Question

Prove Theorem 4.14.

Short answer

Theorem 4.14 is proved.

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 of Theorem 4.14:

If $S$ is a set of all non-constant polynomials which are reducible such that $S\ne 0$and $p\left(x\right)\in S$, then let us assume an irreducible polynomial $f\left(x\right)$, which has a minimal degree in the given set. Then, we have:$f\left(x\right)=c\left(x\right)d\left(x\right)$

Now, we have:

$\begin{array}{l}\mathrm{deg}\left(c\left(x\right)\right)<\mathrm{deg}\left(f\left(x\right)\right)\\ \mathrm{deg}\left(d\left(x\right)\right)<\mathrm{deg}\left(f\left(x\right)\right)\end{array}$

Clearly, this implies that the polynomial $f\left(x\right)$is a product of irreducible factors $c\left(x\right)\mathrm{and}d\left(x\right)$which is a contradiction.

So, $S$ is an empty set that consists of no such polynomial which has irreducible factors.

Now, if $f\left(x\right)=\prod _{i=1}^r{p}_i\left(x\right)=\prod _{i=1}^s{q}_i\left(x\right)$are two factors, such that, $c_1\left(x\right)p_1\left(x\right)=q\left(x\right)$, then:

$c_1\left(x\right)\prod _{i=1}^r{p}_i\left(x\right)=\prod _{i=1}^s{q}_i\left(x\right)$

For $r=s$, we have: $c_i\left(x\right)p_i\left(x\right)=q_i\left(x\right)$

Similarly, for $s>r$, we get: $\prod _{i=1}^r{c}_i\left(x\right)=\prod _{i=s-r+1}^s{q}_i\left(x\right)$

Then, for$i>r$ , the polynomial will be irreducible as well as constant, which is again a contradiction.

So, we finally get $c_i\left(x\right)p_i\left(x\right)=q_i\left(x\right)$for $r=s$.

Hence, Theorem 4.14 has been proved.