proof of Dedekind$-$Mertens lemma

Let $R$ be subring of the commutative ring $T$ and

$$f(X)=f_0+f_{1}X+\mathrm{\dots }+f_m{X}^m\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}g(X)=g_0+g_{1}X+\mathrm{\dots }+g_n{X}^n$$

be arbitrary polynomials in $T[X]$.  We will prove by induction on $n$ that the $R$-submodules of $T$ generated by thecoefficients of the polynomials $f$, $g$, and $fg$ satisfy

$M_f^{n+1}{M}_g=M_f^n{M}_{fg}$(1)

where the product modules are generated by the products of their generators.

The generators of the right hand side of (1) belong obviously to the left hand side,whence only the containment

$M_f^{n+1}{M}_g\subseteq M_f^n{M}_{fg}$(2)

has to be proved.

Firstly, (2) is trivial in the case  $n=0$.  Let now  $n>0$.  Define

and let $G_n$ be the $R$-submodule of $T$ generated by $g_0,g_1,\mathrm{\dots },g_{n-1}$.  We have

where $h_k$ is the coefficient of $X^k$ of the polynomial $fg$, and thus by induction wecan write

$$M_f^n{G}_n\subseteq M_f^{n-1}(M_{fg}+g_n{M}_f)\subseteq M_f^{n-1}{M}_{fg}+M_f^n{g}_n.$$

This implies the containment

$$f_i{M}_f^n{G}_n\subseteq M_f^n{M}_{fg}+M_f^n{f}_i{g}_n$$

for every $i$.  In addition, we have

$$f_i{g}_n\in M_{fg}+f_{i+1}{G}_n+M_{i+2}{G}_n+\mathrm{\dots }+f_n{G}_n,$$

whence

$$f_i{M}_f^n{G}_n\subseteq M_f^n{M}_{fg}+f_{i+1}{M}_f^n{G}_n+\mathrm{\dots }+f_n{M}_f^n{G}_n.$$

From this we infer that

$$f_i{M}_f^n{G}_n\subseteq M_f^n{M}_{fg}$$

is true for each  $i$ $=$ $m$, $m-1,\mathrm{\dots },\mathrm{\hspace{0.17em}0}$.  Thus also (2) is true.