Talk:Semisimple module

Simple ring versus semisimple ring

the statement: "One should beware that despite the terminology, not all simple rings are semisimple." is really confusing. Isn't a simple ring by definition the sum of simple submodules?

• No, a simple ring R is a ring where the only 2-sided ideals are 0 and R. See Simple_ring 64.5.88.115 02:00, 27 March 2007 (UTC)

• It's confusing but it's true: we need some extra condition to conclude a simple ring is semisimple. Typically people assume the ring is Artinian$D_{i}$. This product is unique up to permutation of the factors.

References

Subdirect products

Jacobson's book Basic Algebra II defines semisimple ring as being a subdirect product of simple rings. This definition is likely not equivalent to the definition in here. What do you think of this? —Preceding unsigned comment added by 132.77.4.129 (talk) 21:27, 6 July 2008 (UTC)

Those are precisely the Jacobson semisimple rings. A ring is (wikipedia) semisimple if and only if it is an Artinian, Jacobson semisimple ring. I suggest on wikipedia always using the form "artinian semisimple ring". This article actually needs a lot of basic wikignome work. Thanks for bringing it back to my attention. JackSchmidt (talk) 03:01, 7 July 2008 (UTC)

A ring with jacobson radical equal to zero is equivalent to a ring which is a subdirect product of primitive rings (i.e. semiprimitive). However, what I talked about was a ring which is a subdirect product of simple rings. Is there some proof that subdirect product of simple rings is equivalent to subdirect product of primitive rings when the rings are not artinian? —Preceding unsigned comment added by 132.77.4.129 (talk) 12:44, 7 July 2008 (UTC)

You are right, this is not clear. For reference, the weird Jacobson definition is on p. 203, Ch 4.4. For commutative rings, primitive=field=simple, so subdirect of primitive is the same as subdirect of simple. Every simple ring is primitive, so every subdirect of simples is semiprimitive. What we want is a subdirectly irreducible primitive ring; it will be semiprimitive but not a subdirect of simples. If a primitive ring has a nonzero socle (as either a left or right module), then it is subdirectly irreducible.

I think a ring is a subdirect of simples if and only if its Brown-McCoy radical is 0, where the Brown-McCoy radical is the intersection of the maximal two-sided ideals. An example then is the endomorphism ring of an infinite dimensional vector space. It should be semiprimitive (indeed von Neumann regular), but the intersection of the maximal two-sided ideals should be the set of those elements whose images are finite dimensional, so it is not a subdirect product of simples. JackSchmidt (talk) 13:29, 7 July 2008 (UTC)

Equivalence of definitions

Could anybody give me a detailed clear proof of the statement "It can be shown that the complement of P is irreducible(i.e. simple) "(written in the hint of the proof of (3) implies (2)) ? — Preceding unsigned comment added by 182.156.226.126 (talk) 14:38, 4 April 2017 (UTC)

One of the definitions offered was this one: Every submodule of M is a direct summand: for every submodule N of M, there is a complement P such that $M=N\oplus P$. Consider an infinite dimensional vector space. From what I know, it cannot be expressed as a direct sum, only as a direct product. However, every subspace of it has a complement. —Preceding unsigned comment added by 132.77.4.129 (talk) 21:12, 11 July 2008 (UTC)

The equivalence is true. Let me address two different aspects of your concern.

• I believe there are modules which are direct products of simples, but which are not direct sums of simples (vector spaces actually are not such an example, but this requires the axiom of choice). Thus, had the equivalence required N or P to be simple, there could be problems.
• I think you are concerned with direct sums of infinitely many modules, not just two. The equivalence is only concerned with two submodules. For instance, take M to be a countably infinite dimensional vector space with basis e1, e2, ..., and take N to be the subspace spanned by e1. Since M is semisimple, there must be a direct complement P to N. One can take P to be the submodule generated by e2, e3, .... Then N+P contains the entire basis, so N+P=M. Also P contains no nonzero multiple of e1, so N intersect P = 0. Hence M is the direct sum of N and P.

Does this answer your concern? JackSchmidt (talk) 21:36, 11 July 2008 (UTC)

Oh I misread your concern. Yes, every vector space is a direct sum of simple modules. Take a basis of the vector space (using the axiom of choice), and then the cyclic submodule generated by a single basis element is simple. The vector space is the module direct sum of these cyclic submodules. This is a very special thing about vector spaces (or fields).

For instance, the direct sum of infinite cyclic groups is always a free abelian group, but the (cartesian or unrestricted) direct product of infinitely many infinite cyclic groups is not free. In other words, the Z-module formed as a countable direct product of copies of Z is not a direct sum of copies of Z. For an example with simple modules, take Z to be the ring, and the simple modules to be Z/pZ, varying over the primes p, one module for each prime p. The direct sum of simple Z-modules is torsion, but the direct product of Z/pZ over all p is not torsion. JackSchmidt (talk) 21:40, 11 July 2008 (UTC)