MHB Help with Problem 2(a), Rings & Modules, Paul E. Bland

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need help with Problem 2(a) of Problem Set 2.1 ...

Problem 2(a) of Problem Set 2.1 reads as follows:
View attachment 8049I am unsure of my solution to problem 2(a) and need help in the following way ...

... could someone please confirm my solution is correct and/or point out errors and shortcomings ...

... indeed I would be grateful if someone could critique my solution ...
My attempted solution to problem 2(a) is as follows:... we have to show that $$\prod_\Delta A_\alpha$$ is a right ideal of $$\prod_\Delta R_\alpha$$ ...

To demonstrate this we have to show that $$\prod_\Delta A_\alpha$$ is closed under addition and closed under multiplication on the right by an element of $$\prod_\Delta R_\alpha$$ ...So ... let $$(x_\alpha), (y_\alpha) \in \prod_\Delta A_\alpha$$ and $$(r_\alpha) \in \prod_\Delta R_\alpha$$

Then $$x_\alpha, y_\alpha \in A_\alpha$$ for all $$\alpha \in \Delta$$

$$\Longrightarrow x_\alpha + y_\alpha \in A_\alpha$$ since $$A_\alpha$$ is a right ideal of $$R_\alpha$$ for all $$\alpha \in \Delta$$ ...

$$\Longrightarrow (x_\alpha) + (y_\alpha) \in \prod_\Delta A_\alpha$$

$$\Longrightarrow \prod_\Delta A_\alpha$$ is closed under addition ...
Now ... $$(x_\alpha) \in \prod_\Delta A_\alpha , (r_\alpha) \in \prod_\Delta R_\alpha$$

$$\Longrightarrow x_\alpha \in A_\alpha , r_\alpha \in R_\alpha$$ for all $$\alpha \in \Delta$$ ...

$$\Longrightarrow x_\alpha r_\alpha \in A_\alpha$$ since $$A_\alpha$$ is a right ideal of $$R_\alpha$$ ...

$$\Longrightarrow ( x_\alpha r_\alpha ) \in \prod_\Delta A_\alpha$$Thus $$\prod_\Delta A_\alpha$$ is a right ideal of $$\prod_\Delta R_\alpha$$ ...

Hope the above is correct ...

Peter

 

[steenis]

Correct, but one remark:

Because $x_\alpha + y_\alpha \in A_\alpha$

we have $(x_\alpha + y_\alpha) \in \prod_\Delta A_\alpha$

and $(x_\alpha + y_\alpha) = (x_\alpha) + (y_\alpha)$

and therefore, and so on

 

[Math Amateur]

Correct, but one remark:

Because $x_\alpha + y_\alpha \in A_\alpha$

we have $(x_\alpha + y_\alpha) \in \prod_\Delta A_\alpha$

and $(x_\alpha + y_\alpha) = (x_\alpha) + (y_\alpha)$

and therefore, and so on

Thanks steenis ... appreciate the help and guidance ...

Peter