This problem deals with polynomials in Z_{2}\left[X\right]. Remember these

arneguet9k

arneguet9k

Answered question

2021-11-12

This problem deals with polynomials in Z2[X]. Remember these have 01 coefficients. You use the usual algorithms, but with all arithmetic done mod 2.
a) It was observed in class that the constant term of an irreducible polynomial is nonzero. Show further that the Hamming weight (number of nonzero coeffs) of an irreducible polynomial of degree 1 is odd. [Hint: Use the remainder theorem.]
b) Only one degree 2 polynomial f meets the conditions of a). What is it? Give the addition and multiplication tables for polynomial arithmetic mod f.
c) Using similar ideas, find all the irreducibles of degrees 3 and 4. (Note: For degree 4, you will need to consider a possible quadratic factor.)

Answer & Explanation

Nancy Johnson

Nancy Johnson

Beginner2021-11-13Added 17 answers

a) Suppose that any polynomial p(x) in Z2[X] has even number of nonzero coefficient then p(1) is even (because the coefficient are eithes 0 or 1. So nonzero coefficient sum of coefficient which are even) and p(r) is even 1 is root of p(x) with operation modZ If 1 is root of p(x)p(x) is reducible. So, if a polynomial is irreducible with a constant term than the Hamming weight (No of nonzero coefficient) is add.
(If constant term is not included then x=0 is a root andpolynomial is reducible)
b) We want a polynomial of degree Z with constant term which is irreducible ie no of no of nonzero coefficients is add. Since polynomial of degree with constant term has at last two nonzero coefficient (one coefficient of x2 and 2nd constant term). Se, for total coefficient to be add coefficient of x must be nonzero. So f(x)=x2+x+1=f(x).
Under mod f operation any polynomial has degree at most 1. So, polynomials under mod f are
{1,x,1+x,0}
x01x1+x00000101x1+xx0x1+x11+x01+x1x
+01x1+x001x1+x1101+xxxx1+x011+x1+xx10 c) Irreducible polynomials of degree 3.
No of nonzero coefficient "with constant term" are add in an irreducible polynomial with constant term.
So, For degree 3, irreducible polynomial x4+x3+1, x4+x2+1 (For a poiynomunal of degree with constant term x3 and constants has coefficient 1: So remainingterms x2 and x; For nonzero coefficient to be add we can add any one of x2 and x)
For polynomial of degree 4, polynomial with add no of nonzero coefficients and constant terms are
x4+x3+1, x4+x2+1, x4+x+1, x4+x3+x2+x+1
Now, if any polynomial f(x) of degree 4 can be factorized into f1(x) and f2(x) of degree 2, and f(x) has hamming no add.
(If hamming no is even then f(x) is reducible.)
Now f(x) has hamming no add. So f1(x) and f2(x) also has hamming no add. Since any of f1(x), f2(x) has hamming no even then 1 is root of f1(x) o rf2(x)1 is root of

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