 Solution

Noetherian means every ideal is finitely generated, equivalently every ascending ideal chain stabilises. Quotients preserve this; for example $\mathbb{Z}[\sqrt{-5}]$ is Noetherian as a quotient of $\mathbb{Z}[X]$ but not a UFD. A UFD need not be Noetherian (a polynomial ring in infinitely many variables is a counterexample). Stabilisation of kernels proves every surjective endomorphism of a Noetherian ring injective. The shift $k[x_1,x_2,\dots ]\to k[x_1,x_2,\dots ]$, $x_1\mapsto 0$, $x_{i+1}\mapsto x_i$, is surjective noninjective; $x\mapsto x^2$ on $k[x]$ shows injective need not mean surjective. Finally $f_n(X)=\left(\genfrac{}{}{0ex}{}{X}{n}\right)$ is integer-valued. Finite differences prove uniquely that every integer-valued polynomial is an integral linear combination of the $f_n$, so they are a $\mathbb{Z}$-basis. Int$(\mathbb{Z})$ is not Noetherian: the denominators in $f_p$ yield an ideal chain requiring new prime denominators.