Module over a Ring

Let $R$ be a Ring. A left R-module is an Abelian Group $M$ equipped with an operation $\cdot:R\times M\to M$ such that:

• $(r+s)\cdot x=r\cdot x+s\cdot x$
• $(rs)\cdot x=r\cdot(s\cdot x)$
• $1\cdot x=x$
• $r\cdot(x+y)=r\cdot x+r\cdot y$

Dually, a right R-module is an Abelian Group $M$ equipped with an operation $\cdot:M\times R\to M$ such that:

• $x\cdot(r+s)=x\cdot r+x\cdot s$
• $x\cdot(rs)=(x\cdot r)\cdot s$
• $x\cdot 1=x$
• $(x+y)\cdot r=x\cdot r+y\cdot r$

If $R$ is a Commutative Ring, then left and right R-modules coincide.

Note that we could ignore the left/right distinction, and simply think about modules over a ring $R$ and modules over the Opposite Ring $R^{mathrm{op}}$.

Motivation

There are 2 possible sources of motivation for R-modules:

1. They generalize Vector Spaces from Fields to (potentially non-commutative) rings.
2. They are the ring-theoretic analogs of (co)presheaves.

Properties

• A left module $M$ over $R$ is equivalent to the data of a Ring Homomorphism $(-\,\cdot-):R\to\mathrm{End}(M)$ into the Endomorphism Ring of $M$. This gives us a couple of immediately useful properties:

  • $0\cdot x=0$
  • $(-r)\cdot x=-(r\cdot x)$

  We shall present the proof for the first case for the sake of clarity:

  	$\displaystyle 0\cdot x$	$\displaystyle=(0\,\cdot\,-)(x)$	$\displaystyle(\text{Modules are ring homomorphisms})$
  		$\displaystyle=(\lambda x.\;0)(x)$	$\displaystyle(\text{$(-\,\cdot\,-)$ is a ring homomorphism})$
  		$\displaystyle=0$	$\displaystyle(\text{$\beta$ reduction})$