Is $\langle x^2+1\rangle$ a maximal ideal in $Z[x]$? What about $\langle x^2+1,5\rangle$?

1.3k Views Asked by At

I am thinking that $\langle x^2+1\rangle$ is not a maximal ideal in $Z[x]$ but am having trouble coming up with a solid argument. Also I am confused about what/how to read the second term $\langle x^2+1,5\rangle$ and how that is related. Thanks!

1

There are 1 best solutions below

0
On

The ideal $(x^2+1)$ is not maximal, because $\mathbf Z[x]/(x^2+1)\simeq \mathbf Z[i]$, which is an integral domain, but not a field.

The ideal $(x^2+1,5)$ is not maximal either, because \begin{align} \mathbf Z[x]/(x^2+1,5)&\simeq\mathbf Z/5\mathbf Z[x]/(x^2+1)=\mathbf Z/5\mathbf Z[x]/(x-2)(x+2)\\&\simeq\mathbf Z/5\mathbf Z[x]/(x-2)\times\mathbf Z/5\mathbf Z[x]/(x+2)\simeq\mathbf Z/5\mathbf Z\times\mathbf Z/5\mathbf Z, \end{align} which is not even an integral domain.