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Principal Ideal, Polynomial generators

  1. Sep 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose R is an integral domain and I is a principal ideal in R[x], and I [itex]\neq[/itex] {0}

    a) Show I = <g(x)> for some g(x)[itex]\in[/itex]R[x] that has minimal degree among all non-zero polynomials in I.

    b) Is it necessarily true that I = <g(x)> for every g(x)[itex]\in[/itex]R[x] that has minimal degree among all non-zero polynomials in I?

    2. Relevant equations

    Theorems, the division algorithm.

    3. The attempt at a solution

    For (a) we can take a g(x) of minimal degree as our generator; if deg(g) = 0, then g(x) is a unit in I and I = R[x], so we can assume deg(g)[itex]\geq[/itex]1

    Now we let f(x)[itex]\in[/itex]I and apply the division algorithm.

    f = qg + r, where q,r[itex]\in[/itex]R[x] and deg(r)<deg(g)

    This means that r = f - qg which is an element of I. By our choice of g having minimal degree, this means that r = 0. Otherwise r would be an element of I, and could be written as r = pg, p[itex]\in[/itex]R[x]. (I'm not quite sure how best to explain the reasoning behind the fact that r must equal 0)

    finishing the proof after this is fine.

    b) I'm confused for this question, but I assume its a contradiction type answer, although I can't imagine a g(x) that wouldn't generate a principal ideal...

    Please help!
     
  2. jcsd
  3. Sep 10, 2011 #2

    micromass

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    Why??

    Did you prove that the division algorithm holds when R is simply an integral domain?

    Consider [itex]\mathbb{Z}[/itex] and take the ideal generated by X+2. Then 2X+4 is in the ideal and is of minimal degree.
     
  4. Sep 11, 2011 #3
    Hey,

    regarding that deg(g)=0 bit, I guess I just assumed that a constant would be a unit in I. This would only be the case if R was a field, right?

    Now I'm really confused for this question too; don't know how to prove the division algorithm, let alone for an integral domain.

    I think I understand what you're saying for b)

    Z is an integral domain so we can use that for R. b) is not true because <x+2> generates I and is of minimal degree, but <2x+4> can't generate I (it can't generate x+2, as Z doesn't contain fractions) but is also of minimal degree. Hence not every g(x) in R[x] of minimal degree generates I.
     
  5. Sep 11, 2011 #4

    micromass

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    The trick for (a) is that you already know that I is principal. That is, you already know that I=<g(x)> for some g. The only thing you need to prove is that g is of minimal degree.
     
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