TITLE: Ideal, not finitely generated
QUESTION [0 upvotes]: This exercise is driving me insane. I think there might be a mistake in it.

Consider the ring $R$ of matrices of the form:
$ R = \lbrace \begin{pmatrix}
  z & q_{1}  \\
  0 & q_{2}  \\
 \end{pmatrix} , z\in\mathbb{Z}, q_{1},q_{2}\in\mathbb{Q} \rbrace$
I have to find a left ideal of $R$ which is not finitely generated. 

I write:
$R = R_{1} \times R_{2}$ where
$R_{1} = \lbrace \begin{pmatrix}
  z & q_{1}  \\
  0 & 0  \\
 \end{pmatrix} , z\in\mathbb{Z}, q_{1}\in\mathbb{Q} \rbrace$ 
$R_{2} = \lbrace \begin{pmatrix}
  0 & 0  \\
  0 & q_{2}  \\
 \end{pmatrix} , z\in\mathbb{Z}, q_{2}\in\mathbb{Q} \rbrace $
Every ideal of $R$ must have the form (since both $R_{1}, R_{2}$ contain an identity element):
$I\times J$ where $I$ is an ideal of $R_{1}$, $J$ is an ideal of $R_{2}$. 
The ideals of $R_{1}$:
$ R_{1} = \lbrace \begin{pmatrix}
  z & q_{1}  \\
  0 & 0  \\
 \end{pmatrix} , z\in\mathbb{Z}, q_{1}\in\mathbb{Q} \rbrace$ 
$ \lbrace \begin{pmatrix}
  0 & q_{1}  \\
  0 & 0  \\
 \end{pmatrix} , q_{1}\in\mathbb{Q} \rbrace$ 
$0 = \lbrace \begin{pmatrix}
  0 & 0  \\
  0 & 0  \\
 \end{pmatrix} \rbrace$ 
Let $n\in\mathbb{Z}$:
$\lbrace \begin{pmatrix}
  n\cdot z & q_{1}\\
  0 & 0  \\
 \end{pmatrix} ,  q_{1}\in\mathbb{Q}, z\in\mathbb{Z} \rbrace$ 
The ideals of $R_{2}$
$R_{2} = \lbrace \begin{pmatrix}
  0 & 0  \\
  0 & q_{2} \\
 \end{pmatrix} ,  q_{2}\in\mathbb{Q} \rbrace$ 
$0 = \lbrace \begin{pmatrix}
  0 & 0  \\
  0 & 0  \\
 \end{pmatrix}   \rbrace$ 
When I check the possible constructions I obtain that they are all finitely generated. Is there a mistake in this exercise (maybe a right ideal of $R$ instead of left??) or where is the mistake?

REPLY [2 votes]: The left ideal $I=\{\begin{pmatrix} 0 & q_{1} \\ 0 & 0\end{pmatrix} | q_1 \in \mathbb{Q} \}$ is not finitely generated. This follows by the fact that $\mathbb{Q}$ is not a finitely generated $\mathbb{Z}$-module. (What happens when you multiply elements of $I$ by elements in $R$?)
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