show that ${ x^4+x+1 \in GF(2^2)[x] }$

Question

show that ${ x^4+x+1 \in GF(2^2)[x] }$ is the product of tow quadratic polynomials .

"i tried to factor ${ x^4+x+1 }$ but don't know "

I did this as an example in the last part of this answer. – Jyrki Lahtonen Dec 30 '19 at 05:10 — Jyrki Lahtonen, Dec 30 '19 at 05:10

score 2 · Accepted Answer · answered Dec 30 '19 at 01:01

2

In this case, it isn't too hard to do this by direct computation. Since $x^4+x+1$ is monic, suppose that $a,b,c,d\in\mathrm{GF}(4)$ with $$x^4+x+1=(x^2+ax+b)(x^2+cx+d).$$ Expand the RHS, compare coefficients and solve for $a,b,c,d$. It will be helpful to remind yourself of what the elements of $\mathrm{GF}(4)$ look like.

answered Dec 30 '19 at 01:01

Thorgott

11,682

4

The constant term should be $bd$. Now, sort the terms based on the power of $x$, then equate coefficients on the RHS and LHS. – Thorgott Dec 30 '19 at 01:14

score 0 · Answer 2 · answered Jan 01 '20 at 05:50

0

i tried to solve it like ${ (x^2+bx+1)(x^2+dx+1)=x^4+x+1 }$

x-terms→b+d=1

x³-terms→b+d=0

contradiction !

not factorisable !

answered Jan 01 '20 at 05:50

Ibrahim Ahmed

189

show that ${ x^4+x+1 \in GF(2^2)[x] }$

2 Answers2