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Thread: Principle Ideal

  1. #1
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    Principle Ideal

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

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

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

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