Question

Question:

(a) Show that ${\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$is irreducible in${\mathit{Z}}_{\mathbf{2}}\left[x\right]$ and construct a field of the order8.

(b) Show that ${\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$is irreducible${\mathit{Z}}_{\mathbf{3}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ in and construct a field of order 27.

(c) Show that${\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$ is irreducible in ${\mathit{Z}}_{\mathbf{4}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$and construct a field of the order16.

Short answer

Answer:-

1. Thus, the construct of a field of order is .
2. Thus, the construct of a field of order is .
3. Thus, the construct of a field of order is .

Step-by-step solution

(a)Step 1: Concept

In the given question Z is a field.

$\therefore x^3+x+1$is irreducible.

If it has no root in.

Now,

$Z_2=\left\{0,1\right\}$

$\therefore 0^3+0+1\ne 0$ in Z₂.

0 is not a root in Z₂.

For,

$1^3+1+1=3=1$$\left(in{Z}_2\right)$

Not zero

Hence, 1 is also not a root.

$f\left(x\right)=x^3+x+1$is irreducible over Z₂.

Construct of the field of order 8.

Concept

We know that it $f\left(x\right)$is polynomial over a field F which has degree 2 or 3 .

Then

$f\left(x\right)$is reducible$\iff f\left(x\right)$ has root in F.

i.e

$f\left(x\right)$is reducible $\iff f\left(x\right)$does not have root in F .__________(A)

We know that,

Here we construct the field of order 8.

$\frac{Z_p\left(x\right)}{\text{ideal}\text{generated}\text{by}\text{an}\text{irreducible}\text{polynomial}\text{over}{Z}_p\text{of}\text{degree}n}=\text{field}\text{of}\text{order}{p}^n$

Calculation

Here , we Construct the field of order 8.

The field of order is

, here P=2,n=3

(b)Step 4: explanation

We have to show that $f\left(x\right)=x^3-x+1$is irreducible.

By above result (A) check it has root in or not.

$$\begin{gathered} Z_3=\left\{1,2,3\right\} \\ f\left(0\right)=0-0+1=1\ne 0 \\ f\left(1\right)=1-1+1=1\ne 0 \\ f\left(2\right)=8-2+1=7 \\ 7=1\text{in}{z}_3\ne 0 \end{gathered}$$

Hence we show that ,

$f\left(x\right)=x^3-x+1$has no root in$z_3$ .

$f\left(x\right)$is irreducible in $z_3\left[x\right]$.

Solve

Here we construct the field of order 27.

$\frac{Z_p\left(x\right)}{\text{ideal}\text{generated}\text{by}\text{an}\text{irreducible}\text{polynomial}\text{over}{Z}_p\text{of}\text{degree}n}=\text{field}\text{of}\text{order}{p}^n$

So , we get the field of order 27 is ,

here $p=3,n=3$.

(c)Step 6: Explanation

Given in the question$f\left(x\right)=x^4+x+1$ .

Here put the value,x=0

$f\left(0\right)=1,$

If put x = 1

$f\left(1\right)=1+1+1$

$=3=1\left(in{\text{Z}}_2\right)$

$f\left(x\right)$has no root in ${\text{Z}}_2$.

Let $f\left(x\right)$is reducible over${\text{Z}}_2$.

factor of $f\left(x\right)$does not contain linear factor .

Possible factor of 4 degree polynomial is

$$\begin{gathered} 4=1.3 \\ =1,1,2 \\ =1,1,1,1 \end{gathered}$$

These factor not possible because$f\left(x\right)$ has not root .

$4=2,2$

Here we get possible factor .

i.e if is $f\left(x\right)$reducible then$f\left(x\right)$ is reduced into 2 degree polynomial .

concept

Let$f\left(x\right)=\left(a{x}^2+bx+c\right)\left(d{x}^2+ex+f\right)$

$\Rightarrow x^4+x+1=ad{x}^4+\left(ae+bd\right)x^3+\left(bf+ce\right)x+\left(af+cd\right)x^2+cf$_________(A)

Comparing coefficient both sides

$$\begin{gathered} ad=1,a,d\in z_2 \\ \therefore a=1,d=1 \\ cf=1,c,f\in z_2 \\ \therefore c=1,f=1 \end{gathered}$$

Form (A)

$x^4+x+1=x^4+\left(e+b\right)x^3+\left(b+e\right)x+0.x^2+cf+1=0$

Comparing coefficient of .

$b+e=0$__________(1)

Comparing coefficient of x,

$b+e=1$_________(2)

From (1) and (2)

1 = 0

Which is a contradiction .

Hence ,$f\left(x\right)=x^4+x+1$ is irreducible over$z_2$ .

Calculation

By the above result (B) which is used in (a) part .

Field of order 16 is ,

, here$p=2,n=4$ .