Proving HomR(F;M) isomorphic to M^n for Free Modules of Rank n < 1

In summary, HomR(F;M) is the set of R-linear maps from the free module F to the module M, indicating functions that preserve the module structure between the two. When HomR(F;M) is isomorphic to M^n, there exists a bijective linear map between the two sets, showing they have the same size and structure. To prove this isomorphism, a bijective linear map must be constructed that preserves the module structure. Proving this isomorphism provides a deeper understanding of abstract algebra and linear algebra. Any free modules of rank n < 1 can be isomorphic to M^n, as long as the two sets have the same size and structure.
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View attachment Exampleoflatex.pdf

Let R be a commutative ring with 1. If F is a free module of rank n < 1, then show that
HomR(F;M) is isomorphic to M^n, for each R-module M.

I was thinking about defining a map
Psi : HomR(F;M)--> M^n by psi(f) = (f(e1); f(e2); ... ; f(en))
where F is free on (e1; ... ; en) and
show Psi is an isomorphism. But I am having difficulties showing it is onto.
 
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  • #2
Oh, I think I can use the Universal property of free modules to get the onto part.
 

1. What is HomR(F;M)?

HomR(F;M) is the set of all R-linear maps from the free module F to the module M. In other words, it is the set of all functions that preserve the module structure between F and M.

2. What does it mean for HomR(F;M) to be isomorphic to M^n?

When HomR(F;M) is isomorphic to M^n, it means that there exists a bijective linear map (a map that preserves the module structure) from HomR(F;M) to M^n. This indicates that the two sets have the same size and structure.

3. How do we prove that HomR(F;M) is isomorphic to M^n?

To prove that HomR(F;M) is isomorphic to M^n, we need to show that there exists a bijective linear map between the two sets. This can be done by constructing a map that is both injective (one-to-one) and surjective (onto) and showing that it preserves the module structure.

4. What is the significance of proving HomR(F;M) isomorphic to M^n?

Proving that HomR(F;M) is isomorphic to M^n allows us to understand the structure of free modules of rank n < 1. It also provides a deeper understanding of the concept of isomorphism and its applications in abstract algebra and linear algebra.

5. Can HomR(F;M) be isomorphic to M^n for any free modules of rank n < 1?

Yes, HomR(F;M) can be isomorphic to M^n for any free modules of rank n < 1. This is because the definition of isomorphism depends on the structure and size of the two sets, and not on the specific elements within the sets. As long as the two sets have the same size and structure, they can be proven to be isomorphic.

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