# Module homomorphism

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Flesh out, deal with unital modules, so forth
See Homomorphism for a list of other morphism types, and see morphism for a categorical overview.

## Definition

Let [ilmath](R,+,*,0)[/ilmath] be a ring with or without unity and let [ilmath]A[/ilmath] and [ilmath]B[/ilmath] be (left) [ilmath]R[/ilmath]-modules. A homomorphism of left [ilmath]R[/ilmath]-modules is[1]:

• A mapping, [ilmath]\varphi:A\rightarrow B[/ilmath], such that:
1. [ilmath]\forall x,y\in M[\varphi(x+y)=\varphi(x)+\varphi(y)][/ilmath] and
2. [ilmath]\forall r\in R,\forall x\in M[\varphi(rx)=r\varphi(x)][/ilmath][Note 1]

## Auxiliary structure

Morphisms of [ilmath]R[/ilmath]-modules can be added pointwise:

• Let [ilmath]f,g:A\rightarrow B[/ilmath] be module homomorphisms, then:
• [ilmath](f+g):A\rightarrow B[/ilmath] by [ilmath](f+g):a\mapsto f(a)+g(a)[/ilmath]
Claim 1: this is indeed a homomorphism

I also expect we can multiply morphisms too, eg:

• [ilmath](rf):A\rightarrow B[/ilmath] by [ilmath](rf):a\mapsto rf(a)[/ilmath]

Caution:But maybe not! This is certainly true with vector spaces, perhaps not here - NOT MENTIONED in Grillet's abstract algebra - at least not on page 321.

## Proof of claims

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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The message provided is:
easy and routine

This proof has been marked as an page requiring an easy proof

## Types of homomorphism

There are also (following standard terminology)

• Automorphism - an isomorphism of the form [ilmath]\varphi:M\rightarrow M[/ilmath]

TODO: List more

TODO: This style should be duplicated across other homomorphism pages

## Notes

1. A homomorphism of right modules is the same but this rule (rule #2) becomes:
• [ilmath]\forall r\in R,\forall x\in M[\varphi(xr)=\varphi(x)r][/ilmath] - as ought to be expected.