Northcott - Sums and Products of Ideals

  • MHB
  • Thread starter Math Amateur
  • Start date
  • Tags
    Sums
In summary, Northcott explains that the product of two ideals is an ideal, which means it is closed under addition. Additionally, the finite sum of ideals in a ring is also an ideal, which is why $(A_1 + \cdots + A_m)(B_1 + \cdots + B_n)$ is an ideal of $R$. This helps to understand why $\sum_{i,j} A_iB_j \subseteq (A_1 + \cdots + A_m)(B_1 + \cdots + B_n)$.
  • #1
Math Amateur
Gold Member
MHB
3,998
48
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

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

and therefore

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

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

Peter
 
Physics news on Phys.org
  • #2
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 $R$.
 
  • #3
Euge said:
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
 

FAQ: Northcott - Sums and Products of Ideals

1. What is the purpose of Northcott's paper on sums and products of ideals?

The purpose of Northcott's paper is to provide a systematic study of the sums and products of ideals in a commutative ring. This allows for a better understanding of the algebraic structure of commutative rings and their ideals.

2. How does Northcott define sums and products of ideals?

Northcott defines the sum of two ideals as the set of all elements that can be written as the sum of an element from each ideal. The product of two ideals is defined as the set of all elements that can be written as a finite sum of products of elements from each ideal.

3. What is the significance of Northcott's paper?

Northcott's paper is significant because it provides a comprehensive and rigorous understanding of the algebraic properties of sums and products of ideals. This has applications in many areas of mathematics, including algebraic geometry and number theory.

4. How does Northcott's paper relate to other mathematical theories?

Northcott's paper builds upon existing theories in commutative algebra and extends them to the study of sums and products of ideals. It also has connections to other areas of mathematics, such as ring theory and homological algebra.

5. What are some applications of Northcott's results?

Some applications of Northcott's results include the study of polynomial rings, algebraic curves, and algebraic number theory. These results also have applications in coding theory, cryptography, and computer science.

Similar threads

Replies
4
Views
2K
Replies
4
Views
1K
Replies
23
Views
1K
Replies
4
Views
2K
Replies
4
Views
1K
Back
Top