Math Help - Principle Ideal

1. 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].

2. Originally Posted by mathman88
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 $g(x)=(a_0x^{m-n}+u(x))f(x) \in(f(x)).$