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Yes, this is a quite basic answer, but I have to admit to be absolutely confused about this notion.

Searching on the web, I managed to found two possible definition of trivial modules, referring actually to two different mathematical objects.

The first one is just the singleton set with the only possible module structure, also called the zero module.

The second one, that is the definition that is leading me to be confused and a bit stuck after so much hours of studying, is the following:

Let $A$ be a ring, $M$ an abelian group. $M$ is called a trivial module if it is a module endowed with the trivial action.

But...what exactly is a trivial action? Yes, of course the first things that I think is the trivial $ax=x$ for each $x \in M$, but there is something wrong with it, because directly from the axioms of modules I have:

for each $x \in M, (1+1)x = x$ (because the action is trivial), and $(1+1)x = 1x+1x=x+x$ that implies $x = 0$, i.e. $M$ is the group $0$.

Please, could you help me in understanding what am I missing? Thank you very much!!!

Ps: this question is related to this one (From $G$-mod to $\mathbb{Z}G$-mod and a related question.) where I believed to have understood this definition :)

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    "Let A be a ring, M an abelian group. M is called a trivial module if it is a module endowed with the trivial action." Where exactly is this said like this? – quid Jun 08 '15 at 13:38
  • Yes, it is no literally written, but my confusion raised from wikipedia (https://en.wikipedia.org/wiki/Trivial_module). I didn't understand clearly the explanation about trivial action because it seems to refers to some facts from representation theory that I don't know yet (unfortunately...). –  Jun 08 '15 at 13:42
  • The trivial action of a group is as identity (not $0$). I do not think it is common to consider this for rings instead of groups. – quid Jun 08 '15 at 13:48

2 Answers2

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Trivial action of ring $A$ on Abelian group $M$:

$am=0$ for all $a\in A$ and $m\in M$.

If $A$ has an identity element (and the axiom $1x=x$ is posed), then this forces $m=1m=0$ for all $m$, hence $M=\{0\}$.

Berci
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    This is based on the trivial ring homomorphism $A\to\operatorname{End}(M)$ – Hagen von Eitzen Jun 08 '15 at 09:59
  • Can't we just say it acts by the zero endomorphism and avoid the ridiculous confusion the word 'trivial' generates? Plus, the original point of the OP's question is to understand if the trivial $G$-module $A$ (i.e. the action of $G$ fixing all elements of $A$) corresponds to the trivial $\mathbb{Z} G$-module $A$ (i.e. the action of the group ring $\mathbb{Z} G$ which sends all elements of $A$ to the zero element) which does not seem to be the case. He wants to understand how this discrepancy is compatible with the identification of categories of $G$-modules and $\mathbb{Z}G$-modules. – guest Jun 08 '15 at 10:49
  • @guest your comment is a bit confusing to me. But there is no such discrepancy, because "trivial $\mathbb{Z}G$-module" there just does not mean that there is only the $0$ element in that modulo, but rather that the $\mathbb{Z}G$-action is trivial. This is evidence by the fact that on linked thread it says the trivial $\mathbb{Z}G$-module $\mathbb{Z}$. – quid Jun 08 '15 at 12:34
  • Dear users, first of all thanks to everyone for your interest and helpful suggestions! Berci, your definition seems to work and be reasonable, but it implies that I must have misunderstood what is happening in the post I linked at the end of my OP, that is the real reason for my question. As guest correctly says, I want to understand the equivalence between the category $G$-mod and $\mathbb{Z}G$-mod following the explanation that user @DietrichBurde kindly gave me. I am thinking exactly what written by guest, in other words: –  Jun 08 '15 at 13:21
  • Let $G$ be a group and $\mathbb{Z}$ a trivial $G$-mod (so we have the action $gn=n$ for each $n \in \mathbb{Z}$, and there is no any particular problem). Then, considering the equivalence between $G$-mod and $\mathbb{Z}G$-mod explained in the link at the end of my OP, we can consider the same $\mathbb{Z}$ as a $\mathbb{Z}G$-module (where now $\mathbb{Z}G$ is a ring, so we use the "standard" definition of a module over a ring). Now: why is $\mathbb{Z}$ a trivial $\mathbb{Z}G$-module? It seems not to have the action that @Berci defined before. –  Jun 08 '15 at 13:26
  • Furthermore: let's suppose that $\mathbb{Z}$ is actually a trivial $\mathbb{Z}G$-module. But $\mathbb{Z}G$ has the unit $1g_{id_G}$, and so our previous argumentation should implies that $\mathbb{Z}=0$...what am I missing? :-)

    Thank you a lot for everything, and sorry for my long and probably tedious comments. Cheers

     –  Jun 08 '15 at 13:31
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    @user233650 When we say that $M$ is a trivial $\mathbb{Z}G$ module, it means that the action of $G$ is trivial, that is $gm=m$ for all $g\in G$ and all $m\in M$. It does not mean that $M$ is trivial as a $R$-module for $R=\mathbb{Z}G$ (which is $M=0$). This definition of trivial is NOT categorical. – Roland Jun 08 '15 at 14:27
  • @user113969 ...ohhhh!!!! Really?!?!? It explains so much things...yes, of course I need a bit to think on it. Thank you immensely for having pointed out this absolutely crucial detail!!! –  Jun 08 '15 at 17:00
  • Yes, I think you are right. Thanks to everyone! :-) –  Jun 08 '15 at 18:17
  • @quid as user233650 says, the "trivial $\mathbb{Z}G$-module" is NOT the "trivial R-module for R=$\mathbb{Z}G$", which as he/she emphasizes does not correspond in the equivalence of the two categories. It is not an action that fixes all points, either; only points of $G$ do that, the $\mathbb{Z}$ factor acts as it should to make the correspondence categorical. Plus, $\mathbb{Z}G$ is a ring, and "trivial action" in the sense of zero action of a ring also makes sense, confusing the OP. This chaos is what I mean by discrepancy and this is where my complaint lies about the overuse of 'trivial'. – guest Jun 08 '15 at 20:46
  • @quid Correction: it was user113969 who made this remark, also implicit in your comment. – guest Jun 08 '15 at 20:55
  • @guest thanks for the explanation. I think we basically meant the same thing all along, but approached the details differently. – quid Jun 08 '15 at 21:56
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You seem to be just confusing different notions of module.

  • $G$-module would be an abelian group $M$ with the action of a group $G$ compatible with addition. (This can well be the trivial action.)

  • $R$-module, or just module, where $M$ is an abelian group and one has scalar multiplication with the elements from the ring $R$ similarly to case of vector-spaces. (This can also be understood as an instance of the former.)

To make matter worse it is also not uncommon to have an $R$-module with an additional group action.

quid
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  • Thank you, but I actually know the difference between the definitions. My main problem is essentially with the equivalence between the category $G$-mod (where here $G$ is a group, as you say) and $\mathbb{Z}G$-mod (where here $\mathbb{Z}G$ is a ring, and we have the traditional structure of module over a ring). If you are so polite to be still interested in helping me, I suggest you to read the comments to the other answer, and in the link at the end of my OP (in order to avoid redundancy). Again, of course, thank you very much for your interest!!! Cheers –  Jun 08 '15 at 13:34
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    I do not think D. Burde meant "trivial" in that sense. It is the trivial ZG module in that the G-action is trivial. – quid Jun 08 '15 at 13:53
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    Dear @quid, yes, I think you're right. Thank you for your help! :-) –  Jun 08 '15 at 18:17