# zero times an element is zero in a ring

###### Lemma 1.

Let $R$ be a ring with zero element $\mathrm{0}$ (i.e. $\mathrm{0}$ is the additive identity of $R$). Then for any element $a\mathrm{\in}R$ we have $\mathrm{0}\mathrm{\cdot}a\mathrm{=}a\mathrm{\cdot}\mathrm{0}\mathrm{=}\mathrm{0}$.

###### Proof.

$0\cdot a$ | $=$ | $(0+0)\cdot a,\text{by definition of zero}$ | ||

$=$ | $0\cdot a+0\cdot a,\text{by the distributive law}$ |

Thus $0\cdot a=0\cdot a+0\cdot a$. Let $b$ be the additive inverse of $0\cdot a\in R$. Hence:

$b+0\cdot a=b+(0\cdot a+0\cdot a)$ | ||

$(b+0\cdot a)=(b+0\cdot a)+0\cdot a$ | ||

$0=0+0\cdot a$ | ||

$0=0\cdot a$ |

as claimed. The proof of $a\cdot 0=0$ is done analogously. ∎

Title | zero times an element is zero in a ring |
---|---|

Canonical name | ZeroTimesAnElementIsZeroInARing |

Date of creation | 2013-03-22 14:13:57 |

Last modified on | 2013-03-22 14:13:57 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 8 |

Author | alozano (2414) |

Entry type | Theorem |

Classification | msc 20-00 |

Classification | msc 16-00 |

Classification | msc 13-00 |

Synonym | $0\cdot a=0$ |

Related topic | 1cdotAA |

Related topic | AbsorbingElement |