Surjection between noncommutative Artinian modules Let $R=\mathbb C\langle x,y,z\rangle$ be the free $\mathbb C$-algebra on three non-commuting generators $x$, $y$ and $z$. Suppose $R_x=\mathbb C\langle x,y,z,x^{-1}\rangle$ is the localization at the variable $x$. Let $I$ be a left ideal such that $R/I$ is a $\mathbb C$-vector space of dimension $n$, and consider the commutative diagram
$$\require{AMScd}
\begin{CD}
R @>>> R/I \\
@V\lambda V V @VV\\\overline{\lambda} V\\
R_x @>>> R_x/I_x
\end{CD}
$$
where the horizontal maps are the natural surjections. So $R/I$ is an $n$-dimensional $\mathbb C$-vector space, in particular a left Artinian $R$-module.

Is it the case that $\overline{\lambda}:R/I\to R_x/I_x$ is surjective?

This would be true if $R$ were commutative (say, $R=\mathbb C[x,y,z]$), as it can be seen by writing $R/I$ and $R_x/I_x$ as products of their localizations at maximal ideals, and then using that every maximal ideal in $R_x/I_x$ is the extension of a maximal ideal in $R/I$.
In geometric terms: in the commutative setup, the localization $R\to R_x$ would correspond to an open immersion $D(x)\to \textrm{Spec }R$ and the surjection $R/I\twoheadrightarrow R_x/I_x$ would correspond to the inclusion $V(I)\cap D(x)\subset V(I)$, which is not just open but also closed by the Artinian ($=$ finiteness) condition. I wonder if the same statement holds in this (very special) non-commutative case.
 A: Yes, $\overline{\lambda}$ is surjective.  In brief, the argument is: inverting $x$ on $R/I$ can be achieved by just taking the quotient that makes $x$ injective, since an injective endomorphism of a finite-dimensional vector space is surjective.
Here's the detailed argument.  The module $R_x/I_x$ can be described as the module obtained by taking $R/I$ and then freely making $x$ act invertibly.  Define a sequence $(M_k)$ of submodules of $R/I$ as follows.  First, let $M_0=0$.  Given $M_k$, define $M_{k+1}$ as the submodule generated by the set of $a\in R/I$ such that $x^ma\in M_k$ for some $m\in\mathbb{N}$.  Let $M=\bigcup_k M_k$ and $N=(R/I)/M$.
I now claim that $N$ has the universal property of $R_x/I_x$.  First, note that multiplication by $x$ is injective on $N$.  Indeed, if $xa\in M$ for some $a\in R/I$, then $xa\in M_k$ for some $k$, and then $a\in M_{k+1}$, so $a\in M$.  Now since $N$ is finite-dimensional, that means $x$ actually acts invertibly on $N$.  On the other hand, if $f:R/I\to L$ is any $R$-module homomorphism where $x$ acts invertibly on $L$, it is easy to see by induction that $f$ must vanish on each $M_k$ and hence on $M$, so $f$ factors uniquely through the quotient $R/I\to N$.
Thus $R_x/I_x\cong N$, with the canonical map $\overline{\lambda}:R/I\to R_x/I_x$ corresponding to the quotient map $R/I\to N$.  In particular, $\overline{\lambda}$ is surjective.
