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.