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Factor Rings of Polynomials Over a Field

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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 [TEX] A \ne 0 [/TEX] 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 [TEX] A \ne 0 [/TEX], 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

[This has also been posted on MHF]
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
BUT! why must A contain monic polynomials??
Suppose $p(x)=a_nx^n+\ldots+a_1x+a_0\in A$ and $a_n\ne 0$. As $A$ is an ideal of $F[x]$, $q(x)=\dfrac{1}{a_n}p(x)$ belongs to $A$ and $q(x)$ is monic.