- #1
kobulingam
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Z_2/<u^4+u+1> isomorphism Z_2/<u^4+u^3+u^2+u+1>
How to figure an isomorphism from
Z_2/<u^4 + u +1> to Z_2/<u^4 + u^3 + u^2 + u + 1>
What I can now show (after a page and a half of work) is that the two polynomials generating the ideals are irreducible over Z_2.
I've been able to prove that the elements creating the ideas are both irreducible polynomials.
I can show proof that the ideals are irreducible, but I don't think we need to reuse that part in remaining solution. Essentially u^4 + u +1 has no linear factors by factor theorem (neither 0 nor 1 root), so only possibility is that it could be factored into 2 irreducible quadratics, and there is only once such quadratic in Z_2. Squared this quadratic and didn't get u^4 + u +1. Thus u^4 + u +1 is irreducible.
Similarly, u^4 + u^3 + u^2 + u + 1 has no linear factors (neither 0 nor 1 is a root), so only possibility is that it's the product of an irreducible quadratic and irreducible cubic. There are only 2 possible such cubics. Multiplying each of these cubics with the irreducible quadratic does not give u^4 + u^3 + u^2 + u + 1. Thus u^4 + u^3 + u^2 + u + 1 irreducible over Z_2. I am guessing this is the easy part of the answer, yet this itself stretched me fully...
Homework Statement
How to figure an isomorphism from
Z_2/<u^4 + u +1> to Z_2/<u^4 + u^3 + u^2 + u + 1>
What I can now show (after a page and a half of work) is that the two polynomials generating the ideals are irreducible over Z_2.
Homework Equations
I've been able to prove that the elements creating the ideas are both irreducible polynomials.
The Attempt at a Solution
I can show proof that the ideals are irreducible, but I don't think we need to reuse that part in remaining solution. Essentially u^4 + u +1 has no linear factors by factor theorem (neither 0 nor 1 root), so only possibility is that it could be factored into 2 irreducible quadratics, and there is only once such quadratic in Z_2. Squared this quadratic and didn't get u^4 + u +1. Thus u^4 + u +1 is irreducible.
Similarly, u^4 + u^3 + u^2 + u + 1 has no linear factors (neither 0 nor 1 is a root), so only possibility is that it's the product of an irreducible quadratic and irreducible cubic. There are only 2 possible such cubics. Multiplying each of these cubics with the irreducible quadratic does not give u^4 + u^3 + u^2 + u + 1. Thus u^4 + u^3 + u^2 + u + 1 irreducible over Z_2. I am guessing this is the easy part of the answer, yet this itself stretched me fully...