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I am aware that this question has been asked in other ways such as here:

(1) Show that $R \otimes_R M\cong M$ is isomorphic to $M$, for every left $R$-module $M$, $R$ a ring

(2) Prove that $R \otimes_R M \cong M$

And a related (more general) question is here:

(3) Show that $Hom_R(R^n,M) \cong M^n$ for R-modules

However, I some of the subtleties still seem a little hand-wavy to me, and I would very much appreciate if anyone could shed some light on them. In particular, my approach to try proving this was very similar to what is briefly mentioned (but not actually given in detail) in link (3) above. I said let $f: Hom_{R-Mod}(R,M) \rightarrow M$ be defined by $f(\phi) = \phi(1)$, for $\phi \in Hom_{R-Mod}(R,M)$. I proved f was an injective R-module homomorphism, but proving surjectivity is where I am not quite sure I am satisfied. It is relatively easy to prove that f is surjective onto the group M, but why does this imply it is surjective onto the the R-module M?

I suppose I am a little unclear about just what the "R-module structure" actually IS. For instance, in Aluffi's book, this structure is given as a ring homomorphism $R \rightarrow End_{Ab}(M)$, but yet this doesn't seem to matter when dealing with the surjectivity question? In the event that the order of the group M is finite, say |M|=n, is the set underlying the module M also of the same cardinality n? Why don't we care about what the size of the set $End_{Ab}(M)$ is (since this is what is actually giving M its module structure)?

I'm sorry if this question seems trivial, but I just can't seem to grasp why exactly we can "gloss over" what seems to be the the essential difference between M as a group and as a module.

Thanks for any help!

wanderingmathematician
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  • What books do you follow? – Ninja Nov 23 '16 at 19:08
  • I am using Aluffi's Algebra: Chapter 0 – wanderingmathematician Nov 23 '16 at 19:11
  • Can you sense the difference between $\mathbb R\times \mathbb R$ as a group and as an $\mathbb R$-vector space? Modules are basically the same thing, they're generalizations of vector spaces. – rschwieb Nov 23 '16 at 19:13
  • @rschwieb I actually thought about that after seeing this question, but if I am not mistaken the cardinality of a vector space is defined to be the cardinality of a basis for the vector space, which in the case of $\mathbb{R} \times \mathbb{R}$ is far from the same as the cardinality of the set (and group) $\mathbb{R} \times \mathbb{R}$, so it is not really helping me think about the surjectivity question or what the "cardinality" of the module M would be, and if that is the same as the group M. – wanderingmathematician Nov 23 '16 at 19:17
  • I know modules also are talked about in terms of generating sets, but this still doesn't clarify (and perhaps even worsens) my confusion :( – wanderingmathematician Nov 23 '16 at 19:19
  • @user334137 Cardinality is a poor measure of structure. It only gives information about the underlying sets and ignores the group and vector space structures. A vector space structure is an additional layer of organization on top of that of the group structure. Consider: all subspaces of $\mathbb R^2$ are subgroups of $\mathbb R^2$, but not all subgroups are subspaces. – rschwieb Nov 23 '16 at 20:28
  • @user334137 For surjectivity, given $m \in M$ can you define an $R$-module homomorphism $\varphi: R \to M$ such that $\varphi(1)=m$? I think understanding why a module homomorphism from $R$ to $M$ is completely determined by the image of $1$ will help you understand why we can't ignore the module structure on $R$ and $M$. – Exit path Nov 26 '16 at 21:48

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