Question

Show that$x^2+x$ can be factored in two ways in${\mathrm{\mathbb{Z}}}_6\left[x\right]$ as the product of non-constant polynomials that are not units and not associates of x or x+1.

Short answer

Hence, the given statement is proved.

Step-by-step solution

Find two different factorizations of x2+x in ℤ6x

Consider the given polynomials,

$x^2+x$can be factorized in two different ways in ${\mathrm{\mathbb{Z}}}_6\left[x\right]$.

Note that:

localid="1654523249471" $$\begin{gathered} \begin{array}{rcl}& & \left(x^2+x=\left(5x+3\left)\right(5x+2\right) \\ =\left(x+3\right)\left(x+4\right)\right)\end{array} \end{gathered}$$

The above factors are not unit or associate of x or x+1.

Now, prove x+1 that 5x+3 and are not associate.

Let’s suppose that x+1=d (5x+3) for non-zero constant d. Then, it must be d=3 and d=5 but $3\ne 5$in ${\mathrm{\mathbb{Z}}}_6$. Thus, it is proved that x+1 and 5x+3 are not associates.

Constructing the above polynomials 

Now, try to find$a,b,c,d\in {\mathrm{\mathbb{Z}}}_6$so that,

$x^2+x=\left(ax+b\right)\left(cx+d\right)=\left(ac\right)x^2+\left(ad+bc\right)x+bd$

$\left[x^0\right]\left(x^2+x\right)=0$in ${\mathrm{\mathbb{Z}}}_6$, take b=3, d=2 or b=3, d=4.

1. b=3, d=2

To find $a,c\in {\mathrm{\mathbb{Z}}}_6$so that ac=1,2a+3c=1. We can choose a=c=5

Then, it gives $x^2+x=\left(5x+3\right)\left(5x+2\right)$in ${\mathrm{\mathbb{Z}}}_6$,

Constructing the above polynomials 

2. b=3, d=4

To find $a,c\in \mathrm{\mathbb{Z}}$so that ac= 1,4a+3c=1 . We can choose a=c= 1

Then, it gives$x^2+x=\left(x+3\right)\left(x+4\right)$ in ${\mathrm{\mathbb{Z}}}_6$.

Thus, a construction on polynomials is done.