MHB Exact Sequences - Dummit and Foote Ch 10 - Proposition 28

[Math Amateur]

I am reading Dummit and Foote, Section 10.5 : Exact Sequences - Projective, Injective and Flat Modules.

I am studying Proposition 28 (D&F pages 387 - 388)

I need some help in order to fully understand the proof of the last statement of Proposition 28.

Proposition 28 (Ch 10, D&F pages 387-388) reads as follows:

View attachment 2487The proof of the last statement of the proposition reads as follows:View attachment 2488
View attachment 2489

In the proof, after the statement:

"In general, $$ Hom_R (R, X) \cong X $$, the isomorphism being given by mapping a homomorphism to its value on the element $$1 \in R $$"

D&F write:

Taking D = R in (10), the exactness of the sequence:

$$ 0 \longrightarrow L \stackrel{\psi}{\longrightarrow} M \stackrel{\psi}{\longrightarrow} N $$

follows easily.

Can someone please explain how/why this follows easily? ... ... ... that is how/why does taking D=R in (10) lead easily to the result that the sequence $$ 0 \longrightarrow L \stackrel{\psi}{\longrightarrow} M \stackrel{\psi}{\longrightarrow} N $$ is exact?Further, it seems strange to me that in an exact sequence of modules we are putting D = R where (I assume) R is a ring. (Yes, I understand that a ring can be viewed as a module over itself, but it seems it is a special case, which we are using to establish a general result?) Can someone please clarify this issue?Hope someone can help.

Peter

 

[Deveno]

If we assume that:

$0 \to \text{Hom}_R(D,L) \stackrel{\psi'}{\to} \text{Hom}_R(D,M) \stackrel{\phi'}{\to} \text{Hom}_R(D,N)$

for ANY $D$ is exact, then taking $D=R$ and using the fact that $\text{Hom}_R(R,X) \cong X$ for $X = L,M,N$

makes this obvious.

Let's put this in context. If we start with the exact sequence of modules, we can form (for any particular module $D$) the sequence of hom-sets. The sequence of hom-sets has DERIVED properties from the original sequence.

Now, we get a LOT of sequences from this (one for every different $D$). The "special case" $D = R$ picks out a particular sequence that mimics our original sequence in every detail.

In other words, to make the "return" trip of a one-to-many correspondence, we ought to pick a "special" (distinguished) element in the image.

 

[Math Amateur]

If we assume that:

$0 \to \text{Hom}_R(D,L) \stackrel{\psi'}{\to} \text{Hom}_R(D,M) \stackrel{\phi'}{\to} \text{Hom}_R(D,N)$

for ANY $D$ is exact, then taking $D=R$ and using the fact that $\text{Hom}_R(R,X) \cong X$ for $X = L,M,N$

makes this obvious.

Let's put this in context. If we start with the exact sequence of modules, we can form (for any particular module $D$) the sequence of hom-sets. The sequence of hom-sets has DERIVED properties from the original sequence.

Now, we get a LOT of sequences from this (one for every different $D$). The "special case" $D = R$ picks out a particular sequence that mimics our original sequence in every detail.

In other words, to make the "return" trip of a one-to-many correspondence, we ought to pick a "special" (distinguished) element in the image.

Oh ... neat ... yes, solution was staring at me! ...

Thanks so much for the help ...

Peter