MHB Demonstrating that a mapping is injective

[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)

In the latter stages of the proof of Proposition 28 we find the following statement (top of page 388):

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

I am having some trouble in demonstrating the isomorphism involved in the relationship $$ Hom_R (R, X) \cong X $$.

To demonstrate the isomorphism I proceeded as follows:

Let $$f, g \in Hom_R (R,X) $$ so $$ f,g : \ R \to X $$

Consider $$ \theta \ : \ Hom_R (R,X) \to X $$

where $$ \theta (f) = f(1_R) $$

To show $$ \theta $$ is a homomorphism we proceed as follows:

$$ \theta (f + g) = (f + g)(1_R) = f(1_R) + g(1_R) $$

$$ = \theta (f) + \theta (g) $$

and

$$ \theta (rf) = (rf) = rf (1_R) = r \theta (f) $$ where $$ r \in R $$

Then I need to show $$ \theta $$ is injective and surjective.

BUT ... I am having problems in demonstrating that $$ \theta $$ is injective ... can someone help me with this task?Note that I suspect one would proceed as follows:

Suppose we have $$ f, g \in Hom_R (R,X) $$ such that:

$$ \theta (f) = f(1_R) $$ and $$ \theta (g) = f(1_R) $$

Now we have, of course, by definition of g, that $$ \theta (g) = g(1_R) $$

So $$ f(1_R) = g(1_R) $$ ... but how do we proceed from here to show that f = g?

Hope someone can help.

Peter

 

[Deveno]

Suppose $f \in \text{ker }\theta$.

By definition, this means:

$f(1_R) = 0_X$.

Hence, for any $r \in R$, we have:

$f(r) = f(r\cdot1_R) = r\cdot f(1_R) = r\cdot 0_X= 0_X$

which show the kernel consists solely of the 0-map, and is thus injective.

It seems to me surjectivity is more of a problem, we need to prove there IS $f \in \text{Hom}_R(R,X)$ with:

$f(1_R) = x$, for every $x \in X$. I suggest looking at the submodule generated by $x$.

 

[Math Amateur]

Suppose $f \in \text{ker }\theta$.

By definition, this means:

$f(1_R) = 0_X$.

Hence, for any $r \in R$, we have:

$f(r) = f(r\cdot1_R) = r\cdot f(1_R) = r\cdot 0_X= 0_X$

which show the kernel consists solely of the 0-map, and is thus injective.

It seems to me surjectivity is more of a problem, we need to prove there IS $f \in \text{Hom}_R(R,X)$ with:

$f(1_R) = x$, for every $x \in X$. I suggest looking at the submodule generated by $x$.

Thanks Deveno ...

You are certainly right that proving $$ \theta $$ is surjective is a problem ... I am having problems proving it ... even given your suggestion ...

Can you help?

Peter

 

[Deveno]

There is a reason that I suggested the submodule generated by $x$.

Note that if $Y \subseteq X$ is a submodule, then $\text{Hom}_R(R,Y) \subseteq \text{Hom}_R(R,X)$.

The obvious candidate is: $g(r) = r\cdot x$, which maps $R \to \langle x\rangle$.

 

[Math Amateur]

There is a reason that I suggested the submodule generated by $x$.

Note that if $Y \subseteq X$ is a submodule, then $\text{Hom}_R(R,Y) \subseteq \text{Hom}_R(R,X)$.

The obvious candidate is: $g(r) = r\cdot x$, which maps $R \to \langle x\rangle$.

Thanks Deveno ... Reflecting on your post now ... ...

Peter