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Is there a nontrivial example of a non-unital module over a ring with identity? By trivial, I mean modules with $rm = 0$ for all $r$ and $m$.

Just an idle question. (Why is there no tag for that?)

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    $\begingroup$ what do you mean by non-unital module? $\endgroup$ – Manos Aug 10 '13 at 1:45
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    $\begingroup$ One that does not satisfy the axiom $1m = m$ for all $m$ in the module. $\endgroup$ – Stephen Herschkorn Aug 10 '13 at 1:49
  • $\begingroup$ @Stephen I've answered your question below by explaining the underlying principle behind it. $\endgroup$ – Amitesh Datta Aug 10 '13 at 5:59
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Let $A$ denote the ring $\mathbb{Z}$, and let $M$ denote the abelian group $\mathbb{Z}\times\mathbb{Z}$ considered as an $A$-module with the scalar multiplication map $A\times M\to M$ defined by $$(a,(m_1,m_2))\mapsto (am_1,0).$$ This is a non-trivial $A$-module structure, and it is non-unital because multiplication by $1\in A$ is not the identity map on $M$.

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In essence there is no counterexample because of the following lemma:

Lemma: Every (not necessarily unital) module is a direct sum of a unital module and a trivial module.

Proof: Let $V$ be the module. Then we claim that $V=1_AV\oplus \operatorname{ker}\lambda_{1_A}$, where $\lambda_{1_A}$ is left multiplication with $1_A$. Indeed let $v$ be in their intersection, then $v=1_Aw$ and $0=1_Av=1_A^2w=1_Aw=v$, hence the intersection is zero and furthermore $v=1_Av+(v-1_Av)$ with the former summand belonging to $1_AV$, the latter to $\operatorname{ker}\lambda_{1_A}$.

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I'd just like to add to Zev's excellent counterexample:

If $M$ is an abelian group, then $\text{End}(M)$ denotes the set of endomorphisms of $M$. (An endomorphism of the abelian group $M$ is a group homomorphism $M\to M$.) In fact, $\text{End}(M)$ is a ring under the operations of pointwise summation (addition in the ring) and function composition (multiplication in the ring).

Exercise 1: Prove that $\text{End}(M)$ is a ring if $M$ is an abelian group. (Check the ring axioms!)

Now, specifying an $R$-module $M$ is equivalent to specifying a ring homomorphism $R\to \text{End}(M)$. How so? Well, if we have an $R$-module $M$, then we just map $r\to \phi_r$ where $\phi_r:M\to M$ is defined by the rule $\phi_r(m)=rm$ for $m\in M$. Thus, we have a map $R\to \text{End}(M)$.

Exercise 2: If $M$ is an $R$-module, then check that this is a ring homomorphism $R\to \text{End}(M)$. Check also that specifying such a ring homomorphism is equivalent to specifying an $R$-module $M$.

Now, to say that $M$ is a unital $R$-module is to say that this ring homomorphism $R\to \text{End}(M)$ is unital; i.e., the unity of $R$ is mapped to the unity of $\text{End}(M)$. (The unity of $\text{End}(M)$ is just the identity homomorphism $M\to M$ in case you didn't do Exercise 1. If this isn't obvious, then I suggest you do Exercise 1!)

Exercise 3: Prove that every ring is isomorphic to $\text{End}(M)$ for some abelian group $M$.

This is the analogue of Cayley's theorem in ring theory and is the converse of Exercise 1. (You might draw inspiration from the proof of Cayley's theorem in solving Exercise 3.) So, in fact, finding the example you seek is equivalent to finding a ring homomorphism $R\to S$ that doesn't take the unity of $R$ to the unity of $S$. I.e., finding a non-unital ring homomorphism $R\to S$ of rings. That's what Zev has done. But thinking about the problem in this way makes it quite easy to construct counterexamples. What are examples of non-unital ring homomorphisms of which you know? I know a few, e.g.,

Exercise 4: Prove that $\mathbb{R}\to M_2(\mathbb{R})$ defined by the rule $x\to \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$ is a non-unital ring homomorphism.

Can you find others?

I hope this helps!

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