Exercise 2.3: Reducible and Irreducible Polynomials

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Polynomials of degree  $m = 2$,  $m = 3$,  $m = 4$

Important prerequisites for understanding channel coding are knowledge of polynomial properties.

In this exercise we consider polynomials of the form

$$a(x) = a_0 + a_1 \cdot x + a_2 \cdot x^2 + \hspace{0.1cm}... \hspace{0.1cm} + a_m \cdot x^{m} \hspace{0.05cm},$$

• where for the coefficients  $a_i ∈ {\rm GF}(2) = \{0, \, 1\}$  holds  $(0 ≤ i ≤ m)$

• and the highest coefficient is always assumed to  $a_m = 1$. 

One refers to  $m$  as the  "degree"  of the polynomial.

Adjacent ten polynomials are given where the polynomial degree is either 

• $m = 2$  (red font), 
• $m = 3$  (blue font)  or 
• $m = 4$  (green font).

A polynomial  $a(x)$  is called  reducible  if it can be represented as the product of two polynomials  $p(x)$  and  $q(x)$  of lower degree:

$$a(x) = p(x) \cdot q(x)$$

If this splitting is not possible,  that is,  if for the polynomial

$$a(x) = p(x) \cdot q(x) + r(x)$$

with a residual polynomial  $r(x) ≠ 0$  holds,  then the polynomial is called  irreducible.  Such irreducible polynomials are of special importance for the description of error correction methods.

⇒   The proof that a polynomial  $a(x)$  of degree  $m$  is irreducible requires several polynomial divisions  $a(x)/q(x)$,  where the degree of the respective divisor polynomial  $q(x)$  is always smaller than  $m$.   Only if all these divisions  $($modulo $2)$  always yield a remainder  $r(x) ≠ 0$  it is proved that  $a(x)$  is indeed an irreducible polynomial.

⇒   This exact proof is very complex.  Necessary conditions for  $a(x)$  to be an irreducible polynomial at all are the two conditions  $($in nonbinary approach  "$x=1$"  would have to be replaced by  "$x≠0$"$)$:

• $a(x = 0) = 1$,

• $a(x = 1) = 1$.

Otherwise,  one could write for the polynomial under investigation:

$$a(x) = q(x) \cdot x \hspace{0.5cm}{\rm resp.}\hspace{0.5cm}a(x) = q(x) \cdot (x+1)\hspace{0.05cm}.$$

The above conditions are necessary,  but not sufficient,  as the following example shows:

$$a(x) = x^5 + x^4 +1 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}a(x = 0) = 1\hspace{0.05cm},\hspace{0.2cm}a(x = 1) = 1 \hspace{0.05cm}.$$

Nevertheless,  this polynomial is reducible:

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

Hint:  This exercise belongs to the chapter  "Extension field".

Questions

1

How many polynomial divisions  $(N_{\rm D})$  are required to prove exactly that a  ${\rm GF}(2)$ polynomial  $a(x)$  of degree  $m$  is irreducible?

$m = 2 \text{:} \hspace{0.35cm} N_{\rm D} \ = \ $

$m = 3 \text{:} \hspace{0.35cm} N_{\rm D} \ = \ $

$m = 4 \text{:} \hspace{0.35cm} N_{\rm D} \ = \ $

2

Which of the degree–2 polynomials are irreducible?

	$a_1(x) = x^2 + x$,
	$a_2(x) = x^2 + x + 1$.

3

Which of the degree–3 polynomials are irreducible?

	$a_3(x) = x^3$,
	$a_4(x) = x^3 + 1$,
	$a_5(x) = x^3 + x$,
	$a_6(x) = x^3 + x + 1$,
	$a_7(x) = x^3 + x^2 + 1$.

4

Which of the degree–4 polynomials are irreducible?

	$a_8(x) = x^4 + 1$,
	$a_9(x) = x^4 + x^3 + 1$,
	$a_{10}(x) = x^4 + x^2 + 1$.

Solution

Solution

(1)  The polynomial  $a(x) = a_0 + a_1 \cdot x + a_2 \cdot x^2 + \hspace{0.1cm}... \hspace{0.1cm} + a_m \cdot x^{m}$

• with  $a_m = 1$

• and given coefficients  $a_0, \ a_1, \hspace{0.05cm}\text{ ...} \hspace{0.1cm} , \ a_{m-1}$  $(0$  or  $1$  in each case$)$

is irreducible if there is no single polynomial  $q(x)$  such that the modulo  $2$  division  $a(x)/q(x)$  yields no remainder.  The degree of all divisor polynomials  $q(x)$  to be considered is at least  $1$  and at most  $m-1$.

