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As far as I understand, a semiprimitive ring can be fully 'explored' by its simple modules, in the sense that a semiprimitive ring is the subdirect product of its simple modules (for brevity, I'll use 'module' to mean 'left module' throughout.) This is great, and it clearly tells us a lot about the ring. However, we can only claim to fully understand a ring when we know ALL of its modules, not just its simple ones. So can we use this subdirect product decomposition to help us characterize all of the modules for the ring, beyond the ones which are products of simple modules? (Note that a semiprimitive ring is not necessarily semisimple.)

EDIT: It seems I made a mistake in the original post, it's not true that a semiprimitive ring is a subdirect product of its simple modules. But it IS always the semidirect product of SOME list of primitive rings. (Is a minimal such list of primitive rings uniquely defined?) So, the question is to what extent understanding the representation theory of these primitive rings helps you with understanding the representation theory of the original ring.

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You want the direct sum of the simple modules for your faithful semisimple modules, not the direct product. (Think about what happens when $R=\mathbb{Z}$.) –  Robin Chapman Jul 3 '10 at 18:18
    
In the case of Banach algebras, and if we replace "module" by "Banach module" in the most obvious way, then the answer is no: when $R$ is a commutative Banach algebra then the map $f$ is just the Gelfand transform, and the problem is somehow that $R$ (or even $f(R)$) possesses many more modules than the direct product of simples. –  Yemon Choi Jul 3 '10 at 18:18
    
just to add to my last comment: my intuition (which is gained from the Banach setting, and hence could well be in error in this setting) is that $f(R)$ might not be a very big subalgebra of $\prod_i R_i$, and so knowing the module theory of the big guy is insufficient for understanding the module theory of the subalgebra –  Yemon Choi Jul 3 '10 at 18:21
    
I still don't understand your revised version of the question. The algebra of continuous complex-valued functions on $[0,1]$ is semiprimitive. What are the primitive rings you're thinking of in this case? That's before we get onto examples such as the disc algebra (algebra of all continuous functions on closed unit disc which are analytic in the interior). –  Yemon Choi Jul 3 '10 at 19:38
    
In that case, an example of a list of primitive rings which together give rise to a subdirect product for the function ring are the values $V_x$ of the functions at a given point $x$ in [0,1], which are one-dimensional vector spaces. So we have an injective ring homomorphism $f: ([0,1],C) \to \Pi_x V_x$. Clearly in this case $([0,1],C)$ is indeed not a very big subalgebra, as you say. –  Jamie Vicary Jul 4 '10 at 1:14

2 Answers 2

You write

these simple modules can be obtained by taking a quotient of the ring by maximal left ideals, and so these simple modules are themselves also rings

but in general the quotient of a ring by a left ideal isn't a ring (it is when the ideal is two-sided but that won't usually be the case when the ring is noncommutative).

In the commutative case, then yes, a semiprimitive $R$ does embed into $S=\prod_j R_j=\prod_j R/I_j$ where $I_j$ runs over the maximal ideals of $R$. But alas, I cannot see a concrete question here. As a simple example, consider $R=\mathbb{Z}$. Then $S=\prod_p(\mathbb{Z}/p\mathbb{Z})$ but I cannot see how looking at $S$ helps with understanding the $\mathbb{Z}$-modules $\mathbb{Z}$ or $\mathbb{Q}$ or the plethora of modules in between these two.

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Hi Robin, thanks for your comment. The concrete question is simply whether knowledge of the subdirect decomposition of a semiprimitive ring can be helpful in understanding the full representation theory of the ring. This question has been more than a little prompted by the Wikipedia article --- not that I give Wikipedia articles more weight than they are due! --- which states "a semiprimitive ring is a type of ring ... where simple modules still provide enough information about the ring". –  Jamie Vicary Jul 4 '10 at 1:24

No! The Weyl algebra is simple. The problem of classifying the finite dimensional representations is wild. This was discussed in Is there a machinery describing all the irreducible representations ?

In case this is too succinct. The Weyl algebra is the algebra of linear differential operators with polynomial coefficients. It is generated by $x$ and $D$ with defining relation $Dx-xD=1$. This acts on polynomials in $x$ with $D=d/dx$. This is a faithful simple representation.

The problem of finding finite dimensional representations is the problem of solving linear differential equations.

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Thanks very much for your comment! I understand that the representation theory of simple algebras can be wild. My question is to what extent the representation theory of a general semiprimitive algebra is helped by the knowledge of its subdirect product in terms of primitive rings. If an algebra is itself primitive, then it makes sense that this decomposition does not help in understanding its representation theory. –  Jamie Vicary Jul 4 '10 at 1:19
    
What do you mean by 'finite dimensional'? Do you mean 'finitely generated over $\mathbb C[x]$'? (The Weyl algebra doesn't have any representations that are finite dimensional over $\mathbb C$ - if it did, you could take trace of both sides of the matrix equation $Dx-xD = 1$ and get $0 = n$.) –  Peter Samuelson Jun 16 '13 at 19:43

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