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Finite dimensional simple algebras are semisimple.

I've found two different answers and got a bit confused. Any clarification would be of great help!

A simple algebra that is not semisimple

https://en.wikipedia.org/wiki/Semisimple_algebra#:~:text=A%20finite%2Ddimensional%20algebra%20is,are%20A%20and%20%7B0%7D.

Thank you

scsnm
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    What's the question? A simple finite dimensional unital algebra over a field is semisimple (indeed a simple Artinian ring is semisimple). An infinite-dimensional simple algebra need not be semisimple. – Angina Seng Aug 31 '20 at 15:05
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    Yes! Is it because of the Wedderburn's theorem? $A$ is simple artinian iff $A \cong M_n(D)$ for some division ring. And this means $A$ is semisimple by Wedderburn's theorem? – scsnm Aug 31 '20 at 15:12
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    What is causing confusion? Finite dimensional simple -> Artinian simple -> semisimple. Qiaochu's example is an infinite dimensional simple ring. – rschwieb Aug 31 '20 at 19:29

1 Answers1

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It's an unfortunate fact about the terminology that for rings (unlike for, say, modules or Lie algebras) "simple" doesn't imply "semisimple," so let's be clear:

  • A semisimple ring is a ring which is semisimple as a left module over itself. Other definitions are possible, for example "Artinian and trivial Jacobson radical," or "every short exact sequence of left modules splits," and then one has to do some work to show that they're equivalent.
  • A simple ring is a ring with no nontrivial two-sided ideal.
  • Every division ring is simple (exercise). If $R$ is a simple ring, then so is $M_n(R)$ (exercise). So any matrix algebra over a division ring is simple.
  • Every division ring is semisimple (exercise). If $R$ is a semisimple ring, then so is $M_n(R)$ (exercise). If $R$ and $S$ are semisimple, then so is $R \times S$ (exercise). So any finite product of matrix algebras over division rings is semisimple.
  • The Artin-Wedderburn theorem asserts the converse: every semisimple ring has the above form.
  • Every finite-dimensional algebra over a field $k$ is Artinian, hence is semisimple iff it has trivial Jacobson radical. The Jacobson radical is a two-sided ideal, so any simple ring has trivial Jacobson radical, and so any finite-dimensional simple $k$-algebra is semisimple, and more generally a simple ring is semisimple iff it's Artinian iff (by Artin-Wedderburn) it's a matrix algebra over a division ring.
  • The Weyl algebra, as stated in the first answer you linked to, is a simple ring which is not Artinian, so it's not semisimple.

For a proof that "semisimple" as defined above is equivalent to "Artinian and has trivial Jacobson radical" see this blog post.

Qiaochu Yuan
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  • Here's an old math.SE answer where I show that if $R$ is simple then so is $M_n(R)$ and also show that the Weyl algebra is simple: https://math.stackexchange.com/questions/2540758/an-example-of-a-non-commutative-ring-with-multiplicative-identity-1-in-which-the/2540826#2540826 – Qiaochu Yuan Sep 01 '20 at 03:13