Answer to Question #137508 in Abstract Algebra for Ram

Question

does the ring Z2[x]/<(x^8)+1> have nilpotent elements? justify.

Answer 1

An element $x$ of a ring $R$ is called nilpotent if there exists some positive integer $n$ such that $x^n=0$.

Polynomial remainder theorem implies that each element of $\mathbb Z_2[x]/\langle x^8+1\rangle$ is of the form $[f(x)]=f(x)+\langle x^8+1\rangle=f(x)+(x^8+1)\mathbb Z_2[x]$, where $\deg f(x)<\deg(x^8+1)=8$. Thus, $[0]=\langle x^8+1\rangle=(x^8+1)\mathbb Z_2[x].$ Since $1+1=0$ in the ring $\mathbb Z_2[x]$ and $x^8+1\in\langle x^8+1\rangle=[0]$, $[(x^4+1)]^2=[(x^4+1)^2]=[x^8+x^4+x^4+1]=[x^8+(1+1)x^4+1]=[x^8+1]=[0].$

Consequently, $[x^4+1]$ is a nilpotent element.