MHB Generating/spanning modules and submodules .... .... Blyth Theorem 2.3

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Modules Theorem
Click For Summary
The discussion centers on the proof of Theorem 2.3 from T. S. Blyth's book on module theory, specifically regarding the definition of linear combinations in the context of submodules. The user questions whether a linear combination of elements from the union of submodules should include scalar multipliers, as defined in the text. A hint suggests that if an element belongs to a module, then its scalar multiples also belong to that module, which supports the inclusion of scalars in linear combinations. However, it is noted that the set used in the proof is not a module, complicating the application of this definition. Clarification on these points is sought to better understand the theorem's implications.
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading T. S. Blyth's book: Module Theory: An Approach to Linear Algebra ...

I am focused on Chapter 2: Submodules; intersections and sums ... and need help with the proof of Theorem 2.3 ...

Theorem 2.3 reads as follows:View attachment 8157In the above proof we read the following:

" ... ... A linear combination of elements of $$\bigcup_{ i \in I }$$ is precisely a sum of the form $$\sum_{ j \in J } m_j$$ for some $$J \in P(I).$$ ... ... "But ... Blyth defines a linear combination as in the text below ...https://www.physicsforums.com/attachments/8158So ... given the above definition wouldn't a linear combination of elements of $$\bigcup_{ i \in I } M_i$$ be a sum of the form $$\sum_{ j \in J } \lambda_j m_j$$ ... and not just $$\sum_{ j \in J } m_j$$ ... ... ?
Hope someone can help ...

Peter
 
Physics news on Phys.org
Hint: if $m \in M$ and $M$ is a module, then also $\lambda m \in M$, for $\lambda \in R$.

Thus $\Sigma_{j \in J} \lambda_j m'_j = \Sigma_{j \in J} m_j $ for $m_j = \lambda_j m'_j \in M_j$

However, the set $S$ used above, is not a module ...
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K