Question

Let p(x) = 24 + 23 +1€ Z2[2] and let a = [z] in the field E = Z2[z]/(p(x)), so a is a root of p(x). (a) (15 points) Write the following elements of E in the form aa+ba+ca+d, with a,b,c,d € Z2. i. a“, a, a6, and a 10 ii. a5 +a+ + a2 + 1 iii. (a? + 1)4 (b) (5 points) The set of units E* = E-{0} of the field E is a group of order 15 under multiplication. Determine the order of a as an element of the group of units E* of E.

Answer 1

Let $\small p(x)=x^4+x^3+1\in \mathbb Z_2[x]$ and let $\small \alpha = [x]\in E = \mathbb Z_2[x]/\langle p(x) \rangle$ , so $\small \alpha$ is a root of $\small p(x)$

So, we must have

$\small p(\alpha)=0 \Rightarrow \alpha^4+\alpha^3+1=0$.........................(1)

(a) we use the above expression for calculating the rest of the questions:

(i)
$\small \alpha^4 = -\alpha^3-1 \equiv (\alpha^3+1)\mod 2$

Hence the required form is

$\small \alpha^4 \equiv_2 \alpha^3+1$..........answer

Multiplying both sides by alpha we get

$\small \alpha^5 \equiv_2 \alpha^4+\alpha \equiv_2 -\alpha^3-1+\alpha \equiv_2 \alpha^3+\alpha+1$................answer

Again multiplying by alpha we have

$\small \alpha^6 \equiv_2 \alpha^4+\alpha^2+\alpha\equiv_2 -\alpha^3-1+\alpha^2+\alpha \equiv_2 \alpha^3+\alpha^2+\alpha+1$.............answer

Multiplying the above by the expression for alpha^4 we have

$\small \alpha^{10} \equiv_2 (\alpha^3+\alpha^2+\alpha+1)(\alpha^3+1)\equiv_2 \alpha^6 +\alpha^5+\alpha^4+2\alpha^3+\alpha^2+\alpha +1$

Now, we put in all the values from the previous answers into this to get

$\small \equiv_2 (\alpha^3+\alpha^2+\alpha+1)+(\alpha^3+\alpha +1)+(\alpha^3+1)+0\alpha^3+\alpha^2+\alpha +1$

Simplifying,

$\small \equiv_2 3\alpha^3+2\alpha^2+3\alpha+4 \equiv_2 \alpha^3+\alpha$

Hence,

$\small \alpha^{10} \equiv_2 \alpha^3+\alpha$....................................answer

(ii)

Again, using the values of alpha^5 and alpha^4 we have

$\small \alpha^5+\alpha^4+\alpha^2+1 \equiv_2 (\alpha^3+\alpha+1)+(\alpha^3+1)+\alpha^2+1$

Simplifying

$\small \equiv_2 2\alpha^3+\alpha+\alpha^2+3\equiv_2 \alpha^2+\alpha+1$.......................answer

(iii) We have, using alpha^4 from the part (i)

$\small (\alpha^2+1)^4=((\alpha^2+1)^2)^2=(\alpha^4+2\alpha^2+1)^2\equiv_2 (\alpha^3+1+2\alpha^2+1)^2$

$\small \equiv_2 (\alpha^3+2\alpha^2+2)^2 \equiv_2 (\alpha^3)^2\equiv_2 \alpha^6$

Now, again using the expression for alpha^6 from part (i) we have

$\small (a\lpha^2+1)^4\equiv_2 \alpha^3+\alpha^2+\alpha+1$............................answer