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I am studying Semi-Simple Rings from "A first course in Non-Commutative Rings " by T.Y.Lam .

But hardly I am finding any examples to the definitions that are being taught.There is no example of a semisimple ring only the definition is provided i.e "A ring is semisimple if it is simple as a module over itself"

When I read rings ,fields there were so many examples but here I get none.Why ??

Also how to verify whether the popular rings such as $\mathbb Z$ ,$M_n(\mathbb R)$ are simple or not?

Learnmore
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1 Answers1

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You mean semisimple as a module over itself. Examples include

  • Any field. More generally, any division ring.
  • If $R$ is semisimple, then so is $M_n(R)$ (exercise). Hence, for example, $M_n(\mathbb{R})$ is semisimple.
  • If $R$ and $S$ are semisimple, then so is $R \times S$ (exercise).

It follows that any ring which is a finite product of matrix rings over division rings is semisimple. By the Artin-Wedderburn theorem, this exhausts all possibilities. In particular, $\mathbb{Z}$ is not semisimple: it is not semisimple as a module over itself because it has nontrivial submodules, such as $2 \mathbb{Z}$, which don't have complements.

In practice, a common source of examples of semisimple rings come from finite groups: if $k$ is a field and $G$ is a finite group of order not divisible by the characteristic of $k$, then the group ring $k[G]$ is semisimple by Maschke's theorem: equivalently, the category of $G$-representations over $k$ is semisimple.

See this blog post for a longer discussion with proofs.

Qiaochu Yuan
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  • @learnmore: this should be somewhere in Lam. A submodule $N$ of a module $M$ has a complement $C$ if $M$ can be written as a direct sum $M \cong N \oplus C$. – Qiaochu Yuan Jun 09 '15 at 05:42
  • @learnmore: there are lots of ways to see this. One is that $C$, if it exists, must be the quotient $M/N$, so a necessary condition is that $M$ must contain a submodule isomorphic to $C$. Here the quotient $\mathbb{Z}/2\mathbb{Z}$ is the cyclic group of order $2$, but $\mathbb{Z}$ is torsion-free so it doesn't contain a submodule isomorphic to this quotient. – Qiaochu Yuan Jun 09 '15 at 05:44
  • Another is that if $M \cong N \oplus C$ is a nontrivial direct sum decomposition, then $\text{End}(M)$ contains an idempotent corresponding to projection onto the first factor (and conversely idempotents give direct sum decompositions). But $\text{End}(\mathbb{Z}) \cong \mathbb{Z}$ has no nontrivial idempotents, hence $\mathbb{Z}$ has no nontrivial direct sum decompositions. A semisimple module with this property must be simple, but $\mathbb{Z}$ isn't. – Qiaochu Yuan Jun 09 '15 at 05:45
  • That's not too important; the important thing is that $\mathbb{Z}$ doesn't have any submodules isomorphic to $\mathbb{Z}/2\mathbb{Z}$. Do you see this? – Qiaochu Yuan Jun 09 '15 at 05:47
  • Well, the definition of a semisimple ring involves a certain condition on a module, so it's natural that to understand it you have to know some things about modules. If you aren't comfortable working with direct sums of modules, quotient modules, and submodules yet, then that's a more basic issue that you can resolve before going back to understanding semisimple rings. Any textbook on commutative algebra will teach this material; for example Atiyah-Macdonald. – Qiaochu Yuan Jun 09 '15 at 05:51
  • Forgive me for saying so, but if you aren't comfortable with "direct sum decomposition" and "projection onto the first factor," then you aren't comfortable with direct sums yet. A direct sum decomposition of a module $M$ is an isomorphism $M \cong N \oplus C$ for two other modules $N$ and $C$. "Projection onto the first factor" is the map $N \oplus C \to N$ given by $n \oplus c \mapsto n$. It extends to an idempotent endomorphism $N \oplus C \to N \oplus C$ given by $n \oplus c \mapsto n \oplus 0$. – Qiaochu Yuan Jun 09 '15 at 05:56
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    I don't know why you keep deleting all of your comments; it makes it look like I'm just talking to myself. – Qiaochu Yuan Jun 09 '15 at 06:05