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A simple module is a semisimple module .

A module $M$ is called semisimple if every submodule is a direct summand of $M$

Since a simple module has $\{0\}$ and $M$ as its submodules so it is semisimple.

But why is a simple ring not semisimple because a

A ring is semisimple if it is semisimple as a module over itself

How to construct such a ring?When will the result hold?

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    $\begingroup$ Simpel rings have no ideals but they can have left-ideals (i.e., submodules under left multiplication) $\endgroup$ – user45878 Jun 26 '15 at 16:08
  • $\begingroup$ yes ,you are right ,can you give an example @user45878 $\endgroup$ – Learnmore Jun 26 '15 at 16:22
  • $\begingroup$ "When will the result hold?" What result? $\endgroup$ – rschwieb Jun 26 '15 at 20:30
  • $\begingroup$ A simple ring is semisimple if and only if it is artinian (on one side, hence on both). $\endgroup$ – egreg Jun 26 '15 at 20:50
  • $\begingroup$ @egreg You mean left artinian implies right artinian for simple rings? Is that obvious? $\endgroup$ – Tobias Kildetoft Jun 26 '15 at 21:05
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In different categories (such as the category of rings or the category of modules) different substructures are important. Ideals are the natural substructures of rings, and submodules are the natural substructures of modules. These substructures are special because they are suitable for forming quotient objects in their categories (that is why I am not mentioning 'subrings,' for example.)

"Simple" has come to mean "having no nontrivial substructure," (or as suggested in the comments "admitting no nontrivial quotient") so in the case of rings this means it has no nontrivial ideals.

If you want to talk about right and left ideals, then you must realize that we are switching gears and thinking of the ring as a module over itself and no longer like a ring. Notice how the definition you gave for semisimplicity of a ring is entirely in terms of its semisimplicity as a module.

Anyhow, this gives you the answer to your question. The definition of a "simple ring" refers only to ring ideals, and that is not directly connected to its identity as a module.

How to construct such a ring?

The Artin-Wedderburn theorem completely classifies all semisimple rings. They are exactly finite direct products of matrix rings over division rings. This gives you every possible example.

You can make simple rings out of any ring by taking the quotient by a maximal ideal. To get an example where the result isn't semisimple, you'll just have to be careful to make sure the quotient isn't Artinian. One such example is to take the ring of linear transformations of an infinite dimensional vector space and take the quotient by a maximal ideal. This is a non-semisimple simple ring.

There is even a simple ring without nonzero zero divisors. One such example is that of Weyl algebras. Semisimple rings usually have nonzero zero divisors: the only semisimple rings without nonzero zero divisors are division rings.

When will the result hold?

A simple ring will be semisimple iff any if the following conditions hold:

  1. it has a minimal left ideal (or minimal right ideal)

  2. It is right artinian (or left artinian)

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    $\begingroup$ I generally feel that the right notion simple should be "no nontrivial quotients", since ideals are not necessarily the right substructure for rings (subrings might as well be). This also works nicely for groups for example. $\endgroup$ – Tobias Kildetoft Jun 26 '15 at 20:36
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    $\begingroup$ @TobiasKildetoft I was taking for granted the viewpoint that substructures are suitable for quotients, so there is no conflict here (except in opinion of exposition.) But what you're saying is clearer, so I'll incorporate some, if you dont' mind. $\endgroup$ – rschwieb Jun 26 '15 at 20:38
  • $\begingroup$ What I mean is that in the category of rings and ring homomorphisms, the way to recognize which monomorphisms correspond to ideals is to see if they have a cokernel. I don't think I have ever tried studying the category of rings with only monomorphisms coming from inclusion of ideals. $\endgroup$ – Tobias Kildetoft Jun 26 '15 at 20:41
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    $\begingroup$ @rschwieb Subgroups are not in general suitable for quotients of groups; and ideals are not substructures of rings (if we assume they're unital). Simplicity, both for rings and groups is defined in terms of quotients. More generally, quotient of algebras, in the sense of universal algebras, are generally not associated to any “substructure”. Besides this, your answer is very good. $\endgroup$ – egreg Jun 26 '15 at 20:47
  • $\begingroup$ I meant that when will a simple ring besemisimple? $\endgroup$ – Learnmore Jun 27 '15 at 1:49

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