I'm reading the T. Y. Lam's A First Course in Noncommutative Rings, Theorem (3.1), page 32. In the proof, Lam asserts that the following fact is "easily seen":

Let $R$ be a ring with identity (not necessarily commutative) and let $I$ be an ideal in the usual ring $M_n(R)$ of the $n\times n$ matrices with entries in $R$. Show that $\mathfrak{A}=\{a_{11}\in R\,:\, (a_{ij})\in I\}$ is an ideal in $R$.

But I cannot understand it. If $a_{11}$ and $b_{11}\in \mathfrak{A}$, then, ok, $a_{11}+b_{11}\in \mathfrak{A}$ since $a_{11}+b_{11}$ is a $(1,1)$-entry of a matrix in $I$, namely, that matrix which is sum of the matrices who yield both $a_{11}$ and $b_{11}$. But how do I know that $a_{11}b_{11}\in \mathfrak{A}$? When I do the usual product of both matrices, I only know that $$c_{11}=\sum_{i=1}^na_{1i}b_{i1}\in \mathfrak{A}...$$ but not that the isolated term $a_{11}b_{11}\in \mathfrak{A}$...

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    $\begingroup$ You don't want $a_{11}b_{11}$ in your ideal, you want $ca_{11}$ and $a_{11}c$ in your ideal for all $c\in R$. Don't forget that these are the top left entries of $(cI)A$ and $A(cI)$. $\endgroup$ – Lord Shark the Unknown Jul 8 '17 at 18:23

First of all, you need to show that $a_{11}r\in\mathfrak{A}$ (and also $ra_{11}\in\mathfrak{A}$) for every $r\in R$, not just $r\in\mathfrak{A}$. And in order to show this, you get to use the fact that $I$ is an ideal, so that if $A\in I$, then $AB\in I$ (and $BA\in I$) for every $B\in M_n(R)$. As you say, when you multiply $A$ and $B$ you will get that $\sum_{i=1}^na_{1i}b_{i1}\in\mathfrak{A}$. But the key here is that you can choose $B$ to be any matrix at all, so you can choose the entries $b_{i1}$ to be any elements of $R$ at all. So you're free to just choose $b_{11}=r$ and $b_{1i}=0$ for $i\neq 1$, to get $a_{11}r\in \mathfrak{A}$. Similarly, you can choose $B$ so that $BA\in I$ gives you $ra_{11}\in\mathfrak{A}$.


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