# A simple ring which is not semisimple

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.

• Why don't you assume $J$ is a two-sided ideal? Commented Mar 28, 2021 at 22:26
• @Berci If I assume that $J$ is two - sided then $I$ is maximal with respect to the two - sided ideals but am I able to conclude that $R/I$ is simple ? What if there exists a left ideal $J'$ with $I\subsetneq J'\subsetneq R$? Commented Mar 29, 2021 at 5:45
• Being simple exactly means that. Otherwise you are trying to prove that the quotient does not have any left ideal, and that means it is a division ring (which is much stronger, and obviously false). Commented Mar 29, 2021 at 5:52
• @CaptainLama Oh okay, so I have confused the definitions. I thought that being simple means that if I see $R/I$ as an $R/I$ - module with multiplication given by $(r+I,x+I)\mapsto r\cdot x+I$ then $R/I$ is semisimple iff $R/I$ does not have any non trivial submodules, where in this case submodules of $R/I$ correspond to $J/I$ for every left ideal $J$ with $I\subseteq J\subseteq R$. Commented Mar 29, 2021 at 5:57
• @dem0nakos Perhaps you could self-answer your question now that you understand the solution and your mistake? Commented Mar 29, 2021 at 14:24

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!

• Claim 1 case 2 is absurd. V cannot have a finite basis by assumption. Remember the assumption is that V has a countable infinite basis. when $V$ is finite dimensional the construction won’t work at all. Commented Apr 12, 2021 at 17:14
• @rschwieb J could be finite. For example if $(e_n)$ is the countable basis of $V$ then for some $f$ might for example happen that $im f =span(e_2,e_3,...)$. Then in this case we have $|J|=1$. The index set $J$ does not corresponds to the basis of $V$ but rather on the basis of the complement of $im f$. Commented Apr 12, 2021 at 21:31
• ah yes, sorry I thought J was the extended basis at first rather than the complement. Really I think it should be possible to do both cases at once... Commented Apr 12, 2021 at 22:45