= Solution
Let $K=k(z)$, where $z$ is an indeterminate, and form the <right Artinian triangular ring that is not left Noetherian>
$$
\boxed{T=\begin{pmatrix}k&K\\0&K\end{pmatrix}.}
$$
Multiplication uses the natural $k$–$K$ bimodule structure on the upper-right entry. With the two diagonal <idempotents> $e_1,e_2$, the right regular <module> is $e_1T\oplus e_2T$. The off-diagonal submodule of $e_1T$ is one-dimensional over the bottom-right <field> $K$, hence simple. Its quotient is one-dimensional over the top-left <field> $k$, hence simple. Also $e_2T$ is a simple right <module> over $K$. Thus $T_T$ has a composition series of length three, and in particular $T$ is right <Artinian>.
On the left, multiplication of an off-diagonal matrix gives
$$
\begin{pmatrix}a&m\\0&b\end{pmatrix}
\begin{pmatrix}0&u\\0&0\end{pmatrix}
=\begin{pmatrix}0&au\\0&0\end{pmatrix}.
$$
Every $k$-subspace $U\subseteq K$ therefore yields a <left ideal> of off-diagonal matrices with entries in $U$. The subspaces $U_m=\operatorname{span}_k\{1,z,\ldots,z^m\}$ are strictly increasing because $z$ is transcendental. Their associated <left ideals> violate the <ascending chain condition>. Hence \b[$T$ is not left Noetherian], even though it is right Artinian and right Noetherian.
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