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In the book I've been reading recently (Algebra by Thomas W. Hungerford) the following definition is given:

A left module $A$ over a ring $R$ is simple provided that $RA\neq 0$ and $A$ has no proper submodules. A ring $R$ is simple if $R^2\neq 0$ and $R$ has no proper two-sided ideals.

My question is why don't we define a simple ring to be a ring which is simple as a module over itself (as in many definitions for rings that arise from definitions for modules). Clearly a ring which is simple as a module over itself is also simple, however the converse is not true (take the ring of square matrices over a simple ring).

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A simple ring means simple in the category of rings. –  Frank Murphy Sep 15 '11 at 21:18
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The simple here means that there are no proper homomorphic images. In the case of a ring $R$, it has a proper homomorphic image $R\rightarrow S$ with non zero kernel $K$ and then to keep the structure of the ring, $K$ must be a two sided ideal. In the case of a module, the kernel $K$ only needs to be a module, because you only need to worry about the multiplication from one side by the ring.

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Because the notion "simple as a (left, say) module over itself" is not very useful. A ring with that property, assuming it has a $1$, has no proper non-zero left ideals, and then it is a division ring.

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What if we don't assume it has 1? (What I mean is that there are other results or definitions that become trivial once we have 1.) –  Pandora Aug 8 '11 at 15:12
    
I don't know. I rarely care about such things... –  Mariano Suárez-Alvarez Aug 8 '11 at 15:14
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