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Thread: Polynomial Over a Field

  1. #1
    Nov 2009

    Polynomial Over a Field

    Hi, I was preparing for my final and I was working a practice question and got very lost. Here's the problem.

    Suppose F = Z_5, and F is a field since 5 is prime.

    i.) Show the polynomial x^4+1 is irreducible over the field.
    ii.) Show that the principal ideal <x^4+1> of F[x] generated by x^4+1 is maximal and thus L=F[x]/<x^4+1> is a field.
    iii.) How many elements are in L? We can use the Euclidean algorithm (division algorithm) to show that for any g(x) in F[x], that
    g(x) + <x^4+1>=(ax+b) + <x^4+1>, where x, b are in Z_5.
    iv.) Produce addition and Cayley table.

    Attempt at the solution.

    i.) This part was pretty easy. I showed. Z_5 = {0,1,2,3,4}.
    p(x) = x^4+1
    so, p(0) = 1
    p(1) = 2
    p(2) = 17 = 2
    p(3) = 82 = 2
    p(4) = 257 = 2 (I think that calculation is right).
    But overall, this polynomial is irreducible over the Field because for each of the values there is no zero value.

    ii.) This part of the problem I have no idea where to start. In order for an ideal to be maximal, then the ideal, we'll call M is not a proper subset of any ideal except R itself. There's another definition as well, but not sure how to type it....but here goes.

    An ideal M is a maximal ideal of R iff M is a subset of I is a subset of R. (If you're not sure what I am saying here feel free to ignore it.)

    If you can help me with ii then I think I can get a better handle on the next two, but if you're feeling generous feel free to help with all . But really, it's mainly the second one.

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  2. #2
    Super Member
    Apr 2009
    For i) you still have to prove that the polynomial is not a product of two quadratic polynomials.

    For ii) use that if $\displaystyle p \in R$ is irreducible then $\displaystyle \langle p \rangle$ (the ideal generated by this element) is prime and if $\displaystyle R$ is a principal ideal domain then every prime ideal is maximal.
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