# proof of product of left and right ideal

###### Theorem 1

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

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

Clearly if $a_{1},a_{2}\in\mathfrak{a},b_{1},b_{2}\in\mathfrak{b}$, then $a_{1}b_{1}-a_{2}b_{2}\in\mathfrak{ab}$ by definition.

If $a\in\mathfrak{a},b\in\mathfrak{b},r\in R$, then

 $\displaystyle r\cdot ab=(r\cdot a)b\in\mathfrak{ab}\text{ since }\mathfrak{a}% \text{ is a left ideal}$ $\displaystyle ab\cdot r=a(b\cdot r)\in\mathfrak{ab}\text{ since }\mathfrak{b}% \text{ is a right ideal}$

and thus $\mathfrak{ab}$ is a two-sided ideal. This proves the first statement.

If $\mathfrak{a},\mathfrak{b}$ are two-sided ideals, then $ab\in\mathfrak{a}$ since $b\in R$; similarly, $ab\in\mathfrak{b}$. This proves the second statement.

Title proof of product of left and right ideal ProofOfProductOfLeftAndRightIdeal 2013-03-22 17:41:25 2013-03-22 17:41:25 rm50 (10146) rm50 (10146) 6 rm50 (10146) Proof msc 16D25