Question #206175

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.


Expert's answer

Since M is finitely generated then there exists mi∈Mm_i \in M such that for each m∈Mm:=∑i=1nrimi,ri∈Rm \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(∑i=1nrimi)=∑i=1nf(rimi)=∑i=1nrif(mi)since f(m)∈N,thenf(m)=n,and,f(mi)=niThenn=∑i=1nrimiwhich shows that for each n∈N,there exist ni∈N,such that n=∑i=1nrimi    ri∈RHenceN is also finitely generatedf(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}


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