Suppose N is a submodule of M which is generated by n elements all over the PID R. We prove by induction on n that N is also finitely generated.
n= 1 case: So M = (m) for some m. Let I = {a | a m is in N}. Then I is an ideal in R, say generated by a0 and N = (a0 m).
induction step: M = ( m_1, ..., m_n+1). Let P = all linear combinations of m_1, ..., m_n for which there is some a_n+1 such that the linear combination plus a_n+1 m_n+1 is in M. Then by induction P is finitely generated, say by p_1,..., p_k . Let I = { a_n+1 | p + a_n+1 m_n+1 is in N for some p in P}, an ideal generated by say a_n+1.Then N = is generated by p_1, ..., p_k+1, a_n+1 m_n+1.
Answers & Comments
Verified answer
Suppose N is a submodule of M which is generated by n elements all over the PID R. We prove by induction on n that N is also finitely generated.
n= 1 case: So M = (m) for some m. Let I = {a | a m is in N}. Then I is an ideal in R, say generated by a0 and N = (a0 m).
induction step: M = ( m_1, ..., m_n+1). Let P = all linear combinations of m_1, ..., m_n for which there is some a_n+1 such that the linear combination plus a_n+1 m_n+1 is in M. Then by induction P is finitely generated, say by p_1,..., p_k . Let I = { a_n+1 | p + a_n+1 m_n+1 is in N for some p in P}, an ideal generated by say a_n+1.Then N = is generated by p_1, ..., p_k+1, a_n+1 m_n+1.