Direct product of right ideals is a right ideal

[Math Amateur]

Homework Statement

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:

I 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 ...

Homework Equations

The definition of a right ideal is relevant to this problem ... Bland's definition of a right ideal is as follows:

Also relevant is the definition of a direct product ... Bland's definition for a direct product for modules follows ... simply adjust for rings and ideals ..

The Attempt at a 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

 

[fresh_42]

Yes, it is correct. I don't see any problems in this exercise, as the componentwise definition of the operations will do the job in all of them.

 

[Math Amateur]

Thanks fresh_42 ... that helps !

Peter