Answer to Question #138420 in Abstract Algebra for Ruth

Question

Show that the ideal <x^2+1> is not prime in Z2[x]

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 $ab\in P$, then $a\in P$ or $b\in P$;

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.