Is it possible that the socle of a ring (with identity) is cyclic as a left ideal but not finitely generated as a right ideal !?
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2$\begingroup$ Does the matrix ring $\begin{pmatrix}\mathbb Q&\mathbb R\\0&\mathbb R\end{pmatrix}$ do? $\endgroup$– Mariano Suárez-ÁlvarezCommented May 23, 2014 at 23:35
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1$\begingroup$ What do you mean by the socle of a ring? The left socle (sum of all simple left ideals) and right socle (sum of all simple right ideals) are different in general, which makes a difference since your question is not left/right symmetric. Mariano's example has right socle cyclic as a left ideal and two-generator as a right ideal, and left socle infinitely generated as a left ideal and cyclic as a right ideal. (So strictly speaking doesn't answer the question, but the opposite ring does if you mean right socle, but not if you mean left socle.) $\endgroup$– Jeremy RickardCommented May 24, 2014 at 11:07
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The example given by Mariano in comments (or rather, its opposite ring) is a counterexample if "socle" means "right socle". I think the following example works for both the right and left socle.
Let $k$ be a field with an endomorphism $\alpha$ such that $k$ is an infinite degree field extension of $\alpha(k)$. For example, let $k=\mathbb{Q}(t_1,t_2,\dots)$ with $\alpha(t_i)=t_{i+1}$.
Let $V$ be the $k$-bimodule where $V=k$ as an abelian group, with $k$ acting on the left by multiplication and on the right via $\alpha$, so that as a left $k$-module $V$ is a one-dimensional vector space and as a right $k$-module is an infinite-dimensional vector space.
Let $A$ be the trivial extension ring $k[V]$. I.e., $A=k\oplus k$ as an abelian group, with multiplication $(u,v)(x,y)=(ux,uy+x\alpha(v))$.
Then the left and right socle are both equal to $V=0\oplus k$, which is cyclic as a left module but infinitely generated as a right module.
answered May 27, 2014 at 16:51
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$\begingroup$ I had a stock example in mind, but I think it might be isomorphic to this one :) Tacking it on for a second view: form the twisted polynomial ring $k[x;\alpha]$ where $xa:=\alpha(a)x$ for all $a\in k$. Then the ring $R=k[x,\alpha]/(x^2)$ has $soc(R_R)=soc(_RR)=(x)$, and it is simple as a left module but infinitely generated as a right module. $\endgroup$– rschwiebCommented May 27, 2014 at 17:53
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$\begingroup$ @Jeremy Rickard Is your ring a clean ring? $\endgroup$ Commented Mar 16, 2017 at 16:27
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1$\begingroup$ @karparvar yes, obviously: my example is a local ring. $\endgroup$– rschwiebCommented Mar 16, 2017 at 17:19
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1$\begingroup$ @karparvar Jeremy's is also obviously local, so also clean. $\endgroup$– rschwiebCommented Mar 16, 2017 at 17:43