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well-definedness of product of finitely generated ideals

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Ler $R$ be of a commutative ring with nonzero unity.\, If
\mathfrak{a} \;=\; (a_1,\,\ldots,\,a_m) \;=\; (\alpha_1,\,\ldots,\,\alpha_\mu)
\mathfrak{b} \;=\; (b_1,\,\ldots,\,b_n) \;=\, (\beta_1,\,\ldots,\,\beta_\nu)
are two finitely generated ideals of $R$, both with two \PMlinkescapetext{generating systems}, then the ideals
$$\mathfrak{c} \;:=\; (a_1b_1,\,\ldots,\,a_ib_j,\,\ldots,\,a_mb_n)$$
$$\mathfrak{d} \;:=\; (\alpha_1\beta_1,\,\ldots,\,\alpha_i\beta_j,\,\ldots,\,\alpha_\mu\beta_\nu)$$
are equal.\\

\emph{Proof.}\, By (1) and (2), for every $i,\,j$, there are elements $r_{ik},\,s_{jl}$ of $R$ such that
a_i \;=\; r_{i1}\alpha_1\!+\ldots+r_{i\mu}\alpha_\mu, \quad b_j \;=\; s_{j1}\beta_1\!+\ldots+s_{j\nu}\beta_\nu.
Multiplying the equations (3) we see that
$$a_ib_j \;=\; 
whence the generators $a_ib_j$ of $\mathfrak{c}$ belong to $\mathfrak{d}$ and consecuently\, 
$\mathfrak{c} \subseteq \mathfrak{d}$.\, The reverse containment is seen similarly.