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In Maclachlan and Reid's The Arithmetic of Hyperbolic 3-Manifolds, when proving that quaternion algebras are simple, they make use of the fact that $M_2(K)$, where $K$ is an algebraically closed field, is simple.

Could someone point me to a reference or proof of this?

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$M_n(k)$ is simple for any field $k$, with no algebraic closure hypothesis. There are various ways to prove this. One is to prove the following more general result: if $R$ is a ring, then the two-sided ideals of $M_n(R)$ are $M_n(I)$ where $I$ is a two-sided ideal of $R$. See this math.SE question for more details.

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Qiaochu Yuan
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  • Is it true that if $k$ is a field, then every simple $k$-algebra is isomorphic to $M_n(k)$ for some $n \in \mathbb{Z}_{\geq 1}$? Perhaps with some finite-dimensionality assumptions or something like that? – goblin GONE Jan 03 '16 at 08:49
  • I know about the Artin-Wedderburn theorem but its precise meaning is a little confusing to me... – goblin GONE Jan 03 '16 at 08:54
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    @goblin: the Artin-Wedderburn implies that every Artinian simple $k$-algebra is isomorphic to $M_n(D)$ where $D$ is a division algebra over $k$. In the non-Artinian case there are other possibilities, like the Weyl algebra. – Qiaochu Yuan Jan 03 '16 at 18:09
  • Being semisimple is meant to generalize being simple, right? Why would that be confusing? – goblin GONE Jan 11 '16 at 05:07
  • @goblin: ah, sorry, I meant "simple" does not imply "semisimple." – Qiaochu Yuan Jan 11 '16 at 16:45
  • Wikipedia writes that "an algebra $A$ is called simple if it has no proper ideals and ${ab \mid a,b∈A}≠0$. As the terminology suggests, simple algebras are semisimple." If I catch your drift correctly, you're suggesting that the correct definition of "simple" should really be "has no proper ideals" and that, with this modification, simple no longer implies semisimple. Is that right? – goblin GONE Jan 11 '16 at 20:24
  • @goblin: that second condition is vacuous if $A$ is unital (and for me algebras and rings are always unital). The first condition is ambiguous; it should be "no proper two-sided ideals." Whether this implies semisimple depends on your definition of semisimple. My preferred definition is that a ring is semisimple if its category of modules is semisimple. Some authors define semisimple to mean semiprimitive (vanishing Jacobson radical). – Qiaochu Yuan Jan 11 '16 at 21:37
  • Hmm okay. I'm still not 100% clear on what your preferred definitions are. According to nLab, a semisimple category is defined by the condition that each object is a direct sum of finitely-many simple objects, and that all such direct sums exist. If so, then I think you mean: "a ring is semisimple iff its category of finitely-generated modules is semisimple." Is that right? On a different topic: do you know of a book that, in your opinion, deals with these kinds of concepts "properly"? i.e. a book in which the definitions and proofs feel as natural and modern as possible. – goblin GONE Jan 12 '16 at 01:42
  • @goblin: no, for me a category (say abelian and cocomplete) is semisimple if each short exact sequence splits. For categories of modules this is equivalent to every module being a direct sum of possibly infinitely many simple objects. I don't know a good reference. – Qiaochu Yuan Jan 12 '16 at 16:00