We know that , If F is a finite field , then
<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] is a domain (has no zero divisiors)
⟺<f(x)> is a prime ideal
⟺f(x)=0 or f(x) is a prime elements of the ring (Because we are in a commutative PID)
⟺f(x) is irreducible
(Because in a PID prime elements are same as irreducible)
Now in your case f(x)=x8−1 and F=Z2
Therefore f(x) is not irreducible because 1 is a root of f(x) .
Now under mod 2 , x8−1=x8+1
Now x8+1=(x4)2+2x4+12
=(x4+1)2
Again, x4+1=(x2)2+2x2+12
=(x2+1)2
But ,x2+1=x2+2x+12
=(x+1)2
∴x8+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∈R such that
either ab=0 or ba=0 .
In our case , a zero divisor should be divisior of x8−1 and not equal to x8−1 .
Therefore the zero divisor of
<x8−1>Z2[x] are (x−1)+<x8−1>,(x−1)2+<x8−1>
,(x−1)3+<x8−1>,(x−1)4+<x8−1>
,
(x−1)5+<x8−1>,(x−1)6+<x8−1>
,(x−1)7+<x8−1>
An element r(x)+<x8−1> is nilpotent if and only if there exist a n∈Z>0 such that (x8−1)∣rn(x) but x8−1 does not divide r(x).
Therefore the nilpotent element are all those element of <x8−1>Z2[x] which have zero divisor as a factor.
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