Is the following ring left Noetherian? Let $S$ be a ring with identity (but not necessarily commutative) and $f:M_{2}(\mathbb R)\to S$ a non zero ring homomorphism ($M_{2}(\mathbb R)$ is the ring of all $2\times2$ matrices). Is $S$ a left Noetherian ring?
 A: If $f$ is surjective, then the answer is YES since every quotient ring of a left noetherian ring is left noetherian. Otherwise, take $S=M_2(\mathbb R)\times T$, where $T$ is not left noetherian. In this case $S$ is not left noetherian.
Edit. Since in the example above the canonical ring homomorphism $f:M_2(\mathbb R)\to M_2(\mathbb R)\times T$ does not send the unit element to the unit element (N.B. The OP didn't ask this, but in the comments appeared this as an objection!), one can change it by choosing $T$ an $\mathbb R$-algebra which is not noetherian and taking the tensor product (over $\mathbb R$) instead of the direct product. This means that we have $S=M_2(T)$.
A: Not necessarily. Let $S = M_{2} (\mathbb{R})[x_{1}, x_{2}, \ldots]$, the polynomial ring in countably many variables (commuting with everything) with coefficients in $M_{2} (\mathbb{R})$. This clearly admits a non-zero map $M_{2} (\mathbb{R}) \rightarrow S$, but it is not left Neotherian, since there is an increasing, non-stationary chain of two-sided ideals $I_{n} = (x_{1}, \ldots, x_{n})$.
