Question

If $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\in }\mathit{F}\mathbf{\left[}\mathit{x}\mathbf{\right]}$can be written as the product of two polynomials of lower degree, prove that $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is reducible in$\mathit{F}\mathbf{\left[}\mathit{x}\mathbf{\right]}$

Short answer

It is proved thatf is reducible.

Step-by-step solution

Irreducible elements

It has to show that $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

is reducible. For contradiction, consider that$\mathit{f}$is irreducible.

Let $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$be an irreducible element in $\mathit{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$Then, it can be written as the product of two polynomials.

$\mathit{g}$and $\mathit{h}$of a lower degree.

Step 2: Show that f  is reducible

Let three non-constant polynomials $\mathit{f}\mathbf{,}\mathit{g}\mathbf{,}\mathit{h}\mathbf{\in }\mathit{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$

Therefore, g divides fifg is a unit in F or an associate of f:

If g is an associate of fby Theorem 4.2 then $\mathbf{\text{deg}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{=}\mathbf{\text{deg}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}$ that contradicts the first assumption.

If g is a unit element, it can be written as follows:

$\begin{array}{c}\mathbf{\text{deg}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{=}\mathbf{\text{deg}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{-}\mathbf{\text{deg}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\\ \mathbf{=}\mathbf{\text{deg}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{-}\mathbf{0}\\ \mathbf{=}\mathbf{\text{deg}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\end{array}$

Thiscontradictsour second assumption.

Hence, the first assumption is wrong, sof is reducible.