Wikipedia claims that a ring $R$ is Noetherian if and only if all finitely-generated modules $M$ are Noetherian. I can prove the forward direction, but for the converse, I'm wondering whether the ring needs to be unital.
As a counter-example: consider the ring $2\mathbb{Z}$. It does not generate itself, since, for any $x\in2\mathbb{Z}$, $(2\mathbb{Z})x\subseteq 4\mathbb{Z}$. So it's not Noetherian. But any module $M$ over $2\mathbb{Z}$ can be extended to a module over $\mathbb{Z}$, since for any odd number $2n+1$, we can define an action of $(2n+1)m=(2n)m+m$ for $m\in M$. Then any $(2\mathbb{Z})$-submodule $A$ will be finitely-generated over $\mathbb{Z}$. Let $\{a_1,\dots, a_n\}$ be a generating set. Since $M$ is finitely-generated by $(2\mathbb{Z}$), there must be some elements $b_i$ for each $a_i$ such that $2b_i=a_i$. Thus, $\{b_1,\dots,b_n\}$ will generate $A$ as a $(2\mathbb{Z})$-module, since any $a\in A$ can be written as
$$ a=\sum_{i=1}^n n_ia_i = \sum_{i=1}^n n_i(2b_i)=\sum_{i=1}^n (2n_i)b_i$$
for $n_i\in\mathbb{Z}$.
This means that $M$ is Noetherian as a $(2\mathbb{Z})$-module as well. So $2\mathbb{Z}$ is not Noetherian but all of its finitely-generated modules are Noetherian.
Is Wikipedia wrong (it needs to specify that the statement holds for unital rings?) or have I made some error here?