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Z_2 /<u^4+u+1> isomorphism Z_2 /<u^4+u^3+u^2+u+1>

  1. Nov 5, 2008 #1
    Z_2/<u^4+u+1> isomorphism Z_2/<u^4+u^3+u^2+u+1>

    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    I've been able to prove that the elements creating the ideas are both irreducible polynomials.

    3. 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...
  2. jcsd
  3. Nov 5, 2008 #2


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    Re: Z_2/<u^4+u+1> isomorphism Z_2/<u^4+u^3+u^2+u+1>

    Er, you mean two irreducible quadratics, don't you?

    Anyways, don't you know any theorems (or can comptue one) that tell you something about all possible homomorphisms [itex]R[x] / \langle f(x) \rangle \to S[/itex], where R and S are rings?

    (If you need a hint, first consider [itex]R[x] \to S[/itex])

    Incidentally, a useful syntactic tip is to use different indeterminate variables in your two different rings. I.E. write them as
    Z_2/<u^4 + u +1>

    Z_2[v]/<v^4 + v^3 + v^2 + v + 1>.​
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