Compute in a quotient of a polynomial ring

Consider f(x) = x^4 - 16 as an element of E = \mathbb{Z}[x] and use bars to denote the natural projection E \rightarrow E/(f(x)).

  1. Fine a polynomial of degree at most 3 which is congruent to g(x) = 7x^{13} - 11x^9 + 5x^5 - 2x^3 + 3 modulo x^4-16.
  2. Prove that \overline{x+2} and \overline{x-2} are zero divisors in \overline{E}.

  1. Evidently, \overline{g(x)} = \overline{-2x^3 + 25936x + 3}.
  2. Note that x^4-16 = (x+2)(x-2)(x^2+4). By §7.4 #14, \overline{x+2} and \overline{x-2} are zero divisors in \overline{E}.
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  • murphy  On November 29, 2011 at 1:32 am

    can you tell me how you’re doing these? there’s nothing like this in the book

    • nbloomf  On November 29, 2011 at 2:55 am

      There’s a couple of ways to think about this problem.

      Remember that two polynomials p(x) and q(x) are equivalent mod f(x) if f divides p-q. Using the Division Algorithm in the Euclidean domain \mathbb{Q}[x], we can find b and r such that p = bf + r, and the degree of r is less than that of f; this r is the (in this case unique) representative of the coset p(f) with this property.

      Sometimes, it is easier to reduce powers of x directly. For example, in this ring we have x^4 \equiv 16; so x^13 \equiv (x^4)(x^4)(x^4)(x) \equiv 16^3x. In this fashion, we can eliminate powers of x larger than the degree of f.

  • J  On May 10, 2012 at 4:20 pm

    I think there is a typo in part 2. The last term of the factorization of x^4-16 should read (x^2+4).

    • nbloomf  On August 17, 2012 at 11:40 pm

      Fixed. Thanks!

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