• For  $m = 2$  two polynomial divisions  $a(x)/q_i(x)$  are necessary,  namely with

$$q_1(x) = x \hspace{0.5cm}{\rm and}\hspace{0.5cm}q_2(x) = x+1 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}N_{\rm D}\hspace{0.15cm}\underline{= 2} \hspace{0.05cm}.$$

• For  $m = 3$  there are already  $N_{\rm D} \ \underline{= 6}$  divisor polynomials,  namely besides  $q_1(x) = x$  and  $q_2(x) = x + 1$  also

$$q_3(x) = x^2\hspace{0.05cm},\hspace{0.2cm}q_4(x) = x^2+1\hspace{0.05cm},\hspace{0.2cm} q_5(x) = x^2 + x\hspace{0.05cm},\hspace{0.2cm}q_6(x) = x^2+x+1\hspace{0.05cm}.$$

• For  $m = 4$,  besides  $q_1(x), \hspace{0.05cm}\text{ ...} \hspace{0.1cm} , \ q_6(x)$  all possible divisor polynomials with degree  $m = 3$  must also be considered,  thus:

$$q_i(x) = a_0 + a_1 \cdot x + a_2 \cdot x^2 + x^3\hspace{0.05cm},\hspace{0.5cm}a_0, a_1, a_2 \in \{0,1\} \hspace{0.05cm}.$$

For the index, $7 ≤ i ≤ 14 \ \Rightarrow \ N_{\rm D} \ \underline{= 14}$.

(2)  For the first polynomial holds:  $a_1(x = 0) = 0$.  Therefore this polynomial is reducible:

$$a_1(x) = x \cdot (x + 1).$$

On the other hand,  the following is true for the second polynomial:

$$a_2(x = 0) = 1\hspace{0.05cm},\hspace{0.2cm}a_2(x = 1) = 1 \hspace{0.05cm}.$$

This necessary but not sufficient property shows that  $a_2(x)$  could be irreducible.  The final proof is obtained only by two modulo-2 divisions:

• $a_2(x)$  divided by  $x$  yields  $x + 1$,  remainder  $r(x) = 1$,

• $a_2(x)$  divided by  $x + 1$  yields  $x$,  remainder $r(x) = 1$.

The correct solution is therefore the  proposed solution 2.

(3)  The first three polynomials are reducible,  as the following calculation results show:

$$a_3(x = 0) = 0\hspace{0.05cm},\hspace{0.2cm}a_4(x = 1) = 0\hspace{0.05cm},\hspace{0.2cm}a_5(x = 0) = 0\hspace{0.05cm},\hspace{0.2cm}a_5(x = 1) = 0 \hspace{0.05cm}.$$

This could have been found out by thinking:

$$a_3(x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} x \cdot x \cdot x \hspace{0.05cm},$$

$$a_4(x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (x^2 + x + 1)\cdot(x + 1) \hspace{0.05cm},$$

$$a_5(x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} x \cdot (x + 1)\cdot(x + 1) \hspace{0.05cm}.$$

The polynomial  $a_6(x)$  is not equal to  $0$  for both  $x = 0$  and  $x = 1$ . This means that

• "nothing speaks against"  $a_6(x)$  being irreducible,
• division by the irreducible degree-1 polynomials  $x$  and  $x + 1$,  respectively,  is not possible without remainder.

However,  since division by the single irreducible degree-2 polynomial also yields a remainder,  it is proved that  $a_6(x)$  is an irreducible polynomial:

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

Using the same computational procedure,  it can also be shown that  $a_7(x)$  is also irreducible   ⇒   Solutions 4 and 5.

(4)  From  $a_8(x + 1) = 0$  follows the reducibility of  $a_8(x)$.  For the other two polynomials,  however,  holds:

$$a_9(x = 0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1\hspace{0.05cm},\hspace{0.35cm}a_9(x = 1) = 1 \hspace{0.05cm},$$

$$a_{10}(x = 0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}1\hspace{0.05cm},\hspace{0.2cm}a_{10}(x = 1) = 1 \hspace{0.05cm}.$$

So both could be irreducible.  The exact proof of irreducibility is more complicated:

• It is not necessary to use all four divisor polynomials with degree  $m = 2$,  namely $x^2, \ x^2 + 1, \ x^2 + x + 1$,  but it is sufficient to divide by the only irreducible degree-2 polynomial.  One obtains with respect to the polynomial  $a_9(x)$:

$$(x^4 + x^3+1)/(x^2 + x+1) = x^2 + 1 \hspace{0.05cm},\hspace{0.4cm}{\rm remainder}\hspace{0.15cm} r(x) = x\hspace{0.05cm}.$$

Also the division by the two irreducible degree-3 polynomials yields a remainder in each case:

$$(x^4 + x^3+1)/(x^3 + x+1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} x + 1 \hspace{0.05cm},\hspace{0.4cm}{\rm remainder}\hspace{0.15cm} r(x) = x^2\hspace{0.05cm},$$

$$(x^4 + x^3+1)/(x^3 + x^2+1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} x \hspace{0.05cm},\hspace{0.95cm}{\rm remainder}\hspace{0.15cm} r(x) = x +1\hspace{0.05cm}.$$

Finally,  let us consider the polynomial  $a_{10}(x) = x^4 + x^2 + 1$.  Here holds

$$(x^4 + x^2+1)/(x^2 + x+1) = x^2 + x+1 \hspace{0.05cm},\hspace{0.4cm}{\rm remainder}\hspace{0.15cm} r(x) = 0\hspace{0.05cm}.$$

It follows:  Only the polynomial  $a_9(x)$  is irreducible   ⇒   proposed solution 2.