Question #135138
For the Ring, R= Z2[x]/<x^8-1>. Find zero divisors and nilpotent elements if any.
1
Expert's answer
2020-09-28T13:29:37-0400

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

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

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

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

    f(x)\iff f(x) is irreducible

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

Now in your case f(x)=x81f(x)=x^8-1 and F=Z2\mathbb{F}=\Z_2

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

Now under mod 2 , x81=x8+1x^8-1=x^8+1

Now x8+1=(x4)2+2x4+12x^8+1=(x^4)^2+2x^4+1^2

=(x4+1)2=(x^4+1)^2

Again, x4+1=(x2)2+2x2+12x^4+1=(x^2)^2+2x^2+1^2

=(x2+1)2=(x^2+1)^2

But ,x2+1=x2+2x+12x^2+1=x^2+2x+1^2

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

x8+1=(x+1)8=(x1)8\therefore x^8+1=(x+1)^8=(x-1)^8

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

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

either ab=0 or ba=0ab=0 \ or \ ba=0 .

In our case , a zero divisor should be divisior of x81x^8-1 and not equal to x81x^8-1 .

Therefore the zero divisor of

Z2[x]<x81>\frac{ \Z_2[x]}{<x^8-1>} are (x1)+<x81>,(x1)2+<x81>(x-1)+<x^8-1>,(x-1)^2+<x^8-1>

,(x1)3+<x81>,(x1)4+<x81>(x-1)^3+<x^8-1>,(x-1)^4+<x^8-1>

,

(x1)5+<x81>,(x1)6+<x81>(x-1)^5+<x^8-1>,(x-1)^6+<x^8-1>

,(x1)7+<x81>(x-1)^7+<x^8-1>

An element r(x)+<x81>r(x)+<x^8-1> is nilpotent if and only if there exist a nZ>0n\in \Z_{>0} such that (x81)rn(x)(x^8-1)| r^n(x) but x81x^8-1 does not divide r(x).r(x).

Therefore the nilpotent element are all those element of Z2[x]<x81>\frac{\Z_2[x]}{<x^8-1>} which have zero divisor as a factor.



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