Understanding Proposition 2.1.1 in Paul E. Bland's Rings & Modules

In summary, Bland's Proposition 2.1.1 states that any $g$ satisfying the condition $\pi_\alpha \circ g = f_\alpha$ is equal to $f$, and this can be proven by taking a specific $g$ and showing that it is equal to $f$.
  • #1
Math Amateur
Gold Member
MHB
3,996
48
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 another aspect of the proof of Proposition 2.1.1 ...

Proposition 2.1.1 and its proof read as follows:
https://www.physicsforums.com/attachments/8032In the above proof by Paul Bland we read the following:

" ... ... suppose that \(\displaystyle g \ : \ N \rightarrow \prod_\Delta M_\alpha\) is also an \(\displaystyle R\)-linear mapping such that \(\displaystyle \pi_\alpha g = f_\alpha\) for each \(\displaystyle \alpha \in \Delta\). If g(x) = ( x_\alpha ) ... ... "When Bland puts \(\displaystyle g(x) = ( x_\alpha )\) he seems to be specifying a particular \(\displaystyle g\) and then proves \(\displaystyle f = g\) ... ... I thought he was proving that for any g such that \(\displaystyle \pi_\alpha g = f_\alpha\) we have \(\displaystyle f = g\) ... can someone please clarify ... ?Help will be much appreciated ... ...Peter
======================================================================================The above post mentions but does not define \(\displaystyle f\) ... Bland's definition of \(\displaystyle f\) is as follows:
View attachment 8033
Hope that helps ...

Peter
 
Physics news on Phys.org
  • #2
Bland is saying that any $g$ that satisfies the condition $\pi_\alpha \circ g = f_\alpha$ must be equal to $f$.
So take a $g$ that satisfies the condition and prove that $g=f$, that’s all.

So take such a $g$ then $g$ is an R-map $g:N \longrightarrow \prod_\Delta M_\alpha$, take $x\in N$, then $g(x) \in \prod_\Delta M_\alpha$.
This means that $g(x)$ has the form $g(x)=(x_\alpha)$, where $x_\alpha$ is the coordinate of $g(x)$ in $M_\alpha$, for all $\alpha \in \Delta$, and so on.
 

Related to Understanding Proposition 2.1.1 in Paul E. Bland's Rings & Modules

1. What is Proposition 2.1.1 in Paul E. Bland's Rings & Modules?

Proposition 2.1.1 is a mathematical statement in the field of abstract algebra that deals with the properties of rings and modules. It states that if a ring is left Noetherian, then every finitely generated left module over that ring is also Noetherian.

2. Why is Proposition 2.1.1 important?

Proposition 2.1.1 is important because it provides a useful tool for analyzing the structure of rings and modules. It allows us to simplify complex structures and make predictions about their properties based on the properties of the underlying ring.

3. How is Proposition 2.1.1 proven?

Proposition 2.1.1 is proven using a combination of definitions, theorems, and logical reasoning. It relies on the definition of a Noetherian ring and module, as well as theorems about finitely generated modules and their properties.

4. What are some real-world applications of Proposition 2.1.1?

Proposition 2.1.1 has various applications in mathematics, computer science, and other fields. One example is in coding theory, where it is used to construct error-correcting codes. It is also used in algebraic geometry to study the structure of algebraic varieties.

5. Are there any limitations to Proposition 2.1.1?

Yes, there are limitations to Proposition 2.1.1. It only applies to left modules over left Noetherian rings, so it cannot be used in other contexts. Also, it does not provide a complete understanding of the structure of rings and modules, as there are other important theorems and concepts to consider.

Similar threads

  • Linear and Abstract Algebra
Replies
2
Views
988
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
1K
Replies
4
Views
1K
Replies
1
Views
921
  • Linear and Abstract Algebra
Replies
12
Views
2K
Replies
4
Views
1K
Replies
3
Views
1K
Replies
2
Views
2K
Replies
6
Views
1K
Back
Top