16D25-ProofOfProductOfLeftAndRightIdeal.tex

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\begin{document}
\begin{thm} 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}$.
\end{thm}

\textbf{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 $a_1b_1-a_2b_2\in\mathfrak{ab}$ by definition. 

If $a\in\mathfrak{a}, b\in\mathfrak{b}, r\in R$, then
\begin{gather*}
r \cdot ab = (r\cdot a)b\in\mathfrak{ab} \text{ since } \mathfrak{a} \text{ is a left ideal} \\
ab \cdot r = a(b\cdot r)\in\mathfrak{ab} \text{ since } \mathfrak{b} \text{ is a right ideal}
\end{gather*}
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.
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