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so my doubt is that i am studying wedderburn artin theory and it gives structure of simple artinian rings, but if a ring is simple, it has no nonzero proper 2-sided ideals so it satisfies DCC on ideals trivially, so must be artinian, so if every simple ring is artinian, means every simple ring is of the form a matrix ring over a division ring, in T.Y Lam he studies simple left artinian first and then says it is same as simple right artinian and so we can talk about simple artinian straightaway only.

so is there a simple ring which is not artinian?

Bhaskar Vashishth
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2 Answers2

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The Weyl algebra $A_1 := k\{x, y: xy - yx = 1\}$ is simple when char $k = 0$ but is not Artinian: the left ideals $A_1x^i$ form an infinite decreasing sequence of left ideals which is never constant.

Christopher
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Ascending and descending chain conditions for noncommutative rings are defined in terms of left ideals and right ideals, not just two-sided ideals. This makes a world of difference in what is possible.

Take $V$ to be a countable dimensional $k$ vector space over a field $k$, and let $R$ be its ring of linear transformations. $R$ is known to have exactly one nontrivial ideal $I$ made up of transformations whose images have finite rank.

The ring $S=R/I$ is simple, von Neumann regular, but not Artinian on either side and not Noetherian on either side. If $S$ were Artinian, it would have finite $k$ dimension, but clearly it does not.

rschwieb
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  • And why is it not Noetherian on either side? – karparvar Jul 26 '18 at 05:30
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    @karparvar In a VNR. ring, right Noetherian and right artinian are equivalent. – rschwieb Jul 26 '18 at 10:24
  • @rschwieb Is this example written somewhere in detail? Thanks! – 1123581321 Jun 17 '19 at 11:01
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    @giannispapav I'm not aware of where it is written in more detail. These particular facts are usually given as exercises. – rschwieb Jun 17 '19 at 13:13
  • Sorry for commenting several years later, but why $S$ being Artinian would imply finite dimension over $k$? A simple Artinian algebra is a matrix ring over a field extension of $k$. Still, that field extension might be infinite. – user40276 Aug 14 '22 at 00:10
  • Thanks for the reply. Still, I think we are using different definitions. A transcendental extension $k(t)/k$ is a simple (left and right) Artinian (as a module over itself) $k$-algebra and $k(t) = M_1 (k(t))$. Generally, any simple left Artinian $k$-algebra (in my definition) is of the form $M_n (F)$ for $F$ some possibly non-finite and non-commutative field extension of the field $k$. – user40276 Aug 14 '22 at 19:35
  • @user40276 Yes, what I said in the comment is wrong: perhaps even in 2014 I did not find an adequate reason to explain why the quotient is not Artinian... – rschwieb Aug 14 '22 at 19:58