1
$\begingroup$

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 $a_{11}b_{11}\in \mathfrak{A}$...

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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 $ra_{11}\in\mathfrak{A}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.