Answer to Question #137502 in Abstract Algebra for Ram

Question

Is the ideal generated by x^2+1 in Z2[x] a prime ideal of Z2[x]? give reason.

Answer 1

An ideal $P$ of a commutative ring $R$ is prime if it has the following two properties:

1) If $a$ and $b$ are two elements of $R$ such that their product $ab$ is an element of $P$, then $a$ is in $P$ or $b$ is in $P$;

2) $P$ is not the whole ring $R$.

Let $f(x)=x+1\in\mathbb Z_2[x]$ and $g(x)=x+1\in\mathbb Z_2[x]$. Since $1+1=0$ in the ring $\mathbb Z_2$, $f(x)g(x)=(x+1)(x+1)=x^2+x+x+1=x^2+(1+1)x+1=x^2+1$ is in the ideal $I=\langle x^2+1\rangle$ generated by $x^2+1.$ Taking into account that $f(x)\notin I$ and $g(x)\notin I,$ we conclude that the ideal $I=\langle x^2+1\rangle$ generated by $x^2+1$ is not prime.