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Division algorithm in A[x] (A NOT a field!)

  1. Jul 19, 2010 #1
    is it possible? I'm reading a proof where A is a local ring and f is a monic polynomial, v another polynomial in A[x] then the author says there is q,r with v=qf+r and degf<r.

    I thought the algorithm required divison of coefficients?! Maybe it's true we can always do this with f monic?

  2. jcsd
  3. Jul 19, 2010 #2
    I'm assuming you mean to require deg r < deg f (not the other way around).

    No division of coefficients is required as long as f is monic (you don't need that A is local either). If f has degree larger than v you can just let r=f, q=0. Thus you can assume [itex]deg(f) \leq deg(v)[/itex]. Let (v,f) be a pair of polynomials for which no such division exists, and for which the degree of v is minimal.

    If deg v = 0, then deg f = 0 and since f is monic f=1 so we can let q=v, r=0.

    If deg v > 0 let:
    [tex]h(x) = v(x) - Ax^{\deg(v) - \deg(f)}f(x)[/tex]
    where A is the coefficient of the term of highest degree in v(x). Then deg h < deg v and therefore we can divide h by f to give it polynomials q(x) and r(x) with deg r < deg f such that:
    [tex]h(x) = q(x)f(x) + r(x)[/tex]
    [tex]v(x) = (q(x)+Ax^{\deg(v) - \deg(f)})f(x) + r(x)[/tex]
    which shows that division is possible.
  4. Jul 20, 2010 #3
    ahhh, this is a good fact to know, much appreciated!
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