Irreducible Polynomials and Products on Nonzero Polynomials - Nicolson Theorem 11

[Math Amateur]

I am reading W. Keith Nicholson's book: Introduction to Abstract Algebra (Third Edition) ...

I am focused on Section 4.2:Factorization of Polynomials over a Field.

I need some help (with an apparently very simple issue) with the proof Theorem 11 on pages 217 - 218 ... just seem to have a mental block ... :(

The relevant text from Nicholson's book is as follows:https://www.physicsforums.com/attachments/4594
https://www.physicsforums.com/attachments/4595In the above text we read the following:

" ... ... In the second case, \(\displaystyle p = ad, a \in F\), so \(\displaystyle p\) divides \(\displaystyle f\) (because \(\displaystyle d\) divides \(\displaystyle f\)) ... ... "Nicholson argues that because \(\displaystyle d\) divides \(\displaystyle f\) we then have that \(\displaystyle p\) divides \(\displaystyle f\) ... ... but I cannot frame a formal and rigorous argument to show this ...

I must say that I suspect the argument is very simple ... but I would welcome help to get over this sticking point ...

Hope someone can help ...

Peter

 

[Fallen Angel]

Hi Peter,

There is a difference between using simple or strong induction.

Suppose we want to prove some property $P$ for all natural numbers.

If we use simple induction then we will do what you expect, assume $P(n)$ is true and then prove $P(n+1)$.

But, if we use strong induction we will assume $P(i)$ true for all $i<n$ and then prove $P(n)$, and that's what Nicholson is doing here.

He assumes that every polynomial of degree $d<n$ has a unique factorization, and then proves that a polynomial of degree $n$ also has a unique factorization.

EDIT:POSTED IN THE WRONG THREAD

 

[Fallen Angel]

Hi Peter,

$a$ is a non zero element in the field $F$, and we got the two equalities:

$p=ad$, or equivalently $a^{-1}p=d$ (remember $F$ is a field so every non zero element has an inverse)
$f=kd$ for some $k\in F[X]$.

Hence, $f=k(a^{-1}p)$.