global dimension of a subring
Let be a ring with identity and a subring, such that is contained in the center of . In this case is a (left) -module via multiplication. Throughout by modules we will understand left modules and by global dimension we will understand left global dimension (we will denote it by ).
Proposition. Assume that . If is free as a -module, then .
Proof. Let be a -module. Then, there exists exact sequence
of -modules, where each is projective (module is just a kernel of a map ). We will show, that is also projective (and since is arbitrary, it will show that ).
Since is free as a -module, then the extension of scalars is an exact functor from the cateogry of -modules to the category of -modules. Furthermore for any projective -module , the -module is projective (in the category of -modules). Thus we have following exact sequence of -modules
where each is a projective -module. But projective dimension of is at most (since ). Thus is a projective -module (please, see this entry (http://planetmath.org/ExactSequencesForModulesWithFiniteProjectiveDimension) for more details).
Note that the restriction of scalars functor also maps projective -modules into projective -modules. Thus is a projective -module. But is free -module, so
for some index set . Finally we have
|Title||global dimension of a subring|
|Date of creation||2013-03-22 19:05:09|
|Last modified on||2013-03-22 19:05:09|
|Last modified by||joking (16130)|