- #1

Math Amateur

Gold Member

MHB

- 3,998

- 48

I need someone to check my solution to Problem 2(b) of Problem Set 2.1 ...

Problem 2(b) of Problem Set 2.1 reads as follows:

View attachment 8060

My attempt at a solution follows:We claim that \(\displaystyle \bigoplus_\Delta A_\alpha\) is a right ideal of \(\displaystyle \prod_\Delta R_\alpha\) Proof ... Let \(\displaystyle (x_\alpha ) , (y_\alpha ) \in \bigoplus_\Delta A_\alpha\) and let \(\displaystyle (r_\alpha ) \in \prod_\Delta R_\alpha\)Then \(\displaystyle (x_\alpha ) + (y_\alpha ) = (x_\alpha + y_\alpha )\) ...

... further ... if \(\displaystyle (x_\alpha )\) has \(\displaystyle m\) non-zero components and \(\displaystyle (y_\alpha \)) has \(\displaystyle n\) non-zero components then \(\displaystyle (x_\alpha + y_\alpha )\) has at most \(\displaystyle (m+n)\) non-zero components ... that is \(\displaystyle (x_\alpha + y_\alpha )\) has only a finite number of non-zero components ...

... so ... since each \(\displaystyle x_\alpha + y_\alpha \in A_\alpha\) we have that \(\displaystyle (x_\alpha + y_\alpha ) \in \bigoplus_\Delta A_\alpha\) ...

Hence ... \(\displaystyle (x_\alpha ) + (y_\alpha ) \in \bigoplus_\Delta A_\alpha\) ... ... ... ... ... (1)

Also ... we have ...

\(\displaystyle (x_\alpha ) (r_\alpha ) = (x_\alpha r_\alpha)\)

... and assuming \(\displaystyle x_\alpha\) has \(\displaystyle m\) non-zero components, then \(\displaystyle (x_\alpha r_\alpha\)) has at most \(\displaystyle m\) non-zero components ...... and ...\(\displaystyle x_\alpha r_\alpha \in A_\alpha\) since \(\displaystyle A_\alpha\) is a right ideal of \(\displaystyle R_\alpha\)so \(\displaystyle (x_\alpha r_\alpha) \in \bigoplus_\Delta A_\alpha\)and it follows that \(\displaystyle (x_\alpha) ( r_\alpha) \in \bigoplus_\Delta A_\alpha\) ... ... ... ... ... (2)\(\displaystyle (1) (2) \Longrightarrow \bigoplus_\Delta A_\alpha\) is a right ideal of \(\displaystyle \prod_\Delta R_\alpha \)

Can someone please critique my proof either by confirming it to be correct and/or pointing out errors and shortcomings ...

Peter