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 $a b$ 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 $a b$ .

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
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