MHB Why Does \( p \) Divide \( f \) in Nicholson's 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, $$p = ad, a \in F$$, so $$p$$ divides $$f$$ (because $$d$$ divides $$f$$) ... ... "Nicholson argues that because $$d$$ divides $$f$$ we then have that $$p$$ divides $$f$$ ... ... 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, $f=k(a^{-1}p)$.