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Let $K$ be a field and $A$ be a $K$-algebra.

I know, if $A$ is artinain algebra, then by Krull-Schmidt Theorem $A$ , as a left regular module, can be written as a direct sum of indecomposable $A$-modules, that is

$A=\oplus_{i=1}^n S_i$ where each $S_i$ is indecomposable $A$-module

Moreover, each $S_i$ contains only one maximal submodule, which is given by $J_i= J(A)S_i$, and every simple $A$-module is isomorphic to some $A/J_i$.

My question is that, can you, please, tell me an example of a non simisimple algebra, or a ring, such that it has a simple module which does not occur in the regular module.

By occur I mean it has to be isomorphic to a simple submodule of a regular module

Math137
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    Any simple (left) module $S$ for any ring $R$ is a quotient of the regular module, if that's what you mean by "occurs in", since for any $s\neq0$ in $S$, $r\mapsto rs$ is a non-zero homomorphism $R\to S$. –  Jul 29 '13 at 21:29
  • @JeremyRickard, thank you for your reply. By occur I mean it has to be isomorphic to a simple submodule of a regular module. I have also edited my question. – Math137 Jul 29 '13 at 21:43
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    The algebra of upper triangular $2\times 2$ matrices (over a field) has two one-dimensional simple modules, only one of which occurs as a submodule of the regular module. –  Jul 29 '13 at 21:44
  • @JeremyRickard, could you please explain that briefly to me. – Math137 Jul 29 '13 at 21:55

4 Answers4

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For an Artinian example, let $A$ be the algebra of upper triangular $2\times 2$ matrices over a field $K$, let $J$ be the left ideal consisting of all matrices with bottom row zero, and let $S$ be the simple left $A$-module $A/J$. Then it's easy to check that $Ja\neq0$ for every non-zero element $a$ of $A$, but $Js=0$ for every $s\in S$, so $S$ can't be isomorphic to a submodule of the regular module.

Jeremy Rickard
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Consider the algebra $K[T]$. The simple $K[T]$-modules are of the form $K[T]/P(T)$ for some irreducible polynomial $P$. These do not occur as submodules of $K[T]$, since every such submodule contains a free module, and hence is infinite dimensional over $K$.

Kevin Ventullo
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  • Dear Kevin, This is a good answer, except that the OP has the technical requirement of wanting to consider algebras over a field, rather than more general rings. Regards, – Matt E Jul 29 '13 at 23:06
  • Dear Matt, thank you for the comment. I have edited my answer to reflect it. – Kevin Ventullo Jul 29 '13 at 23:38
  • @Kevin Thank you very much for the answer. – Math137 Jul 29 '13 at 23:50
  • Submodules of $K[T]$ (or of any commutative ring for that matter) are the same as ideals, and here ideals are principal ideals, hence isomorphic to $K[T]$. Regards, Todd Trimble. – user43208 Jul 30 '13 at 01:13
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The simple modules for $F[x]$ are $F[x]/(p)$ where $p=p(x)$ is irreducible. These do not occur as submodules of $F[x]$.

Hill
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If $S$ is a simple module over a ring $A$, and $x$ is a non-zero element of $S$, then the annihilator of $x$ is a maximal left ideal $\mathfrak m$ of $A$.

Thus is $S$ embeds into $A$, then $A$ contains a non-zero element annihilated by $\mathfrak m$.

If $A$ is not a simple module over itself (i.e. is not a division algebra), then $\mathfrak m$ will contain non-zero elements, and so if $A$ is an integral domain (i.e. does not contain non-zero zero-divisors), then it will not contain any simple submodules.

The example $K[X]$ of a polyomial algebra that appears in the other answers is a special case of this general observation, since $K[X]$ is an integral domain.

Matt E
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