Is every left primitive ring a division ring? I was solving some exercises about left primitive rings but the "proof" I found for them doesn't use all the assumptions, so I wanted to know if there are left primitive rings wich are not division rings. The exercises were "Let $R$ be a left primitive ring, if either for all $a,b \in R$,  $a(ab-ba) = (ab-ba)a$ or $1+a^2$ is always a unit then $R$ is a division ring."
I have a "proof" (I'm pretty sure it is wrong because it solves both these exercises at the same time), it goes as follows:
(1) If $R$ is a left primitive ring then for all $\forall a \neq 0 \in R,\exists b \in R, ba = 1$.
Proof of (1): Since $R$ is a left primitive ring there exists a simple module $S$ such that $ann_R(S) = \{0\}$. If $a \neq 0 \in R$ then $a \notin ann_R(S) \implies \exists v \in S, av \neq 0$. Since $av \neq 0$ and $S$ is simple, $R(av) = S$. This in particular implies that $\exists b \in R, b(av) = v$, wich in turn implies $b(av) - v = 0 \iff (ba - 1)v = 0$ and since $ann_R(S) = \{0\}$, $ba -1 = 0 \implies ba  = 1.$
Now it is only necessary to check that $b$ is also a right inverse of $a$ but we have that
$$
ab = a\cdot1\cdot b = a \cdot(ba) \cdot b = ab(ab) \implies b = b(ab) \implies 1 = ab
$$
where the implications follow from $a,b \neq 0$ and therefore both have a left inverse.
Where is the mistake in my proof ? Assuming it is wrong what are some examples of left primitive rings which are not division rings?
 A: The flaw in your argument is that $(ba-1)v=0$ implies $ba-1=0$. For this to work, you would have to have that $ba-1$ annihilates all of $S$, but here you only know that it annihilates $v$.
Take a field $k$ and $R=M_2(k)$ the ring of $2\times 2$ matricies. Then $k^2$ is a faithful simple $R$-module. Indeed, if $(0)\subsetneq M\subseteq k^2$ is a non-trivial submodule of $k^2$, let $\begin{pmatrix} a \\ b\end{pmatrix}\in M\setminus\{0\}$. Then multiplying on the right by $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ resp. $\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}$, we obtain that $\begin{pmatrix} a \\ 0\end{pmatrix}\in M$ resp. $\begin{pmatrix} 0 \\ b\end{pmatrix}\in M$. By scaling, and as at least one of $a,b$ is non-zero, we obtain that either $\begin{pmatrix} 1 \\ 0\end{pmatrix}\in M$ or $\begin{pmatrix} 0 \\ 1\end{pmatrix}\in M$ is in $M$. So multiplying on the left by $\begin{pmatrix} x & x \\ y & y\end{pmatrix}$, we obtain that $M$ contains all of $k^2$, i.e. $M=k^2$. So $k^2$ is a simple $R$-module.
Furthermore, $k^2$ is faithful because if $A\begin{pmatrix} 1 \\ 0\end{pmatrix}=A\begin{pmatrix} 0 \\ 1\end{pmatrix}=0$, then $A=0$.
Finally, $R$ is not a division ring, because $\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$ doesn't admit a left nor a right inverse.
A: The mistake is the last implication, concluding that $ba-1=0$. Indeed, you know that $(ba-1)v=0$, but this does not hold for all $v$, but just the one $v$ you picked, and the annihilator requires this equality to hold for all $v$.
I am pretty sure most matrix rings would be left primitive rings that are not division rings. To give you a concrete example, let $R=M_2(\mathbb{R})$. I will be using the equivalent criterion of being left primitive: the ring has a maximal left ideal containing no two-sided ideal.
Look at the maximal left $R$-ideal
$$I =\lbrace \begin{pmatrix} a & 0 \\
b & 0  
\end{pmatrix} \mid a\in\mathbb{R}\rbrace.$$
But
$$\begin{pmatrix} a & 0 \\
b & 0  
\end{pmatrix}\begin{pmatrix} 0 & 1 \\
1 & 0  
\end{pmatrix} = \begin{pmatrix} 0 & a \\
0 & b  
\end{pmatrix},$$
Showing that no non-zero subset of $I$ is a right ideal. This concludes the example.
