Let $V$ be an $\mathbb{F}$ – vector space with a countably infinite basis. Let $R=\text{End}_R V$ the ring of all linear functions $\phi:V\to V$ and $I=\{f\in R:\, \text{dim}\, f<\infty\}$ the two sided ideal of $R$ consisting of the linear functions with finite rank.
I' ve been trying to show that $R/I$ serves as an example of a simple ring which is not semisimple. I have shown that $R/I$ it is not left Artinian and not left Notherian which shows that $R/I$ cannot be semisimple. Although, I am having some trouble to show that $R/I$ is simple.
So far I have proved that if $f\in R\setminus I$ then there exists $x,y\in R$ such that $xfy=I_V$, where $I_V$ is the identity map on $V$. Now I tried to show that $I$ must be a maximal ideal by using the above fact. So I wrote, if $I\subsetneq J$ is a left ideal of $R$ and $f\in J\setminus I$ then there exists $x,y\in R$ such that $xfy=I_V$. Now, $J$ being a left ideal implies that $xf\in J$. I want to somehow prove that $I_V=(xf)y\in J$ to conclude that $J$ must be equal to $R$ but I dont know that $(xf)y\in J$ since $J$ is not necessarily a right ideal.
Am I missing something? Do you have any ideas? Thanks in advance!
Edited Answer: From the comments below, $R/I$ is simple if and only if $R/I$ does not contain any non trivial two sided ideals. Therefore, to show that $R/I$ is simple we need to prove that for a two sided ideal $0\neq \overline{J}\subseteq R/I$ we have $\overline{J}=R/I$. We know that there exists a two sided ideal $I\subsetneq J\subseteq R$ such that $J/I = \overline{J}$. For $f\in J\setminus I$ there exists $x,y\in R$ with $xfy = I_V$. $J$ being a two sided ideal implies that $I_V\in J$. But then $J=R$ and therefore $\overline{J}=R/I$ and $R/I$ does not contain any non trivial two sided ideals.
Best Answer
After a small discusion in the comments I decided to upload a complete answer to the problem, in case if anyone is intrested in the future for this counterexample. I will also point out my misinterpretation which was in the definition of simplicity.
The problem: Find an example of a simple ring which is not semisimple.
Let me clarify the above definitions first. We say that the ring $R$ is simple if $R$ does not have any other two-sided ideals other than $0$ and itself. We say that $R$ is (left) semisimple if for every (left) ideal $I$ there exists a (left) ideal $J$ with $R=I\oplus J$. I had confused the definition of simplicity with the following weaker condition which is known as left simplicity. We say that the ring $R$ is left semisimple if $R$ does not contain any other left ideals other than $0$ and itself.
Solution to the problem: Let $\mathbb{F}$ be a field and $V$ a vector space over $\mathbb{F}$ with a countably infinite basis. Let $R=End_\mathbb{F} V$ be the ring of all linear operators from $V$ to $V$. Furthermore, let $I$ be the two-sided ideal of $R$ which constitutes of all linear operators $x:V\to V$ with finite dimensional image. I.e. $$I=\biggl\{x\in R:\,\text{dim im}x<\infty\biggr\}.$$ We claim that the quotient space $R/I$ is our desired example. The result will follow immediately from the following claims:
Claim 1: For every $f\in R\setminus I$ there exits $x,y\in R$ with $xfy=I_V$, where $I_V$ denotes the identity operator on $V$.
Proof of Claim 1: Let $f\in R\setminus I$. Then $\text{dim im }f=\infty$. Let $(e_n)_{n=1}^{\infty}$ be a countably infinite basis of $\text{im} f$. Then, there exists $u_n$ with $f(u_n)=e_n$. We extend the basis (possibly by using Zorn's lemma) of $\text{im} f$ to a basis of the whole space $V$. Therefore, we can adjust some elements $(v_j)_{j\in J}$ indexed by some set $J$ such that the collection $$\{e_n:\,n\in \mathbb{N}\}\cup \{v_j:\,j\in J\}$$ is a basis of $V$. By our hypothesis that $V$ has a countable basis $J$ would be at most countably infinite. We consider two cases:
Case 1: J is countably infinite. In this case we write $(v_j)_{j\in J}$ as a sequence $(v_n)_{n\in \mathbb{N}}$. We rewrite the basis in a sequence $(e_n')_{n\geq 1}$ with $e_{2n}'=e_n$ and $e_{2n-1}'=v_{n}$ for every $n\geq 1$. To define the operators $x,y$ we only need to determined their values on the elements of the basis. We define $x,y$ by the following relations: $$e_{2n}'\overset{y}{\longrightarrow}u_{2n}\overset{f}{\longrightarrow}e_{2n}=e_{4n}'\overset{x}{\longrightarrow}e_{2n}'$$ $$e_{2n-1}'\overset{y}{\longrightarrow}v_{2n-1}\overset{f}{\longrightarrow}e_{2n-1}=e_{4n-2}'\overset{x}{\longrightarrow}e_{2n-1}'.$$ Note that $y$ has been defined in all elements of the basis, hence it can be extended uniquely to a linear operator $y:V\to V$. We define $x$ on the rest of the basis elements by giving an arbitrary value and we extend $x$ also. Now the extended operators, which we denote again by $x,y$ satisfy $(x\circ f\circ y)(e_n')=e_n'$ for every $n$. Therefore, for every $v\in V$ we have $(x\circ f\circ y)(v)=v$. In other words, $xfy=I_V$.
