 Solution

An ideal of $R/I$ has the form $J/I$ for an ideal $J\supseteq I$ of $R$. If $R$ is Noetherian, writeThenso every ideal of $R/I$ is finitely generated. Hence

The Hilbert basis theorem states that $R[X]$ is Noetherian whenever $R$ is Noetherian. The integers form a principal ideal domain, so every ideal of $\mathbb{Z}$ has one generator and $\mathbb{Z}$ is Noetherian. It follows that $\mathbb{Z}[X]$ is Noetherian.

For a nonsquare integer $d$, evaluation at $\sqrt{d}$ givesIt is surjective and its kernel is $(X^2-d)$. Therefore is Noetherian by the quotient result. This is the noetherianity of a quadratic integer order.