[QUESTION] If $R$ be a ring, what is the meaning of a non-degenerate $R$-module?
In a previous question post at (What is a non-degenerate module?), some experts said that if $M$ is a $R$-module such that $M=RM:=\{\displaystyle{\sum_{i=1}^n}r_im_i\mid r_i\in R,m_i\in M,n\in\mathbb{Z}_{>0}\}$, then $M$ is called non-degenerate $R$-module.
However, I read the lecture note "REPRESENTATION OF p-ADIC GROUPS" by professor Joseph BERNSTEIN (http://www.math.harvard.edu/~gaitsgde/Jerusalem_2010/GradStudentSeminar/p-adic.pdf), "non-degenerate module" first defined in DEFINITION 4 (2) on page 9, where $\mathcal{H}$ is an algebra and $M$ is a $\mathcal{H}$-module. Then $M$ is called a non-degenerate $\mathcal{H}$-module if $\mathcal{H}M=M$. However it seems that $\mathcal{H}M$ is defined as a set $\{hm\mid h\in\mathcal{H},m\in M\}$ instead of a submodule generated. There are at least two evidences in his note. One is in the proof of LEMMA 4 on page 13, and there is a sentence: "each element $M$ has the form $hm$"; the other is in the proof of LEMMA 7 on page 19, there is another sentence: "there exists an $h\in\mathcal{H}$ such that $hw_1=w_2$".
Hence, I shall be grateful if any expert here who read this note before gives any comment on this question. Thanks!
There are certainly omissions in the notes, but $\mathcal{H}M$ is not defined as a set; it's a module. It is, however, equal to the set $\{hm\}$ in this context.
$\mathcal{H}$ is an idempotented algebra, which means that, for any finite collection $h_i\in\mathcal{H}$, there is an idempotent $e\in\mathcal{H}$ with $eh_i = h_i e = h_i$ for all $i$. This last equality doesn't show up in these notes, which is likely to be an error.
Anyway, if $M$ is non-degenerate, and $m\in M$, then we have $m = \sum_i h_i m_i$. But, by the definition, there is an idempotent $e\in\mathcal{H}$ such that $eh_i = h_i$ for all $i$. Then $em = m$, so $m$ is of the required form.