Question

Check whether or not <(x+1)²> is a maximal ideal of Z[x].

Answer 1

The maximal ideals of $Z[x]$ are of the form $(p,f(x))$ where $p$ is a prime number and $f(x)$ is a polynomial in $Z[x]$ which is irreducible modulo $p$.

To prove this let $M$ M be a maximal ideal of $Z[x]$ . Assume that $M\cap Z[x]\neq(0)$ . As $Z/(M\cap Z)$ injects into $Z[x]/M$ , we conclude that $Z/(M\cap Z)$ is a domain, so $M\cap Z$ is a prime ideal of $Z$ so $M\cap Z=(p)$ , where $p$ is a prime number.

In our case

$n=1,\\ f(x)=<(x+1)^2>=(x+1,x+1)=x+1 .$

$n$ is not a prime number, then $<(x+1)^2>$ is not a maximal ideal of $Z[x]$ .