MHB 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 $$I_1 \oplus I_2$$ is a left ideal of $$A$$ ... ...

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 $$I_1 \oplus I_2$$ is a left ideal of $$A$$ ... ...

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 $$I = I_1 \oplus I_2$$ where $$I_1$$ is a left ideal of $$S$$ and $$I_2$$ is a left submodule of $$R \oplus M$$ ...Show $$I$$ is a left ideal of $$A$$
Let $$a \in I$$, then there exists $$a_1 \in I_1$$ and $$a_2 \in I_2$$ such that $$a = (a_1, a_2) \in I $$
[ ... ... actually $$a_2 = (c_1, c_2) \in R \oplus M$$ but we ignore this complication in order to keep notation simple ... ]Similarly let $$b \in I$$ so $$b = (b_1, b_2) \in I$$ ... ...
Now ... if $$I$$ is a left ideal then
$$a, b \in I \ \Longrightarrow \ a - b \in I$$and
$$r \in A$$ and $$a \in I \ \Longrightarrow \ ra \in I$$

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

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

-----------------------------------------------------------------------------------------------------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