Factor Rings of Polynomials Over a Field

On page 222 of Nicholson: Introduction to Abstract Algebra in his section of Factor Rings of Polynomials Over a Field we find Theorem 1 stated as follows: (see attached)

**Theorem 1.** Let F be a field and let $\displaystyle A \ne 0 $ be an ideal of F[x]. Then a uniquely determined monic polynomial h exists exists in F[x] such that A = (h).

The beginning of the proof reads as follows:

**Proof:** Because $\displaystyle A \ne 0 $, it contains non-zero polynomials and hence contains monic polynomials (being an ideal) ... ... etc. etc.

BUT! why must A contain monic polynomials??

Help with this matter would be appreciated!

Peter

Re: Factor Rings of Polynomials Over a Field

Hi,

Suppose $\displaystyle g(x)=a_nx^n+... \text{ is in A with }a_n\ne0$.

Since A is an ideal, $\displaystyle a_n^{-1}g(x)}$ is monic and is in A.