1
$\begingroup$

I would like to know if there is any literature that discusses "transfinite socle series" of a ring module. Below is my attempt at defining the series.

Let $R$ be an associative unital ring and $M$ a unitary left $R$-module. Define the socle of $M$, denoted by $\text{soc}(M)$, to be the sum of all simple $R$-submodules of $M$. We create the transfinite socle filtration $$\text{Soc}(M):=\big(\text{soc}^\alpha(M):\alpha\text{ is an ordinal}\big)$$ of a given $R$-module $M$ as follows. First, define $\text{soc}^0(M):=0$ and $\text{soc}_0(M):=0$. If an ordinal number $\alpha>0$ has an immediate predecessor $\beta$, then $$\text{soc}_\alpha(M):=\text{soc}\big(M\big/\text{soc}^{\beta}(M)\big)\,,$$ and $\text{soc}^\beta(M)$ is the preimage $\pi_\alpha^{-1}\big(\text{soc}_\alpha(M)\big)$ of the socle factor $\text{soc}_\alpha(M)$ under the canonical projection $\pi_\alpha:M\twoheadrightarrow \big(M\big/\text{soc}^{\beta}(M)\big)$. If $\alpha$ is a limit ordinal, then $\text{soc}^\alpha(M)$ to be $\bigcup\limits_{\beta<\alpha}\,\text{soc}^\beta(M)$, and set $\text{soc}_\alpha(M):=0$. Finally, define $\overline{\text{soc}}(M)$ to be the union of all submodules $\text{soc}^\alpha(M)$ of $M$. (Clearly, there exists a smallest ordinal $\mu(M)$ such that $\overline{\text{soc}}(M)=\text{soc}^{\mu(M)}(M)$.)

I do not have a specific question, but I would like to learn about any information regarding transfinite socle series. For example, if $\omega$ is the least infinite ordinal, then does it hold that $\text{soc}_{\omega+1}(M)=0$ for any $R$-module $M$ (in particular, when $R$ is an algebra over a field $\mathbb{K}$, and maybe when $M$ is a countable-dimensional vector space over the same field $\mathbb{K}$)? If this is not true (for a general $R$, or for the case where $R$ is an algebra over a field), what are counterexamples? Any reference, comment, and knowledge about transfinite socle filtration will be greatly appreciated.

$\endgroup$
  • $\begingroup$ I found quite a lot of literature by Googling "transfinite socle series". $\endgroup$ – Jeremy Rickard Jun 13 '20 at 11:16
  • $\begingroup$ @JeremyRickard Not sure what you meant by a lot, but I only found two. (Originally, I called such a filtration simply an "infinite socle series," and Google search was not very helpful. Then, a user suggested that I changed the term to "transfinite socle series." Googling this term results in better references, but there are only two viable links.) $\endgroup$ – Batominovski Jun 13 '20 at 11:36
  • $\begingroup$ I'm not sure what you count as "viable", but most of the Google results I get in (at least) the first three pages of results seem relevant. $\endgroup$ – Jeremy Rickard Jun 13 '20 at 11:43
1
$\begingroup$

As I said in comments, there is a fair amount of literature to be found by Googling "infinite socle series".

More specifically, a module $M$ for which (in the notation of the question) $\overline{\text{soc}}(M)=M$ is called a "semi-artinian module", and a ring $R$ for which every module is semi-artinian (or equivalently for which $R$ is semi-artinian as a module for itself) is called a "semi-artinian ring". There is quite a lot of literature to be found on semi-artinian rings and modules.

For the specific question asked at the end of the question, the following example is adapted from Section 5 of

Nguyen V. Dung; Smith, Patrick F., On semi-artinian $V$-modules, J. Pure Appl. Algebra 82, No. 1, 27-37 (1992). ZBL0786.16002,

although that paper is about a rather specific class of modules, and so it would not surprise me if similar examples were known previously, or if there are simpler examples.

Let $\mathbb{K}$ be a field, let $R_n=\mathbb{K}[t]/(t^n)$ for $n\geq1$, and let $R$ be the subring of $\prod_{n\geq1}R_n$ consisting of elements $(r_n)_{n\geq1}$ such that, for some $a\in\mathbb{K}$, $r_n=a$ for all but finitely many $n$. Then $R$ is a countable dimensional $\mathbb{K}$-algebra, with $\bigoplus_{n\geq1}R_n$ as a codimension one ideal.

Let $M=R$, the regular $R$-module.

For $k\in\mathbb{N}$, $$\text{soc}^kM=\bigoplus_{n\geq1}\text{soc}_{R_n}^kR_n,$$ and so, since $\text{soc}_{R_n}^n R_n=R_n$, $$\text{soc}^\omega M=\bigoplus_{n\geq1}R_n,$$ and so $M/\text{soc}^\omega M=\text{soc}_{\omega+1}M$ is one-dimensional.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.