This is a question from our past exams. I worked it out. Would anyone kind enough to check the correctness? Thanks! (It occurred several times that I somehow skipped crucial steps or overlooked something in algebra problems.)
Let $A$ be a subgroup of $\mathbb{Z}^3$ spanned by $(2,2,4),(1,3,-2),(1,-1,6)$. Let $H=\mathbb{Z}^3/A$ and $R=\text{End}_{\mathbb{Z}}(H)$. Determine whether $R$ is Artinian or Noetherian (as a $\mathbb{Z}-$module).
My attempts
The structure of $H$
Note that $(1,3,-2)+(1,-1,6)=(2,2,4)$, we have $A=\langle(1,3,-2),(1,-1,6)\rangle=\langle(1,3,-2),(0,-4,8)\rangle.$ On the other hand, $$\det\begin{pmatrix} 0 & 0& -1\\ 1 & 3 & -2\\ 0 & -1 & 2 \end{pmatrix}=1.$$ In the new set of generators $(0,0,-1),(1,3,-2),(0,-1,2)$, $\mathbb{Z}^3$ remains itself and $A=\mathbb{Z}\oplus 4\mathbb{Z}.$ Thus, $H=\mathbb{Z}^3/A\simeq \mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}.$
The essential idea here is to work out the Smith normal form of
$$\begin{pmatrix} 2 & 2 &4\\ 1 &3 & -2\\ 1 & -1 & 6 \end{pmatrix}\sim\begin{pmatrix} 1 \\ & 4\\ & & 0 \end{pmatrix}$$
The structure of $R$ as a ring
Write elements in $H$ as $\begin{pmatrix} a \\ [b] \end{pmatrix},$ the elements in $R$ can be viewed as $2\times 2$ matrices: \begin{pmatrix} a & 0\\ [b_1] & [b_2] \end{pmatrix}
Sadly $R$ is not commutative: $$\begin{pmatrix} a_1 & 0\\ [b_1] & [b_2] \end{pmatrix}\begin{pmatrix} a_2 & 0\\ [c_1] & [c_2] \end{pmatrix}=\begin{pmatrix} a_1a_2 & 0\\ [a_2b_1+b_2c_1] & [b_2c_2] \end{pmatrix}$$
The structure of $R$ as a $\mathbb{Z}-$module
Note that $\text{Hom}(\mathbb{Z},\mathbb{Z}/4\mathbb Z)=\mathbb{Z}/4\mathbb{Z}=\text{Hom}(\mathbb{Z}/4\mathbb{Z},\mathbb{Z}/4\mathbb{Z})$, $\text{Hom}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$ and $\text{Hom}(\mathbb{Z}/4\mathbb{Z},\mathbb{Z})=0.$ Hence $R=\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}.$
Since $\mathbb{Z}$ and $\mathbb{Z}/4\mathbb Z$ are Noetherian modules, the so does their direct product $R$. On the other hand, $\mathbb{Z}\supset2\mathbb{Z}\supset2^2\mathbb{Z}\supset\cdots$ is a infinite strictly descending sequence of submodules. Hence $R$ is not Artinian.
Many thanks to @Maty and @user26857.