A note on Noetherian polynomial modules (Q6633408)

Let \(R\) be a commutative ring, \(M\) an \(R\)-module and \(R[X]\) be the polynomial ring over \(R\). The polynomial \(R[X]\)-module \(M[X]\) is the polynomials in an indeterminate \(X\) with coefficients in \(M\) under the usual addition and the scalar multiplication as follows: For \(f =\sum_{i=0}^{m}a_{i}X^{i}\), \(g=\sum_{i=0}^{n}b_{i}X^{i}\), \(a_{i}, b_{i}\in M\) with \(m\geq n\), and \(h=\sum_{i=0}^{s}c_{i}X^{i}\), \(c_{i}\in R\), \(f+g=\sum_{i=0}^{m}(a_{i}+b_{i})X^{i}\) with \(b_{i}=0\) for \(n+1\leq i\leq m\); and \(hf=\sum_{i=0}^{m+s}d_{i}X^{i}\) where \(d_{i}=\sum_{j=0}^{i}c_{j}a_{i-j}\). \N\NIn the paper under review, the author studied Hilbert basis theorem for Noetherian modules. First, he gave a simple prrof of the well-known theorem that if \(M\) is a Noetherian \(R\)-module, then \(M[X]\) is a Noetherian \(R[X]\)-module. He also showed that if \(M[X]\) is a Noetherian \(R[X]\)-module, then \(M\) is a Noetherian \(R\)-module and there exists an element \(e\in R\) such that \(em = m\) for all \(m\in M\). Finally, he proved that if \(M[X]\) is a Noetherian \(R[X]\)-module and ann\(_{R}(M) = (0)\), then \(R\) contains the identity and \(M\) is a unitary \(R\)-module.