well-definedness of product of finitely generated ideals

Ler $R$ be of a commutative ring with nonzero unity.  If

$𝔞=(a_1,\mathrm{\dots },a_m)=({\alpha }_1,\mathrm{\dots },{\alpha }_{\mu })$(1)

and

$𝔟=(b_1,\mathrm{\dots },b_n)=({\beta }_1,\mathrm{\dots },{\beta }_{\nu })$(2)

are two finitely generated ideals of $R$, both with two , then the ideals

$$𝔠:=(a_1{b}_1,\mathrm{\dots },a_i{b}_j,\mathrm{\dots },a_m{b}_n)$$

and

$$𝔡:=({\alpha }_1{\beta }_1,\mathrm{\dots },{\alpha }_i{\beta }_j,\mathrm{\dots },{\alpha }_{\mu }{\beta }_{\nu })$$

are equal.

Proof.  By (1) and (2), for every $i,j$, there are elements $r_{ik},s_{jl}$ of $R$ such that

$a_i=r_{i1}{\alpha }_1+\mathrm{\dots }+r_{i\mu }{\alpha }_{\mu },b_j=s_{j1}{\beta }_1+\mathrm{\dots }+s_{j\nu }{\beta }_{\nu }.$(3)

Multiplying the equations (3) we see that

$$a_i{b}_j=(r_{i1}{s}_{j1})({\alpha }_1{\beta }_1)+(r_{i2}{s}_{j1})({\alpha }_2{\beta }_1)+\mathrm{\dots }+(r_{i\mu }{s}_{j\nu })({\alpha }_{\mu }{\beta }_{\nu }),$$

whence the generators $a_i{b}_j$ of $𝔠$ belong to $𝔡$ and consecuently $𝔠\subseteq 𝔡$.  The reverse containment is seen similarly.