http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2009;task=show_msg;msg=1954.0001
I am specifically referring to this part
Lets prove first that it is not left Artinian. Let $b_1,b_2,\dots$ be a Hamel basis of $\mathbb R$ (the reals) as a vector space over $\mathbb Q$ (the rationals) To make my life easier denote by $[a~ b]$ the matrix with $a,b$ in the first row and zeros in the second. Then the following sequence of ideals never stabilizes. $([1 ~b_1])$ includes $([1~ b_1],[1 ~b_1+b_2])$ includes $([1~ b_1],[1 ~b_1+b_2],[1, ~b_1+b_2+b_3])$ includes ...Check that. Essentially is because you can not get $b_n$ as a linear combination of $b_1,b_2,...b_{n-1}$ over the rationals since they are a basis.
I do not understand what the author is saying. But I understand the second half just fine. More specifically, what does it mean for the sequence (where the matrices are specifically elements of this matrix ring).
$\left(\begin{bmatrix}1 & 0 \\b_1 & 0\end{bmatrix}\right) \subseteq\left(\begin{bmatrix}1 & 0 \\b_1 & 0\end{bmatrix}, \begin{bmatrix}1 & 0 \\b_1 + b_2 & 0\end{bmatrix}\right) \subseteq\left(\begin{bmatrix}1 & 0 \\b_1 & 0\end{bmatrix}, \begin{bmatrix}1 & 0 \\b_1 + b_2 & 0\end{bmatrix}\begin{bmatrix}1 & 0 \\b_1 + b_2 + b_3 & 0\end{bmatrix}\right)$ to never stabilize.
"To stabilize" means that the ideal end up being in an ideal that generates the entire ring. That is how the author is using the word. The sequence of ideals never stabilizes because each of the elements in the Hamel basis is orthogonal from the rest of elements and subsequently cannot be expressed as a linear combination of the rest of the elements.