MHB Northcott - Sums and Products of Ideals

[Math Amateur]

I am reading D.G. Northcott's book: Lessons on Rings and Modules and Multiplicities.

I am currently studying Chapter 2: Prime Ideals and Primary Submodules.

I need help with a result that Northcott quotes and proves on page 80 regarding sums and products of ideals.

The relevant text from Northcott reads as follows:

View attachment 3729

In the above text we read:

" ... ... Accordingly

$$ A_i B_j \subseteq (A_1 + A_2 + \ ... \ + A_m ) ( B_1 + B_2 + \ ... \ + B_n ) $$

and therefore

$$ \sum_{i,j} A_i B_j \subseteq (A_1 + A_2 + \ ... \ + A_m ) ( B_1 + B_2 + \ ... \ + B_n ) $$ ... ... "Can someone explain (formally and rigorously) exactly why it follows that:

$$\sum_{i,j} A_i B_j \subseteq (A_1 + A_2 + \ ... \ + A_m ) ( B_1 + B_2 + \ ... \ + B_n )$$ ... ... ?Hope someone can help ...

Peter

 

[Euge]

Hi Peter,

The reason why it follows that $\sum_{i,j} A_iB_j \subseteq (A_1 + \cdots + A_m)(B_1 + \cdots + B_n)$ is that $(A_1 + \cdots + A_m)(B_1 + \cdots B_n)$ is closed under addition. Remember, the product of two ideals of a ring is an ideal, so in particular, it is closed under addition. Also note that (finite) sum of ideals of a ring is an ideal -- that's why $A_1 + \cdots + A_m$ and $B_1 + \cdots + B_m$ are ideals of $R$.

 

[Math Amateur]

Hi Peter,

The reason why it follows that $\sum_{i,j} A_iB_j \subseteq (A_1 + \cdots + A_m)(B_1 + \cdots + B_n)$ is that $(A_1 + \cdots + A_m)(B_1 + \cdots B_n)$ is closed under addition. Remember, the product of two ideals of a ring is an ideal, so in particular, it is closed under addition. Also note that (finite) sum of ideals of a ring is an ideal -- that's why $A_1 + \cdots + A_m$ and $B_1 + \cdots + B_m$ are ideals of $R$.

Thanks Euge,

Yes, follow that ... most clear and helpful ...

Thanks again,

Peter