12.1.2:

(a) For the first part, define f : Rⁿ --> N by sending e_i to x_i for 1 <= i <=n. There exists such a homomporphism because Rⁿ is free with basis {e₁, ... , e_n}. f is onto because {x₁, ... , x_n} generates N, and the kernel of f is trivial because {x₁, ... , x_n} is linearly independent. For the second part, let y + N be a nonzero element of M/N. Then y is not in N. Since {x₁, ... , x_n} is a maximal independent set, {x₁, ... , x_n, y} is linearly dependent. Then there are ring elements r₁, ... , r_n, r_n+1 such that r₁x₁+ ... r_nx_n + r_n+1y = 0. Moreover r_n+1 is not zero since {x₁, ... , x_n} is linearly independent. Then r_n+1y is in N, hence r_n+1(y + N) equals zero in M/N. Thus y + N is a torsion element of M/N.

(b) Let {x₁, ... , x_n} be a free basis for N, and let {y₁, ... , y_n+1} be a subset of M. Since M/N is torsion, it follows that for each i, 1 <= i <= n, there is a ring element r_i such that r_i y_i is in N. Clearly N has rank n (since it is isomorphic to Rⁿ, and R is commutative). Then {r₁y₁, ... , r_n+1y_n+1} is linearly dependent. Then there are ring elements s₁, ... , s_n+1 such that s₁(r₁y₁) + ... + s_n+1(r_n+1y_n+1) = 0, and this yields a dependence relation among y₁, ... , y_n+1, hence {y₁, ... , y_n+1} is linearly dependent. Thus M has rank n.

12.1.4:

Let {x₁, ... , x_r} be a maximal linearly independent set in the submodule N, and {y₁+N, ... , y_s+N} a maximal linearly independent set in the quotient M/N. First we claim {x₁, ... , x_r,y₁, ... , y_s} is linearly independent in M. Suppose t₁x₁+ ... + t_rx_r + u₁y₁ + ... + u_sy_s = 0. Then u₁y₁ + ... + u_sy_s= - (t₁x₁+ ... + t_rx_r) lies in N, so u₁(y₁ + N) + ... + u_s( y_s+N) = 0 in M/N, so u₁= ... = u_s= 0. Then t₁x₁+ ... + t_rx_r = 0, so t₁= ...= t_r = 0. This proves the claim. Now, suppose x is in M. Then there is some nonzero r such that r(x+N) is in the submodule generated by {y₁+N, ... , y_s+N}. Then rx is in u₁y₁ + ... + u_sy_s+N for some ring elements u₁, ... , u_s. Then rx = u₁y₁ + ... + u_sy_s+ t₁x₁+ ... + t_rx_r for some ring elements t₁, ... , t_r. It follows that {x₁, ... , x_r,y₁, ... , y_s} is a maximal linearly independent subset of M, so that M has rank r + s.