# well-definedness of product of finitely generated ideals

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

 $\displaystyle\mathfrak{a}\;=\;(a_{1},\,\ldots,\,a_{m})\;=\;(\alpha_{1},\,% \ldots,\,\alpha_{\mu})$ (1)

and

 $\displaystyle\mathfrak{b}\;=\;(b_{1},\,\ldots,\,b_{n})\;=\,(\beta_{1},\,\ldots% ,\,\beta_{\nu})$ (2)
 $\mathfrak{c}\;:=\;(a_{1}b_{1},\,\ldots,\,a_{i}b_{j},\,\ldots,\,a_{m}b_{n})$

and

 $\mathfrak{d}\;:=\;(\alpha_{1}\beta_{1},\,\ldots,\,\alpha_{i}\beta_{j},\,\ldots% ,\,\alpha_{\mu}\beta_{\nu})$

are equal.

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

 $\displaystyle a_{i}\;=\;r_{i1}\alpha_{1}\!+\ldots+r_{i\mu}\alpha_{\mu},\quad b% _{j}\;=\;s_{j1}\beta_{1}\!+\ldots+s_{j\nu}\beta_{\nu}.$ (3)

Multiplying the equations (3) we see that

 $a_{i}b_{j}\;=\;(r_{i1}s_{j1})(\alpha_{1}\beta_{1})\!+\!(r_{i2}s_{j1})(\alpha_{% 2}\beta_{1})\!+\ldots+\!(r_{i\mu}s_{j\nu})(\alpha_{\mu}\beta_{\nu}),$

whence the generators   $a_{i}b_{j}$ of $\mathfrak{c}$ belong to $\mathfrak{d}$ and consecuently  $\mathfrak{c}\subseteq\mathfrak{d}$.  The reverse containment is seen similarly.

Title well-definedness of product of finitely generated ideals WelldefinednessOfProductOfFinitelyGeneratedIdeals 2013-03-22 19:12:56 2013-03-22 19:12:56 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 16D25 WellDefined ProductOfIdeals ProductOfFinitelyGeneratedIdeals