Finding a splitting field over Z_13

[PsychonautQQ]

Homework Statement

Describe the splitting field of x^9-1 over Z_13.

Homework Equations

The Attempt at a Solution

Well, seeing if any other of {1,...,12} would be some tricky arithmetic. Is my best bet here moving forward to just divide x^9-1 by x-1 since I know 1 is a root and then go from there?

So basically I'm looking for elements in Z*_13 with orders of 1,3 or 9. But there will be no element of order 9 because 9 does not divide 12 ( the order of Z*_13) so I'm looking for all elements of order 3. This makes checking to see if {1,...,12} work much easier.

Also, there since 3 is prime there is a subgroup of order 3, so there are at least 2 elements such that x^3=1 (mod 13). I have found these elements, 3 and 9, therefore I have 3 out of the 9 roots. What do I do from here? Divide x^9-1 by (x-1)(x-3)(x-9) and then divide the quotient by (x-r)? Is there a simpler way?

 

[fresh_42]

Homework Statement

Describe the splitting field of x^9-1 over Z_13.

Homework Equations

The Attempt at a Solution

Well, seeing if any other of {1,...,12} would be some tricky arithmetic. Is my best bet here moving forward to just divide x^9-1 by x-1 since I know 1 is a root and then go from there?

So basically I'm looking for elements in Z*_13 with orders of 1,3 or 9. But there will be no element of order 9 because 9 does not divide 12 ( the order of Z*_13) so I'm looking for all elements of order 3. This makes checking to see if {1,...,12} work much easier.

Also, there since 3 is prime there is a subgroup of order 3, so there are at least 2 elements such that x^3=1 (mod 13). I have found these elements, 3 and 9, therefore I have 3 out of the 9 roots. What do I do from here? Divide x^9-1 by (x-1)(x-3)(x-9) and then divide the quotient by (x-r)? Is there a simpler way?

Since $\mathbb{Z}_{13}$ is a field, do you expect any differences as if it was $\mathbb{Q}$?
$1$ is a root, as you've said, and you know what $(x^9-1):(x-1)$ is. What happens if you take a primitive 9-th root of unity, say $\xi$ and consider the powers $\xi^i$?

 

[PsychonautQQ]

Since $\mathbb{Z}_{13}$ is a field, do you expect any differences as if it was $\mathbb{Q}$?
$1$ is a root, as you've said, and you know what $(x^9-1):(x-1)$ is. What happens if you take a primitive 9-th root of unity, say $\xi$ and consider the powers $\xi^i$?

Yeah, char(Q) = 0 so there will be no repeated roots, here there will be repeated roots. I'm not exactly sure how to reduce a primitive 9-th root of unity mod 13. I'm pretty sure the only roots of x^9-1 in Z_13 are 1,3 and 9 though?

 

[fresh_42]

Yeah, char(Q) = 0 so there will be no repeated roots, here there will be repeated roots. I'm not exactly sure how to reduce a primitive 9-th root of unity mod 13. I'm pretty sure the only roots of x^9-1 in Z_13 are 1,3 and 9 though?

Then you should compute $q(x)=(x-1)(x-3)(x-9)$ and $(x^9-1):q(x)$ and see what is left.

 

[PsychonautQQ]

Then you should compute $q(x)=(x-1)(x-3)(x-9)$ and $(x^9-1):q(x)$ and see what is left.

q(x) mod 13 is equivalent to x^3-1. (x^9-1)/(x^3-1) = x^6+x^3+1. In hindsight, it really should have been obvious that q(x) would reduce to x^3-1. Anyway, now I'm looking for the roots of f(x)=x^6+x^3+1, and since I've already found all the roots that are in Z*_13, the roots of f(x) must be in a some splitting field. I will now divide f(x) by x-r.

f(x)/(x-r) = x^5+(r)x^4+(r^2)x^3+(r^3)x^2+(r^4)x+r^5. I feel this did not help me very much. Perhaps I should go back and try another strategy at finding the roots of x^6+x^3+1, my gut is telling me something to do with some kind of primitive root of unity. Any insights?

 

[fresh_42]

q(x) mod 13 is equivalent to x^3-1. (x^9-1)/(x^3-1) = x^6+x^3+1. In hindsight, it really should have been obvious that q(x) would reduce to x^3-1. Anyway, now I'm looking for the roots of f(x)=x^6+x^3+1, and since I've already found all the roots that are in Z*_13, the roots of f(x) must be in a some splitting field. I will now divide f(x) by x-r.

f(x)/(x-r) = x^5+(r)x^4+(r^2)x^3+(r^3)x^2+(r^4)x+r^5. I feel this did not help me very much. Perhaps I should go back and try another strategy at finding the roots of x^6+x^3+1, my gut is telling me something to do with some kind of primitive root of unity. Any insights?

You will need only one element $\xi$ to adjoint. To split $x^6+x^3+1$ you could temporarily substitute $y=x^3$ and solve the quadratic equation. Since $2$ is a unit, you'll have to find $2^{-1}$ and use the known formula. Also the tables of $a^2\, , \,a^3$ for $a \in \mathbb{Z}_{13}$ are helpful.

If you'll have the final splitting of $x^9-1$ it would be fine to see it so I can compare it with my solution.