# socles of semiperfect rings

For readers' benefit, a few definitions for a ring $R$.

The left (right) socle of $R$ is the sum of all minimal left (right) ideals of $R$. It may happen that it is zero if no minimals exist.

A ring is semiperfect if all finitely generated modules have projective covers.

Is there a semiperfect ring with zero left socle and nonzero right socle?

Someone asked me this recently, and nothing sprang to mind either way. In any case, I'm interested in a construction method that is amenable for creating imbalanced socles like this.

If you happen to know the answer when semiperfect is strengthened to be 'some side perfect' or 'semiprimary' or 'some side Artinian', then please include it as a comment. (Of course, a ring will have a nonzero socle on a side on which it is Artinian.)

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A right perfect ring $R$ will have nonzero left socle: one of the equivalent characterizations is that $R$ has DCC on principal left ideals, hence has a minimal left ideal. So any such example couldn't be right perfect. That rules out examples that are right artinian (hence semiperfect with nonzero right socle) but with zero left socle, since one-sided Artinian rings are two-sided perfect. –  Manny Reyes Mar 22 '13 at 21:08
Thanks @MannyReyes :) I hadn't filed in my head that one sided perfect rings had nonzero socles, but now it is obvious... –  rschwieb Mar 22 '13 at 21:13

Let $k$ be a field, and let $S$ be the ring of (say) column-finite $\mathbb{N} \times \mathbb{N}$-matrices over $k$. Let $E_{ij}$ denote the usual "matrix units." Consider $J = \operatorname{Span}\{E_{ij} : i > j\}$ be the set of lower-triangular matrices in $S$ that have only finitely many nonzero entries, all above the diagonal.
Let $R = k \cdot 1 + J \subseteq S$. Then $R$ is a local ring with radical $J$, and it's shown to be left perfect in the reference above. As I mentioned in my comment above, this means that $R$ is semiperfect and has nonzero right socle.
I claim that $R$ has zero left socle. Certainly any minimal left ideal will lie inside the unique maximal left ideal $J$. It suffices to show that every $a \in J \setminus \{0\}$ has left annihilator $ann_l(a) \neq J$. For then $Ra \cong R/ann_l(a) \not \cong R/J = k$. Indeed, if $a \in J$ then $a = \sum c_{ij} E_{ij}$ for some scalars $c_{ij}$ that are almost all zero. Let $r$ be maximal such that there exists $c_{rj} \neq 0$. Then there exist finitely many $j_1 < \cdots < j_r$ such that $c_{r j_p} \neq 0$. It follows that $E_{r+1,r} \in J$ with $$E_{r+1,r}a = E_{r+1,r} \sum c_{ij} E_{ij} = \sum c_{ij} E_{r+1,r} E_{ij} = \sum_p c_{i j_p} E_{r+1,j_p} \neq 0.$$ In particular, $E_{r+1,r} \in J \setminus ann_l(a)$ as desired.