Finding a splitting field over Z_13

In summary: 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.
  • #1
PsychonautQQ
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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?
 
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  • #2
PsychonautQQ said:

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##?
 
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  • #3
fresh_42 said:
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?
 
  • #4
PsychonautQQ said:
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.
 
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  • #5
fresh_42 said:
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?
 
  • #6
PsychonautQQ said:
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.
 
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1. What is a splitting field over Z13?

A splitting field over Z13 is a field extension of the finite field Z13 where all irreducible polynomials in Z13[x] can be factored into linear factors. In other words, it is the smallest field where all the roots of a given polynomial in Z13[x] can be found.

2. Why is finding a splitting field over Z13 important?

Finding a splitting field over Z13 is important because it allows us to solve polynomial equations over the finite field Z13. This has applications in many areas of mathematics and science, including coding theory, cryptography, and algebraic geometry.

3. How do you find a splitting field over Z13?

The process for finding a splitting field over Z13 involves finding all the roots of the given polynomial in Z13[x] and then adjoining those roots to Z13 to create a larger field. This process is repeated until all the irreducible factors of the polynomial have been accounted for.

4. Can you give an example of finding a splitting field over Z13?

Consider the polynomial f(x) = x3 + 2x + 5 in Z13[x]. The roots of this polynomial are 2, 4, and 8 in Z13. Adjoining these roots to Z13 gives the field Z13[2]. Factoring out the irreducible polynomial (x-2) from f(x) in this new field gives the polynomial (x-4)(x-8). Adjoining the roots 4 and 8 to Z13[2] gives the final splitting field over Z13.

5. What are the applications of finding a splitting field over Z13?

As mentioned before, finding a splitting field over Z13 has applications in many areas of mathematics and science. For example, in coding theory, it can be used to construct error-correcting codes over finite fields. In cryptography, it can be used to create secure encryption algorithms. In algebraic geometry, it can be used to study curves and surfaces over finite fields. Overall, the ability to solve polynomial equations over finite fields is a valuable tool in many fields of research and application.

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