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I'm considering an algebra to be a ring which is also a vector space over some field $F$, and the algebra $A$ is said to be semisimple if it is semisimple as a ring, i.e., $A$ can be written as a direct sum of simple $A$-submodules (minimal left ideals). My questions are about the relationship between semisimple algebras and semisimple rings.

If $A$ is a non-unitary algebra, does being semisimple mean that we can write it as a direct sum of simple subalgebras, i.e., are the minimal left ideals of an algebra always $F$-subspaces? I know this is the case when $A$ is unitary, but is this true in general? If not, are there any interesting counter-examples? Most of the sources I could find already define algebras as unitary rings, so I was wondering if there are interesting examples of non-unitary algebras with minimal left ideals that are not subalgebras.

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    $\begingroup$ For non unital I think you need to be slightly careful on how you define Semisimple. I would ask it to be a direct sum of simples as both a left and right module. In this case minimal left ideals are F-subspaces. The reason why is your ring will be forced to have local.units and then you can make the usual arguments work $\endgroup$
    – Benjamin Steinberg
    Commented Jan 29 at 19:39
  • $\begingroup$ crossposted from math.se $\endgroup$
    – rschwieb
    Commented Jan 29 at 20:12
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    $\begingroup$ @rschwieb Sorry for the crossposting. Deleted the original math.se post. $\endgroup$
    – Henrique Assumpção
    Commented Jan 29 at 21:14
  • $\begingroup$ Should a (left) semisimple nonunital ring be a finite direct sum of minimal left ideals? (This is automatic in the unital case.) $\endgroup$
    – Jeremy Rickard
    Commented Jan 30 at 8:35

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For a not necessarily unital ring $R$, a left $R$-module $S$ is simple if $RS\neq 0$ and $S$ has no proper submodule. A simple right module is defined dually. For a not necessarily unital ring the following are equivalent:

  1. $R$ is both semisimple as a left and right $R$-module, meaning it can be written as a possibly infinite direct sum of left and also of right $R$-modules.
  2. $R$ is semiprime (it has no nilpotent left (or equivalently right) ideals) and can be written as a sum of minimal left ideals.
  3. $R$ has local units (is a directed union of unital subalgebras under not necessarily unital maps) and can be written as a sum of minimal left ideals.
  4. $R$ is a directed union of unital semisimple subalgebras (by not necessarily identity preserving maps)

Using condition 4, if $L$ is a minimal left ideal and $e$ is an idempotent with the generator of $L$ in $eRe$, then $eL$ is a simple $eLe$ module and so closed under scalar multiplication. From this it easily follows that $L$ is closed under scalar multiplication.

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In this answer, I was assuming the naive definition of a simple module $M$ for a nonunital ring $R$ as one with no proper nonzero submodules. It seems to be common to also insist that $RM\neq0$, in which case the ring I use is not semisimple.

Let $F=\mathbb{F}_{p^2}$, the field with $p^2$ elements, for some prime $p$, and let $A=F$ with zero multiplication.

Then the left ideals of $A$ are just the additive subgroups, and so the minimal left ideals are the one-dimensional $\mathbb{F}_p$-subspaces, which are not $F$-subspaces.

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    $\begingroup$ For nonunital rings R the customary definition of a simple module is a module S such that RS is not 0 and it has no proper submodule. Semisimple usually means that it can be written a sum possible infinite of simple left modules and also as a sum of simple right modules. This forces the ring to have no nilpotnent left or right ideals and every minimal left ideal is actually contained in a unital semsimple sub ring and hence is closed under scalar multiplication. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 30 at 22:25
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    $\begingroup$ @BenjaminSteinberg I didn't know that. I guess it makes sense, although personally I'd have chosen a different name than "simple", as it seems a bit confusing to use "simple module" for anything other than "simple object in the module category". I checked some of the books on my shelf, and your definition does seem quite common. But not universal. Notably, in Jacobson's Lectures in abstract algebra it is set as an exercise to prove that if $M$ is a simple left $R$-module then either $RM=M$ or $RM=0$ with $M$ cyclic of prime order. But I didn't find any very modern references. $\endgroup$
    – Jeremy Rickard
    Commented Jan 31 at 8:27
  • $\begingroup$ @BenjaminSteinberg Anyway, I'll edit my answer to make it clear that I wasn't assuming your definition. $\endgroup$
    – Jeremy Rickard
    Commented Jan 31 at 8:28
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    $\begingroup$ It is quite common in the works on nonunital rings not to use the full module category since this amounts to looking at modules over the unitilization. Quillen and others proposed things like firm modules, which if I recall correctly require the natural map $R\otimes_R M\to M$ to be an isomorphism. For rings with local units, this boils down to $RM=M$ and hence these modules will be the simple modules. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 31 at 11:39

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