Module action sends zero to zero?

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SUMMARY

In the discussion, the user seeks to prove that for any ring R and R-module M, the product of any element r in R with the zero element of M results in zero (r0 = 0). The proof is established by demonstrating that r0 can be expressed as r(0 + 0), which simplifies to r0 + r0. By canceling r0 from both sides, it is confirmed that 0 = r0, thus validating the initial claim. This fundamental property of modules is crucial for understanding module theory in abstract algebra.

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  • Understanding of ring theory and R-modules
  • Familiarity with abelian groups
  • Basic knowledge of algebraic structures and operations
  • Proficiency in mathematical proofs and cancellation laws
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  • Study the properties of R-modules in detail
  • Explore the implications of zero elements in algebraic structures
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  • Investigate the relationship between rings and their modules
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Mathematicians, students of abstract algebra, and anyone interested in the foundational concepts of module theory and ring theory.

geor
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Hello,

Excuse me if I ask something obvious here, but I can't
prove it..

Say R is a ring and M an R-module.
So, 0 \in M as M is an abelian group.

I guess r0=0, no?!

Can you tell me how we prove that?

Again, sorry if I ask something obvious..
 
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It's true:

r0 = r(0 + 0) = r0 + r0. Canceling r0 from both sides gives 0 = r0.
 
adriank said:
It's true:

r0 = r(0 + 0) = r0 + r0. Canceling r0 from both sides gives 0 = r0.

Oh, yes.. :shy:

Thanks a lot!
 

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