Answer to Question #135138 in Abstract Algebra for Ram

Question

For the Ring, R= Z2[x]/<x^8-1>. Find zero divisors and nilpotent elements if any.

Answer 1

We know that , If $\mathbb{F}$ is a finite field , then

$\frac{ \mathbb{F}[x]}{<f(x)>}$ would be a finite dimensional vector space over a finite field and thus would be a finite set .

Now $\frac{\mathbb{F}[x]}{<f(x)>}$ is a domain (has no zero divisiors)

$\iff <f(x)>$ is a prime ideal

$\iff f(x)=0 \ or \ f(x)$ is a prime elements of the ring (Because we are in a commutative PID)

$\iff f(x)$ is irreducible

(Because in a PID prime elements are same as irreducible)

Now in your case $f(x)=x^8-1$ and $\mathbb{F}=\Z_2$

Therefore $f(x)$ is not irreducible because 1 is a root of $f(x)$ .

Now under mod 2 , $x^8-1=x^8+1$

Now $x^8+1=(x^4)^2+2x^4+1^2$

$=(x^4+1)^2$

Again, $x^4+1=(x^2)^2+2x^2+1^2$

$=(x^2+1)^2$

But ,$x^2+1=x^2+2x+1^2$

$=(x+1)^2$

$\therefore x^8+1=(x+1)^8=(x-1)^8$

........................(1)

A nonzero element $a$ of a ring $R$ is said to be zero divisors if there exist a nonzero element $b\in R$ such that

either $ab=0 \ or \ ba=0$ .

In our case , a zero divisor should be divisior of $x^8-1$ and not equal to $x^8-1$ .

Therefore the zero divisor of

$\frac{ \Z_2[x]}{<x^8-1>}$ are $(x-1)+<x^8-1>,(x-1)^2+<x^8-1>$

,$(x-1)^3+<x^8-1>,(x-1)^4+<x^8-1>$

,

$(x-1)^5+<x^8-1>,(x-1)^6+<x^8-1>$

,$(x-1)^7+<x^8-1>$

An element $r(x)+<x^8-1>$ is nilpotent if and only if there exist a $n\in \Z_{>0}$ such that $(x^8-1)| r^n(x)$ but $x^8-1$ does not divide $r(x).$

Therefore the nilpotent element are all those element of $\frac{\Z_2[x]}{<x^8-1>}$ which have zero divisor as a factor.