Editing Free module (section)

=== Another construction ===

The free module {{math|''R''<sup>(''E'')</sup>}} may also be constructed in the following equivalent way.

Given a ring ''R'' and a set ''E'', first as a set we let
: <math>R^{(E)} = \{ f: E \to R \mid f(x) = 0 \text { for all but finitely many } x \in E \}.</math>
We equip it with a structure of a left module such that the addition is defined by: for ''x'' in ''E'',
: <math>(f+g)(x) = f(x) + g(x)</math>
and the scalar multiplication by: for ''r'' in ''R'' and ''x'' in ''E'',
: <math>(r f)(x) = r f(x)</math>

Now, as an ''R''-valued [[Function (mathematics)|function]] on ''E'', each ''f'' in <math>R^{(E)}</math> can be written uniquely as
: <math>f = \sum_{e \in E} c_e \delta_e</math>
where <math>c_e</math> are in ''R'' and only finitely many of them are nonzero and <math>\delta_e</math> is given as
: <math> \delta_e(x) = \begin{cases} 1_R \quad\mbox{if } x=e \\ 0_R \quad\mbox{if } x\neq e \end{cases} </math>
(this is a variant of the [[Kronecker delta]]). The above means that the subset <math>\{ \delta_e \mid e \in E \}</math> of <math>R^{(E)}</math> is a basis of <math>R^{(E)}</math>. The mapping <math>e \mapsto \delta_e</math> is a [[bijection]] between {{math|''E''}} and this basis. Through this bijection, <math>R^{(E)}</math> is a free module with the basis ''E''.