# Do modules have to be defined over rings with unity?

This is definition for left module over a ring $$R$$ given in Wikipedia:

Suppose that $$R$$ is a ring and $$1_R$$ is its multiplicative identity. A left $$R$$-module $$M$$ consists of an abelian group $$(M, +)$$ and an operation $$\cdot\;: R \times M \to M$$ such that for all $$r, s \in R$$ and $$x, y \in M$$, we have:

1. $$r \cdot ( x + y ) = r \cdot x + r \cdot y$$

2. $$( r + s ) \cdot x = r \cdot x + s \cdot x$$

3. $$( r s ) \cdot x = r \cdot ( s \cdot x )$$

4. $$1_R \cdot x = x$$

But then it necessary that the ring $$R$$ should be a ring with unity. Then why it is not mentioned in definition ?

• Module theory is a mess when the ring is non unital. Jun 25, 2015 at 16:03
• Note part 4. It makes no sense without unity, but without it, you will get a lot of pathological examples. Jun 25, 2015 at 16:04

Some authors use the term "ring" to mean "ring with identity." This is neither standard nor nonstandard; there is just no consensus. Authors using this definition would use the term "rng" to denote a ring that possibly does not have an identity.

• @CameronWilliams technically the question is "why is it not mentioned in the definition?" Jun 25, 2015 at 16:07
• What happens if we replace the condition that $1_{R}$ is multiplicative identity of $R$, but continue the other part. That means $1_{R}.x=x$ for all $x \in M$ ? Jun 25, 2015 at 16:20
• @Madhu $\mathbb Z_2$ acts on $\mathbb Z_6$ via multiplication by 3 and this does not satisfy that axiom but otherwise is consistent with the devotion. Jun 25, 2015 at 16:25

When that definition says

Let $R$ be a ring and $1_R$ its multiplicative identity...

they are mentioning (implicitly) that they are requiring $R$ to be unital.

However, it is certainly possible to define modules over a non-unital ring; just throw out statement 4 from that definition.