Answer to Question #206175 in Abstract Algebra for Sujata

Question

f is mapping from M to N . f is an R homomorphism and M is finitely generated then prove that N is also finitely generated.

Answer 1

Since M is finitely generated then there exists $m_i \in M$ such that for each $m \in M\\ m:= \sum_{i=1}^nr_im_i , r_i \in R$

Since f is a R-homomorphism, we have that $f(m)=f(\sum_{i=1}^nr_im_i)=\sum_{i=1}^nf(r_im_i)=\sum_{i=1}^nr_if(m_i)\\ \text{since } f(m) \in N, then f(m)=n , and , f(m_i)=n_i\\ Then \\ n= \sum_{i=1}^nr_im_i\\ \text{which shows that for each }n\in N, \text{there exist }n_i\in N, \text{such that } \\ n=\sum_{i=1}^nr_im_i~~~~r_i \in R\\ Hence\\ \text{N is also finitely generated}$