Triangular Matrix RIngs .... Lam, Proposition 1.17

[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 Part (1) of Proposition 1.17 ... ...

Proposition 1.17 (together with related material from Example 1.14 reads as follows:View attachment 5993
View attachment 5994
View attachment 5995Can someone please help me to prove Part (1) of the proposition ... that is that \(\displaystyle I_1 \oplus I_2\) is a left ideal of \(\displaystyle A\) ... ...

Help will be much appreciated ...

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 Part (1) of Proposition 1.17 ... ...

Proposition 1.17 (together with related material from Example 1.14 reads as follows:

Can someone please help me to prove Part (1) of the proposition ... that is that \(\displaystyle I_1 \oplus I_2\) is a left ideal of \(\displaystyle A\) ... ...

Help will be much appreciated ...

Peter

I have been reflecting on the problem I posed ... here is my 'solution' ... Note: I am quite unsure of this ...Problem ... Let \(\displaystyle I = I_1 \oplus I_2\) where \(\displaystyle I_1\) is a left ideal of \(\displaystyle S\) and \(\displaystyle I_2\) is a left submodule of \(\displaystyle R \oplus M\) ...Show \(\displaystyle I\) is a left ideal of \(\displaystyle A\)
Let \(\displaystyle a \in I\), then there exists \(\displaystyle a_1 \in I_1\) and \(\displaystyle a_2 \in I_2\) such that \(\displaystyle a = (a_1, a_2) \in I \)
[ ... ... actually \(\displaystyle a_2 = (c_1, c_2) \in R \oplus M\) but we ignore this complication in order to keep notation simple ... ]Similarly let \(\displaystyle b \in I\) so \(\displaystyle b = (b_1, b_2) \in I\) ... ...
Now ... if \(\displaystyle I\) is a left ideal then
\(\displaystyle a, b \in I \ \Longrightarrow \ a - b \in I\)and
\(\displaystyle r \in A\) and \(\displaystyle a \in I \ \Longrightarrow \ ra \in I\)

-------------------------------------------------------------------------------------------------To show \(\displaystyle a, b \in I \ \Longrightarrow \ a - b \in I
\)
Let \(\displaystyle a,b \in I\)then \(\displaystyle a - b = (a_1, a_2) - (b_1, b_2)\) where \(\displaystyle a_1, b_1 \in S\) and \(\displaystyle a_2, b_2 \in R \oplus M\) so, \(\displaystyle a - b = (a_1 - b_1, a_2 - b_2) \)But ... \(\displaystyle a_1 - b_1 \in I_1\) since \(\displaystyle I_1\) is an ideal in \(\displaystyle S\)and ... \(\displaystyle a_2 - b_2\) since \(\displaystyle I_2\) is a left sub-module of \(\displaystyle A\) hence \(\displaystyle (a_1 - b_1, a_2 - b_2) = a - b \in I\)

---------------------------------------------------------------------------------------------------------To show \(\displaystyle r \in A\) and \(\displaystyle a \in I \ \Longrightarrow \ ra \in I\)
Now ... \(\displaystyle r \in A\) and \(\displaystyle a \in I \ \Longrightarrow \ ra = r(a_1, a_2) = (ra_1, ra_2)\) [I hope this is correct!]But \(\displaystyle ra_1 \in I_1\) since \(\displaystyle I_1\) is a left ideal ...and \(\displaystyle ra_2 \in I_2\) since \(\displaystyle I_2\) is a left \(\displaystyle R\)-submodule ...Hence \(\displaystyle (ra_1, ra_2) = ra \in I \)

-----------------------------------------------------------------------------------------------------The above shows that I is a left ideal ... I think ...Comments critiquing the above analysis and/or pointing out errors are more than welcome ...Peter