proof of invertible ideals are projective
and, as is onto, there exist such that . For any , , so we can define by
so is indeed a right inverse of , and is projective.
Conversely, suppose that is projective and let generate (this always exists, as we can let include every element of ). Then let be a module with free basis and define by . As is projective, has a right inverse . As freely generate , we can uniquely define by
noting that all but finitely many must be zero for any given . Choosing any fixed nonzero , we can set so that
for all , and must equal zero for all but finitely many . So, we can let be the fractional ideal generated by the and, noting that we get . Furthermore, for any ,
so that , and is the inverse of as required.
|Title||proof of invertible ideals are projective|
|Date of creation||2013-03-22 18:35:51|
|Last modified on||2013-03-22 18:35:51|
|Last modified by||gel (22282)|