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Is there a faster way to do this?

  1. Nov 11, 2007 #1
    Z/p [x] consists of polynomials in x with coefficients in a field Z/p.
    I is an ideal of Z/p [x] generated by some irreducible polynomial i(x) in Z/p [x]
    (let's say I is generated by an irreducible polynomial i(x) of degree 3 in Z/p [x])

    The factor (or quotient)
    (Z/p [x])/I = (Z/p [x])/<i(x)> is a field.
    To verify that it is,
    * (Z/p [x])/<i(x)> is commutative because Z/p [x] is. (ok)
    * i(x) is irreducible so (Z/p [x])/<i(x)> has no divisors of zero. (ok)
    * (Z/p [x])/<i(x)> needs to be a division ring, that is, every nonzero element needs to be invertible. (needs to be verified)

    My (tedious) method is,
    WLOG let f(x) = x^2 + ax + b be a polynomial in Z/p [x]
    need to find some iverse of f(x) (call it g(x)) in Z/p [x] such that
    (f(x).g(x)) + <i(x)> = 1 + <i(x)>

    which can be done by multiplying out f(x).g(x) and then dividing by i(x) to get the remainder (very long) which would be of degree 2 (one less than deg(i(x)) )
    and then solving the coefficient of x^2 and x to be congruent to 0 mod p
    and the constant term to be congruent to 1 mod p
    (three equations and three unkowns)

    and not only does that take very long, if p is not small then in the end I often end up making a mistake somewhere so my calculated "inverse" of f(x) is incorrect
    (end up having f(x).g(x) + <i(x)> different to 1 + <i(x)> )

    Is there a better/cleaner/easier/more efficient way to do these sorts of things rather than having to go through all that mess?
    (Can take p to be a specific prime, say p=5, or 7, or whatever.)
     
  2. jcsd
  3. Nov 11, 2007 #2

    morphism

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    Z/p is a field, so (Z/p)[x] is a principal ideal domain. If i(x) is irreducible over Z/p, then <i(x)> is a prime ideal and consequently (because we're in a PID) a maximal ideal. And what happens when you mod out by a maximal ideal?
     
  4. Nov 12, 2007 #3
    Then the factor is a field! Yay, fantastic! Thanks!
    This shows that I need to do a lot of extra reading on stuff that is not taught in order to fully understand the material.. I've never heard of a pid before, unless it's lying on some fresh page that I've never looked at in our notes, or it's not covered at my level - first course in abstract algebra... (I had to look up maximal and prime ideals too...)
     
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