Notes:Modules

Motivation

Let [ilmath](V,F)[/ilmath] be a vector space over the field [ilmath]F[/ilmath]. Let [ilmath]\tau\in\mathcal{L}(V)[/ilmath] be given. Then for any polynomial, [ilmath]p(x)\in F[x][/ilmath] (where [ilmath]F[x][/ilmath] denotes the space of polynomials over the field [ilmath]F[/ilmath]) the operator:

• [ilmath]p(\tau)[/ilmath] is well defined.
  • For example, consider [ilmath]p(x)=1+2x+x^3[/ilmath] then [ilmath]p(\tau)=i+2\tau+\tau^3[/ilmath] where [ilmath]i:V\rightarrow V[/ilmath] is the identity map, and [ilmath]\tau^3[/ilmath] is the threefold function composition [ilmath]\tau\circ\tau\circ\tau[/ilmath]

Thus using the linear operator [ilmath]\tau:V\rightarrow V[/ilmath] we can define the product of a polynomial, [ilmath]p(x)\in F[x][/ilmath] and a vector, [ilmath]v\in V[/ilmath] by:

• [ilmath]p(x)v:=p(\tau)(v)[/ilmath] meaning [ilmath]p(x)v:=(p(\tau))(v)[/ilmath] of course.

Clearly this "product" satisfies the usual properties of scalar multiplication, namely:

• For [ilmath]r(s),s(x)\in F[x][/ilmath] and [ilmath]u,v\in V[/ilmath]:
  1. [ilmath]r(x)(u+v)=r(x)u+r(x)v[/ilmath]
  2. [ilmath](r(x)+s(x))u=r(x)u+s(x)u[/ilmath]
  3. [ilmath][r(x)s(x)]u=r(x)[s(x)u][/ilmath] and
  4. [ilmath]1u=u[/ilmath]

Thus for a fixed [ilmath]\tau\in\mathcal{L}(V)[/ilmath] we can think of [ilmath]V[/ilmath] as being imbued with operations of addition and multiplication of an element of [ilmath]V[/ilmath] with an element of [ilmath]F[x][/ilmath]!

Of course, [ilmath]\mathbf{F[x]} [/ilmath] is not a field so these operations cannot be the operations of a vector space and so cannot be an alternate vector space structure on [ilmath]V[/ilmath].

This leads us to modules:

Definition

Let [ilmath]R[/ilmath] be a cu-ring, then:

• We may call the elements of [ilmath]R[/ilmath] scalars (but not of a vector space, of a module instead!)

An "[ilmath]R[/ilmath]-module" or "module over [ilmath]R[/ilmath]" is a non-empty set [ilmath]M[/ilmath], together with two operations:

1. Addition: [ilmath]+:M\times M\rightarrow M[/ilmath] by [ilmath]+:(u,v)\mapsto u+v[/ilmath] as discussed, and
2. Multiplication: [ilmath]\times:R\times M\rightarrow M[/ilmath] by [ilmath]\times:(r,v)\mapsto rv[/ilmath] as discussed.

We call [ilmath]R[/ilmath] the base ring of [ilmath]M[/ilmath]

Immediate properties

Immediately we see the following:

1. [ilmath]M[/ilmath] is an Abelian group under addition
2. For all [ilmath]r,s\in R[/ilmath] and [ilmath]u,v\in M[/ilmath] the following hold:
   1. [ilmath]r(u+v)=ru+rv[/ilmath]
   2. [ilmath](r+s)u=ru+su[/ilmath]
   3. [ilmath](rs)u=r(su)[/ilmath]
   4. [ilmath]1u=u[/ilmath]

Examples

---

TODO: Check these for requirements of the ring. See "importance of the base ring" below

---

1. If [ilmath]R[/ilmath] is a ring then the set [ilmath]R^n[/ilmath] of all ordered [ilmath]n[/ilmath]-tuples of elements of [ilmath]R[/ilmath] is an [ilmath]R[/ilmath]-module, with addition and multiplication defined component wise (just as for [ilmath]F^n[/ilmath] - the [ilmath]n[/ilmath]-dimensional vector space of a field)
   • Caution:I think the terminology has changed here, it might mean a ring with unity... I'll check later!
   • For example [ilmath]\mathbb{Z}^n[/ilmath] is the [ilmath]\mathbb{Z} [/ilmath]-module of all ordered [ilmath]n[/ilmath]-tuples of integers.
2. If [ilmath]R[/ilmath] is a ring then the set [ilmath]\mathcal{M}_{m,n}(R)[/ilmath] of [ilmath]m\times n[/ilmath] matrices with elements of [ilmath]R[/ilmath] is an [ilmath]R[/ilmath]-module, with the usual operations.
   • Since [ilmath]R[/ilmath] is a ring we can also take the product of matrices in [ilmath]\mathcal{M}_{m,n}(R)[/ilmath]
   • We can also take [ilmath]R=F[x][/ilmath], where [ilmath]\mathcal{M}_{m,n}(F[x])[/ilmath] is the [ilmath]F[x][/ilmath]-module of all [ilmath]m\times n[/ilmath] matrices whose entries are polynomials
     • Caution:...What... check all of this for any implicit requirements on the rings
3. Any cu-ring is a module over itself. That is [ilmath]R[/ilmath] is an [ilmath]R[/ilmath]-module. Scalar multiplication is just the ring multiplication.

Importance of the base ring

This definition requires that [ilmath]R[/ilmath] be commutative.

Modules of non-commutative rings are a "thing".

I must be very careful when "distilling" this definition