- #1
- 4,807
- 32
How does one prove that for R commutative, a free finitely generated R-module has finite rank?
If R is a field (i.e. in the case of vector space), then we can argue that given a finite generating set S={s1,...,sn}, if S is not linearly independent, then, WLOG, it is that
(*) s1=r2s2+...+rnsn
so we just remove s1 from S and repeat until we are down to a basis of M. But in the general case, equation (*) fails. We can only say that for some non all zero elements r1,...,rn of R,
r1s1=r2s2+...+rnsn.
So we don't know if S-{s1} still generates.
If R is a field (i.e. in the case of vector space), then we can argue that given a finite generating set S={s1,...,sn}, if S is not linearly independent, then, WLOG, it is that
(*) s1=r2s2+...+rnsn
so we just remove s1 from S and repeat until we are down to a basis of M. But in the general case, equation (*) fails. We can only say that for some non all zero elements r1,...,rn of R,
r1s1=r2s2+...+rnsn.
So we don't know if S-{s1} still generates.