# well-definedness of product of finitely generated ideals

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

$\U0001d51e=({a}_{1},\mathrm{\dots},{a}_{m})=({\alpha}_{1},\mathrm{\dots},{\alpha}_{\mu})$ | (1) |

and

$\U0001d51f=({b}_{1},\mathrm{\dots},{b}_{n})=({\beta}_{1},\mathrm{\dots},{\beta}_{\nu})$ | (2) |

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

$$\U0001d520:=({a}_{1}{b}_{1},\mathrm{\dots},{a}_{i}{b}_{j},\mathrm{\dots},{a}_{m}{b}_{n})$$ |

and

$$\U0001d521:=({\alpha}_{1}{\beta}_{1},\mathrm{\dots},{\alpha}_{i}{\beta}_{j},\mathrm{\dots},{\alpha}_{\mu}{\beta}_{\nu})$$ |

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 $\U0001d520$ belong to $\U0001d521$ and consecuently
$\U0001d520\subseteq \U0001d521$. The reverse containment is seen similarly.

Title | well-definedness of product of finitely generated ideals |
---|---|

Canonical name | WelldefinednessOfProductOfFinitelyGeneratedIdeals |

Date of creation | 2013-03-22 19:12:56 |

Last modified on | 2013-03-22 19:12:56 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 6 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 16D25 |

Related topic | WellDefined |

Related topic | ProductOfIdeals |

Related topic | ProductOfFinitelyGeneratedIdeals |