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Math Help - Principle Ideal

  1. #1
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    Principle Ideal

    Let R be a commutative ring. Let A  \neq 0 be an ideal in R[x] such that
    the lowest degree of a nonzero polynomial in A is n   \geq 1 and such that A
    contains a monic polynomial of degree n. Prove that A is a principal ideal
    in R[x].
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  2. #2
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    Quote Originally Posted by mathman88 View Post
    Let R be a commutative ring. Let A  \neq 0 be an ideal in R[x] such that
    the lowest degree of a nonzero polynomial in A is n   \geq 1 and such that A
    contains a monic polynomial of degree n. Prove that A is a principal ideal
    in R[x].
    let f(x) \in A be a monic polynomial of degree n. obviously (f(x)) \subseteq A. now using induction over degrees of elements of A we show that every elment of A is divisible by f(x), which will

    prove that A \subseteq (f(x)). so let h(x)=a_0x^n + \cdots + a_n \in A. then h(x)-a_0f(x) \in A is a polynomial of degree at most n-1 and thus by the minimality of n we have h(x)-a_0f(x)=0.

    thus h(x)=a_0f(x) \in (f(x)). this proves the base of the induction. now let g(x)=a_0x^m + \cdots + a_m \in A, \ m > n, and suppose every polynomial in A of degree at most m-1 is in (f(x)).

    let k(x)=g(x)-a_0x^{m-n}f(x). then k(x) \in A and \deg k(x) \leq m-1. thus by induction hypothesis k(x)=u(x)f(x), for some u(x) \in R[x]. then g(x)=(a_0x^{m-n}+u(x))f(x) \in(f(x)).
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