# proof of product of left and right ideal

###### Theorem 1

Let $\mathrm{a}$ and $\mathrm{b}$ be ideals of a ring $R$. Denote by $\mathrm{a}\mathit{}\mathrm{b}$ the subset of $R$ formed by all finite sums of products $a\mathit{}b$ with $a\mathrm{\in}\mathrm{a}$ and $b\mathrm{\in}\mathrm{b}$. Then if $\mathrm{a}$ is a left and $\mathrm{b}$ a right ideal^{}, $\mathrm{a}\mathit{}\mathrm{b}$ is a two-sided ideal of $R$. If in addition both $\mathrm{a}$ and $\mathrm{b}$ are two-sided ideals, then $\mathrm{a}\mathit{}\mathrm{b}\mathrm{\subseteq}\mathrm{a}\mathrm{\cap}\mathrm{b}$.

Proof. We must show that the difference of any two elements of $\U0001d51e\U0001d51f$ is in $\U0001d51e\U0001d51f$, and that $\U0001d51e\U0001d51f$ is closed under multiplication by $R$. But both of these operations are linear in $\U0001d51e\U0001d51f$; that is, if they hold for elements of the form $ab,a\in \U0001d51e,b\in \U0001d51f$, then they hold for the general element of $\U0001d51e\U0001d51f$. So we restrict our analysis to elements $ab$.

Clearly if ${a}_{1},{a}_{2}\in \U0001d51e,{b}_{1},{b}_{2}\in \U0001d51f$, then ${a}_{1}{b}_{1}-{a}_{2}{b}_{2}\in \U0001d51e\U0001d51f$ by definition.

If $a\in \U0001d51e,b\in \U0001d51f,r\in R$, then

$$r\cdot ab=(r\cdot a)b\in \U0001d51e\U0001d51f\text{since}\U0001d51e\text{is a left ideal}$$ | ||

$$ab\cdot r=a(b\cdot r)\in \U0001d51e\U0001d51f\text{since}\U0001d51f\text{is a right ideal}$$ |

and thus $\U0001d51e\U0001d51f$ is a two-sided ideal. This proves the first statement.

If $\U0001d51e,\U0001d51f$ are two-sided ideals, then $ab\in \U0001d51e$ since $b\in R$; similarly, $ab\in \U0001d51f$. This proves the second statement.

Title | proof of product of left and right ideal |
---|---|

Canonical name | ProofOfProductOfLeftAndRightIdeal |

Date of creation | 2013-03-22 17:41:25 |

Last modified on | 2013-03-22 17:41:25 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Proof |

Classification | msc 16D25 |