitting field. Therefore, K(s) is the splitting field.] Note that x^4 - 2 factors over K as
x^4 - 2 = (x^2+x+2) (x^2+2x+2),
and that both quadratic factors are irreducible over K. [Once again, you need only check that they have no roots in K.] Therefore, s is a root of a K-irreducible polynomial of degree 2. As before, it follows that the splitting field is K(s) = GF(9).
a) Let K = GF(3). In GF(3), x^4 + 2 = x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x+1)(x-1)(x^2 + 1). Therefore the splitting field of x^4 + 2 is the splitting field of x^2 + 1. Note that x^2 + 1 does not split in K. [Since x^2 + 1 has degree 2, this is equivalent to the statement that x^2 + 1 has no roots in K, which you can check by brute force.] Let i denote a root of x^2 + 1. Then the roots of x^2 + 1 are i and -i, so the splitting field is K(i). Note also that K(i) = GF(9). [Since i is a root of an irreducible quadratic polynomial, {1, i} is a basis for K(i) over K. Since K has order 3, it follows that K(i) has order 3^2 = 9.]
b) Let s denote a fourth root of 2, so that s is a root of x^4 - 2. In K(s), x^4 - 2 factors as
x^4 - 2 = (x^2 - s^2)(x^2 + s^2)
= (x^2 - s^2)(x^2 - 2s^2)
= (x^2 - s^2)(x^2 - s^6)
= (x + s)(x - s)(x + s^3)(x - s^3),
This proves that K(s) is the splitting field. [The splitting field must contain K(s), since s is in the splitting field. But K(s) contains all the roots of x^4 - 2, so K(s) contains the splitting field. Therefore, K(s) is the splitting field.] Note that x^4 - 2 factors over K as
x^4 - 2 = (x^2+x+2) (x^2+2x+2),
and that both quadratic factors are irreducible over K. [Once again, you need only check that they have no roots in K.] Therefore, s is a root of a K-irreducible polynomial of degree 2. As before, it follows that the splitting field is K(s) = GF(9).
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First, what does GF(p) refer to? I don't think I've seen this notation, though it seems you're talking about the integers mod p.
Second, why are you using GF(5) anywhere? It has nothing to do with this problem.
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For (a), start with your "naive" approach by writing
x^4 + 2 = (x^2 + 1)(x - 1)(x + 1).
Then notice that x^2 + 1 is irreducible over GF(3), but my adjoining one root, we will get the splitting field. Then we have the field with 3^2 = 9 elements as the splitting field.
Second, why are you using GF(5) anywhere? It has nothing to do with this problem.
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For (a), start with your "naive" approach by writing
x^4 + 2 = (x^2 + 1)(x - 1)(x + 1).
Then notice that x^2 + 1 is irreducible over GF(3), but my adjoining one root, we will get the splitting field. Then we have the field with 3^2 = 9 elements as the splitting field.