This site is being phased out.

Modules

From Mathematics Is A Science
(Redirected from Module)
Jump to navigationJump to search

Informally,

modules are vector spaces over rings

rather than fields, meaning that the coefficients of linear combinations of vectors don't have to be real anymore but might be integers. In fact, our main interest, besides ${\bf R}^n$ of course, will be ${\bf Z}^n$, the grid. We will need it to construct cubical complexes and discrete differential forms.

Definition. Given a commutative ring $R$, a (commutative) $R$-module $M$ consists of an abelian group $(M, +)$ and a scalar product operation $R \times M \rightarrow M$ such that for all $r,s \in R$ and $x, y \in M$, we have:

  • $r(x+y) = rx + ry$,
  • $(r+s)x = rx + sx$,
  • $(rs)x = r(sx)$,
  • $1_Rx = x$, where $1_R$ is the multiplicative identity of $R$.

The scalar multiplication can be written on the left or right.

If $R$ is a field, an $R$-module is a vector space.

A subgroup $N$ of $M$ is a submodule if it is closed under scalar multiplication: for any $n \in N$ and any $r\in R$, we have $rn \in N$.

A group homomorphism $f: M \rightarrow N$ is a module homomorphism (or a linear operator) if it preserves the scalar multiplication: for any $m,n \in M$ and any $r, s \in R$, we have $f(rm + sn) = rf(m) + sf(n)$.

A bijective module homomorphism is an module isomorphism, and the two modules are called isomorphic.

The kernel of a module homomorphism $f : M \rightarrow N$ is the submodule of $M$ consisting of all elements that are taken to zero by $f$. The isomorphism theorems are still valid.

A module $M$ is called finitely generated if there exist finitely many elements $v_1,v_2,...,v_n \in M$ such that every element of $M$ is a linear combination of these elements (with coefficients in $R$).

A module $M$ is called free if it has a basis. The condition is equivalent to this: $M$ is isomorphic to a direct sum of copies of the ring $R$.

Of course, ${\bf Z}^n$ is free and finitely generated and it is our main interest. It behaves very similarly to ${\bf R}^n$ and the main differences lie in these two areas.

First, the quotients may have torsion, such as in ${\bf Z}/ 2{\bf Z} \cong {\bf Z}_2$.

Second, some operators invertible over ${\bf R}$ may be singular over ${\bf Z}$, try $2$.