Exercise 2.3Z: Polynomial Division

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Multiplication and division of polynomials in  $\rm GF(2)$

In this exercise we deal with the multiplication and especially the division of polynomials in the Galois field  $\rm GF(2)$. In the figure the procedure is indicated by a simple and (hopefully) self-explanatory example:

• Multiplying the two polynomials  $x^2 + 1$  and  $x +1$  yields the result  $a(x) = x^3 + x^2 + x + 1$.
• Dividing the polynomial  $b(x) = x^3$  by  $p(x) = x + 1$  gives the quotient  $q(x) = x^2 + x$  and the remainder  $r(x) = x$.
• One can check the latter result as follows:

$$b(x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} p(x) \cdot q(x) + r(x)\hspace{0.05cm}= \big[(x+1) \cdot (x^2+x)\big] +x =\big[x^3+ x^2+x^2+ x\big] +x = x^3\hspace{0.05cm}.$$

Hint:

• This exercise belongs to the chapter  "Extension field".

Questions

1

What is the result  $a(x) = (x^3 + x + 1) \cdot (x^2 + 1)$?

	$a(x) = x^5 + x^3 + x^2 + 1$,
	$a(x) = x^5 + x^2 + x + 1$.
	$a(x) = x^6 + x^3 + x^2 + 1$-

2

Which of the polynomial divisions do not yield a remainder  $r(x)$?

	$(x^5 + x^2 + x + 1)/(x^3 + x + 1)$.
	$(x^5 + x^2 + x + 1)/(x^2 + 1)$,
	$(x^5 + x^2 + x + 1)/(x^2)$,
	$(x^5 + x^2 + x)/(x^2 + 1)$.

3

It is  $a(x) = x^6 + x^5 + 1$  and  $p(x) = x^3 + x^2 + 1$.
Determine  $q(x)$  and  $r(x)$  according to the description equation  $a(x) = p(x) \cdot q(x) + r(x)$.

	$q(x) = x^3 + x^2 + 1, \hspace{0.2cm} r(x) = 0$,
	$q(x) = x^3 + 1, \hspace{0.2cm} r(x) = 0$,
	$q(x) = x^3 + 1, \hspace{0.2cm} r(x) = x^2$.

Solution

Solution

(1)  The modulo 2 multiplication of the two polynomials leads to the result

$$a(x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (x^3+ x+1) \cdot (x^2+1)= x^5+x^3+ x^2+ x^3+x+1 = x^5+ x^2+x+1\hspace{0.05cm}.$$

• Thus the proposed solution 2 is correct.
• The last solution suggestion cannot simmen already alone, since the degree of the product polynomial would be unequal $5$.

Example 1 for polynomial division

(2)  With the abbreviations

$$a(x) = x^5+ x^2+x+1\hspace{0.05cm},\hspace{0.4cm}p(x) = x^3+ x+1\hspace{0.05cm},\hspace{0.4cm}q(x) = x^2+ 1$$

and the result from subtask (1) we get $a(x) = p(x) \cdot q(x)$.

That is:   The divisions $a(x)/p(x)$ and $a(x)/q(x)$ are free of remainders   ⇒   Correct are the solutions 1 and 2.

Even without calculation one recognizes that $a(x)/x^2$ must result in a remainder. After calculation it results explicitly:

$$(x^5 + x^2+x+1)/(x^2) = x^3 + 1 \hspace{0.05cm},\hspace{0.4cm}{\rm Rest}\hspace{0.15cm} r(x) = x+1\hspace{0.05cm}.$$

To the last proposed solution. We use for shortcut $b(x) = x^5 + x^2 + x = a(x) + 1$. This is the given quotient:

$$b(x)/q(x) = a(x)/q(x) + 1/q(x) \hspace{0.05cm}.$$

Example 2 for polynomial division

• The first quotient $a(x)/q(x)$ gives exactly $p(x)$ without remainder, the second quotient $0$ with remainder $1$.
• Thus the remainder of the quotient $b(x)/q(x)$ is equal to $r(x) = 1$, as the calculation in example 1 shows.

(3)  The polynomial division is explained in detail in example 2. Correct is the proposed solution 3.