Two cyclic modules are isomorphic iff they have the same order ideal.

Feb 2009
189
0
Every cyclic R-module V=<v> is R-isomorphic to R/ord(v). Conclude that two cyclic modules are R-isomorphic if and only if they have generators with the same order ideal.

Here ord(v) = {r in R: rv=0}. f: <v> --> R/ord(v), f(rv) = r + ord(v) is well defined and is an R-isomorphism. If ord(v) = ord (u) then <v> isomorphic to <u> as is immediate. Now if <v> isomorphic to <u> by the first part R/ord(v) isomorphic to R/ord(u). If R finite then |R/ord(v)| = |R/ord(u)| and |ord(v)| = |ord(u)| and ord(v) iso to ord(u). But what happens if R infinite? Is there a corresponding problem when R is an abelian group. In that case R is a Z-module.