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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 ?

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

2 Answers 2

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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.

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  • $\begingroup$ @CameronWilliams technically the question is "why is it not mentioned in the definition?" $\endgroup$ Jun 25, 2015 at 16:07
  • $\begingroup$ 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 $x \in M$ ? $\endgroup$
    – Madhu
    Jun 25, 2015 at 16:20
  • $\begingroup$ @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. $\endgroup$ Jun 25, 2015 at 16:25
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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.

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