# Module over a ring with identity versus module over a ring without identity

I'm taking a course which introduces modules this semester. In the notes, it gives the definition of a module (or left module) over a ring (where we generally always assume rings have multiplicative identities) as an abelian group with an extra map (the action of the ring) with the usual four properties (this is the definition you'd find on Wikipedia say).

After giving the definition, it makes this remark:

The axiom $1m = m$ ($1$ being the multiplicative identity of the ring and $m$ being any element of the module) means that the identity element $1$ acts on the module as the identity map. If this were not true, we would be in a state of notational dissonance, which we would have to resolve by renaming the identity element of the ring, say as $e$.

That I'm okay with, but then it continues to say:

But then for all practical purposes it would make no difference to throw a new element called $1$ into our ring, which satisfied $1r = r1 = r$ for all $r$ in the ring, and $1m = m$ for all $m$ in the module. The same would apply if we had allowed the ring not to have a multiplicative identity previously. This is why we are not really losing any generality by assuming that $R$ has an element $1$ and that it acts on the module as the identity.

Here, I don't understand why you could just freely throw a new element in the ring? Would we not have to worry about violating the ring axioms? And even if we ignored them, how do we know the four properties for the action map would still remain intact?

• I always have a hard time understanding why some authors go to such great lengths to justify only using rings with identity, or assuming modules are unital. Such things, which arise naturally in practice, shouldn't really need so much fretting over. Before anyone gets the wrong idea, I am a big fan of both "with identity" and "without identity" theories. – rschwieb Oct 12 '12 at 12:26
• @rschwieb Justifications for doing so would arise naturally in practice, but personally I like a quick, small comment about it at the beginning anyway, just to acknowledge the issue. – M.G. Oct 13 '12 at 10:17

## 1 Answer

$\def\Z{\mathbb Z}$To throw a new element "1" into the ring $A$, means the following:

Let $\tilde A := A \times \Z$ and define operations on $\tilde A$ by (just think of $(a,\lambda) \in \tilde A$ as "$a + \lambda$"): \begin{align*} (a,\lambda) + (b,\mu) &:= (a+b, \lambda + \mu)\\ (a, \lambda)\cdot (b,\mu) &:= (ab + \mu a + \lambda b, \lambda\mu) \end{align*} Then $\tilde A$ is a ring, which has $(0,1)$ as multiplicative identity. If $M$ is an (non unitary) $A$-module, define an $\tilde A$-module operation by $(a,\lambda)m := am + \lambda m$ where the action of $\Z$ on $M$ is the usual of $M$ as an $\Z$-module ($=$ abelian group). Then $M$ with this structure is an $\tilde A$-module on which the "1" in $\tilde A$, that is $(0,1)$ acts as identity.

The claims made above, that $\tilde A$ is a ring and $M$ an $\tilde A$-module can be proved by simple calculations.

• I did this exercise recently, do you know of any shortcuts for proving multiplication is associative/distributive in $\tilde{A}$? – peoplepower Oct 12 '12 at 11:49
• @peoplepower No, I don't. – martini Oct 12 '12 at 11:50
• Thanks a lot @martini, makes sense now. – M.G. Oct 12 '12 at 12:13
• @martini: What happens with the module category of $\tilde{A}$ and that of $A$? are they equivalent? can you study the module category of $A$ through the category of $\tilde{A}$-modules? – Marco Armenta Nov 7 '15 at 4:46