Editing Module homomorphism (section)

== Module structures on Hom ==
In short, Hom inherits a ring action that was not ''used up'' to form Hom. More precise, let ''M'', ''N'' be left ''R''-modules. Suppose ''M'' has a right action of a ring ''S'' that commutes with the ''R''-action; i.e., ''M'' is an (''R'', ''S'')-module. Then
:<math>\operatorname{Hom}_R(M, N)</math>
has the structure of a left ''S''-module defined by: for ''s'' in ''S'' and ''x'' in ''M'',
:<math>(s \cdot f)(x) = f(xs).</math>
It is well-defined (i.e., <math>s \cdot f</math> is ''R''-linear) since
:<math>(s \cdot f)(rx) = f(rxs) = rf(xs) = r (s \cdot f)(x),</math>
and <math>s \cdot f</math> is a ring action since
:<math>(st \cdot f)(x) = f(xst) = (t \cdot f)(xs) = s \cdot (t \cdot f)(x)</math>.

Note: the above verification would "fail" if one used the left ''R''-action in place of the right ''S''-action. In this sense, Hom is often said to "use up" the ''R''-action.

Similarly, if ''M'' is a left ''R''-module and ''N'' is an (''R'', ''S'')-module, then <math>\operatorname{Hom}_R(M, N)</math> is a right ''S''-module by <math>(f \cdot s)(x) = f(x)s</math>.