Case 2: J is finite. We treat this case in similar fashion as the previous one, I will omit this step to make the answer a little more compacted.
Claim 2: $R/I$ is a simple ring.
Proof of Claim 2: Let $0\neq \overline{J}\subseteq R/I$ be a two-sided ideal of $R/I$. Then there exists a two-sided ideal $I\subsetneq J\subseteq R$ with $\overline{J}=J/I$. Now, let $0\neq f+I\in \overline{J}$. Then, $f\in J\setminus I$, thereby $\text{dim im} f=\infty$. By the preceding claim there exists $x,y\in R$ with $xfy=I_V$. Since $J$ is a two-sided ideal, it follows that $xfy\in J$, consequently $I_V+I\in \overline{J}$ and from this we obtain that $\overline{J}=R/I$.
Claim 3: $R/I$ is not left Notherian neither left Artinian.
Proof of Claim 3: For every subspace $U\subseteq V$ let $J_U$ be the left ideal of $V$ given by $$J_U=\{x\in R:\, x(U)=0\}.$$ It is easily seen, that for every two subspaces $W\subseteq U\subseteq V$ we have the following inclusions $J_U\subseteq J_W$. Futhermore, if $\text{dim}_{\mathbb{F}}U/W=\infty$ then $I+J_U\subsetneq I+J_W$. Indeed, first we begin with a basis $(e_i)_{i\in I}$ of $W$ then we extend this basis to a basis $(e_i)_{i\in I}\cup (v_j)_{j\in J}$ of $W$. By the fact that $\text{dim}_{\mathbb{F}}U/W=\infty$ it follows that $J$ is countably infite. At the last step, we extend the basis $(e_i)_{i\in I}\cup (v_j)_{j\in J}$ to a basis $(e_i)_{i\in I}\cup (v_j)_{j\in J}\cup (u_k)_{k\in K}$ of the whole space $V$. We define a linear operator $f:V\to V$ by the relations $f(e_i)=0,\, f(v_j)=v_j$ and $f(u_k)=u_k$. Then, obviously $f\in J_W$ and thereby $f\in I+J_W$. We claim that $f\notin I+J_U$. If not, then $f=x+y$ for some $x\in I$ and $y\in J_U$. Then, on one hand $\text{dim im}(f-y)<\infty$ and on the other hand $x(v_j)=f(v_j)-y(v_j)=f(v_j)=v_j$ and thereby $v_j \in \text{im }x$. But this a contradiction since $\text{dim im} x<\infty$ and $J$ is countably infinite.
Now can prove that $R/I$ is not left Notherian for example. We fix a countable basis $(e_n)_{n=1}^{\infty}$ of $V$. Let a descending sequence $$J_1\supseteq J_2\supseteq ...\supseteq J_n\supseteq J_{n+1}\supseteq ...$$ of infinite subsets of $\mathbb{N}$ with the property that $J_n\setminus J_{n+1}$ is infinite. For every $n$ we define the subspace $W_n=\text{span}\{e_m:\,m\in J_n\}$ of $V$. Then since $(J_n)$ is descending it follows that $W_n\supseteq W_{n+1}$. The fact that $J_n\setminus J_{n+1}$ is infinite implies that $\text{dim}_{\mathbb{F}}W_n/W_{n+1}=\infty$. Consequently, $I+J_{W_n}\subsetneq I+J_{W_{n+1}}$. Therefore, if $\overline{J_n}=(I+W_n)/I$ then every $\overline{J_n}$ is a left ideal of $R/I$ and $$\overline{J_1}\subsetneq \overline{J_2}\subsetneq...\subsetneq \overline{J_n}\subsetneq \overline{J_{n+1}}\subsetneq...$$ and thereby $R/I$ cannot be left Notherian. In similar fashion we show that $R/I$ is not left Artinian too.
Any comments or corrections are appreciated!