Q11.3.11 Artin Algebra 2nd
Let $R$ be a ring, and let $I$ be an ideal of the polynomial ring $R[x]$.
Let $n$ be the lowest degree among nonzero elements of $I$.
Prove or disprove the following:
$I$ contains a monic polynomial of degree $n$ if and only if it
is a principal ideal.
I've proved the first direction, now I want to prove:
a principal ideal contains a monic polynomial of least degree $n$
Is it trivial? I kind of missing the starting point, and can not find any resources on this, any help?
Thanks in advance~
Best Answer
If you are not working over a field, then principal ideals of $R[X]$ may not contain any monic polynomials at all, let alone such polynomials of minimal degree. Principal ideals with a monic generator$~P$ obviously do have such an element (namely $P$, which is monic and clearly of minimal degree for a nonzero multiple of$~P$). But these are in fact all cases where it works. A simple example of failure is the ideal generated by $91X^3$ in $\Bbb Z[X]$ (or by $2X$ if you prefer small examples).
My "these are all cases" claim is just the implication: "if $I$ has a monic polynomial$~P$ among its nonzero elements of minimal degree, then $I$ is the principal ideal generated by$~P$", which you say you have already shown (presumably using Euclidean division by$~P$). Note that not all generators of$~I$ are monic (they could have invertible leading coefficient) nor even do they have to be of minimal degree if $R$ has zero divisors (if $n\in R$ is nilpotent then $1-nX^d\in R[X]$ is invertible for any $d>0$, so the ideal it generates is all of $R[X]$, with monic generator $1$ of degree $0<d$).