Triangular Matrix Rings .... Another Question

[Math Amateur]

I am reading T. Y. Lam's book, "A First Course in Noncommutative Rings" (Second Edition) and am currently focussed on Section 1:Basic Terminology and Examples ...

I need help with yet another aspect of Example 1.14 ... ...

Example 1.14 reads as follows: View attachment 5991
View attachment 5992Near the end of the above text from T. Y. Lam we read the following:" ... ... Moreover \(\displaystyle R \oplus M\) and \(\displaystyle M \oplus S\) are both ideals of \(\displaystyle A\), with \(\displaystyle A / (R \oplus M ) \cong S\) and \(\displaystyle A / ( M \oplus S ) \cong R \) ... ... "


Can someone please help me to show, formally and rigorously that A / (R \oplus M ) \cong S and A / ( M \oplus S ) \cong R ... ... My only thought so far is that the First Isomorphism Theorem may be useful ...

Hope someone can help ... ...

Peter

 

[Math Amateur]

I am reading T. Y. Lam's book, "A First Course in Noncommutative Rings" (Second Edition) and am currently focussed on Section 1:Basic Terminology and Examples ...

I need help with yet another aspect of Example 1.14 ... ...

Example 1.14 reads as follows:
Near the end of the above text from T. Y. Lam we read the following:" ... ... Moreover \(\displaystyle R \oplus M\) and \(\displaystyle M \oplus S\) are both ideals of \(\displaystyle A\), with \(\displaystyle A / (R \oplus M ) \cong S\) and \(\displaystyle A / ( M \oplus S ) \cong R \) ... ... "


Can someone please help me to show, formally and rigorously that A / (R \oplus M ) \cong S and A / ( M \oplus S ) \cong R ... ... My only thought so far is that the First Isomorphism Theorem may be useful ...

Hope someone can help ... ...

Peter

I have been reflecting on my own question posed above ... and believe that indeed it is the case that the First Isomorphism Theorem for Rings can be used to prove Lam's assertions ...
Now ... ... Use of the First Isomorphism Theorem for Rings in order to show that

\(\displaystyle R/ ( R \oplus M ) \cong S\)

would proceed as follows:
Define a surjection \(\displaystyle \phi \ : \ A \rightarrow S\) ... ...... where \(\displaystyle \phi\) is defined as the map \(\displaystyle \begin{pmatrix} r & m \\ 0 & s \end{pmatrix} \ \mapsto \ \begin{pmatrix} 0 & 0 \\ 0 & s \end{pmatrix}\)
\(\displaystyle \phi\) is clearly an epimorphism with kernel ... :\(\displaystyle \text{ ker } \phi = \begin{pmatrix} r & m \\ 0 & 0 \end{pmatrix} \ \cong \ R \oplus M\) So by the First Isomorphism Theorem for Rings we have the following
\(\displaystyle A / \text{ ker } \phi \ \cong \ S\) ...... that is ...\(\displaystyle A / ( R \oplus M ) \ \cong \ S\) ...

Is the above analysis correct?If it is correct ... then \(\displaystyle A / ( M \oplus S ) \ \cong \ R\) ... follows similarly ...[Note: I think we could have proved the above by only invoking the First Isomorphism Theorem for Groups ... ]Any comments critiquing the above analysis are welcome ...Peter