Free finitely generated module has finite rank?

[quasar987]

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,...,s_n}, if S is not linearly independent, then, WLOG, it is that
(*) s₁=r₂s₂+...+r_ns_n
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,...,r_n of R,
r₁s₁=r₂s₂+...+r_ns_n.
So we don't know if S-{s1} still generates.

 

[quasar987]

Ah.. take S a finite generating set and B={x_i} a basis. Every element of S can be written as a finite linear combination of the x_i's, so in particular, finitely many x_i's are required to generate S, which itself generate the whole module. Thus those x_i's are a finite basis.