Question

Prove that every non constant $f\left(x\right)\in F\left[x\right]$ can be written in the form $c{p}_1\left(x\right)p_2\left(x\right).....p_n\left(x\right)$, with $c\in F$ and each $p_i\left(x\right)$ monic irreducible in $F\left[x\right]$. Show further that if $f\left(x\right)=d{q}_1\left(x\right)q_2\left(x\right).....q_m\left(x\right)$ with $d\in F$ and each $q_j\left(x\right)$ monic irreducible in $F\left[x\right]$, then $m=n,c=d$, and after reordering and relabelling if necessary, $p_i\left(x\right)=q_i\left(x\right)$ for each $i$.

Short answer

It is proved that $p_i\left(x\right)\mathrm{and}{q}_i\left(x\right)$ are monic irreducible. And after reordering and relabelling $p_i\left(x\right)=q_i\left(x\right)$.

Step-by-step solution

Theorem 4.14:

If $F$ is any field with a non-constant polynomial $f\left(x\right)$which is the product of some irreducible polynomials $p\left(x\right)$and $q\left(x\right)$, such that:

$\begin{array}{l}f\left(x\right)=p_1\left(x\right)p_2\left(x\right).............p_r\left(x\right)\\ f\left(x\right)=q_1\left(x\right)q_2\left(x\right).............q_s\left(x\right)\end{array}$

Then, $p_j\left(x\right)$is an associate of $q_j\left(x\right)$where $\left\{j=1,2,3,......,r\right\}$.

Proof:

If $f\left(x\right)$ is a non-constant polynomial and $f\left(x\right)=\prod _{i=1}^r{p}_i\left(x\right)$which are irreducible, then for an element $a_i\in F$in $p_i\left(x\right)$, we get:

${a_i}^{-1}{p}_i\left(x\right)$is monic term, for which the desired form is:$f\left(x\right)=\left(\prod _{i=1}^r{a}_i\right)\left(\prod _{i=1}^s{a_i}^{-1}{p}_i\left(x\right)\right)$.

Now, from theorem 4.14, we have:$m=n$, Since, $p_i\left(x\right)\mathrm{and}{q}_i\left(x\right)$ are associates. Both $c\mathrm{and}d$ will be leading terms in the given polynomial.

So, we get, for $c=d$, both$p_i\left(x\right)\mathrm{and}{q}_i\left(x\right)$ are monic, associates and therefore, are equal.

Hence proved, $p_i\left(x\right)\mathrm{and}{q}_i\left(x\right)$ are monic irreducible. And after reordering and relabelling $p_i\left(x\right)=q_i\left(x\right)$.