- #1
Math Amateur
Gold Member
MHB
- 3,998
- 48
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, [itex] Hom_R (R, X) \cong X [/itex], the isomorphism being given by mapping a homomorphism to its value on the element [itex]1 \in R [/itex]"
I am having some trouble in demonstrating the isomorphism involved in the relationship [itex] Hom_R (R, X) \cong X [/itex].
To demonstrate the isomorphism I proceeded as follows:
Let [itex]f, g \in Hom_R (R,X) [/itex] so [itex] f,g : \ R \to X [/itex]
Consider [itex] \theta \ : \ Hom_R (R,X) \to X [/itex]
where [itex] \theta (f) = f(1_R) [/itex]
To show [itex] \theta [/itex] is a homomorphism we proceed as follows:
[itex] \theta (f + g) = (f + g)(1_R) = f(1_R) + g(1_R) [/itex]
[itex] = \theta (f) + \theta (g) [/itex]
and
[itex] \theta (rf) = (rf) = rf (1_R) = r \theta (f)[/itex] where [itex] r \in R [/itex]
Then I need to show [itex] \theta [/itex] is injective and surjective.
BUT ... I am having problems in demonstrating that [itex] \theta [/itex] is injective ... can someone help me with this task?Note that I suspect one would proceed as follows:
Suppose we have [itex] f, g \in Hom_R (R,X) [/itex] such that:
[itex] \theta (f) = f(1_R) [/itex] and [itex] \theta (g) = f(1_R) [/itex]
Now we have, of course, by definition of g, that [itex] \theta (g) = g(1_R) [/itex]
So [itex] f(1_R) = g(1_R) [/itex] ... but how do we proceed from here to show that f = g?
Hope someone can help.
Peter
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, [itex] Hom_R (R, X) \cong X [/itex], the isomorphism being given by mapping a homomorphism to its value on the element [itex]1 \in R [/itex]"
I am having some trouble in demonstrating the isomorphism involved in the relationship [itex] Hom_R (R, X) \cong X [/itex].
To demonstrate the isomorphism I proceeded as follows:
Let [itex]f, g \in Hom_R (R,X) [/itex] so [itex] f,g : \ R \to X [/itex]
Consider [itex] \theta \ : \ Hom_R (R,X) \to X [/itex]
where [itex] \theta (f) = f(1_R) [/itex]
To show [itex] \theta [/itex] is a homomorphism we proceed as follows:
[itex] \theta (f + g) = (f + g)(1_R) = f(1_R) + g(1_R) [/itex]
[itex] = \theta (f) + \theta (g) [/itex]
and
[itex] \theta (rf) = (rf) = rf (1_R) = r \theta (f)[/itex] where [itex] r \in R [/itex]
Then I need to show [itex] \theta [/itex] is injective and surjective.
BUT ... I am having problems in demonstrating that [itex] \theta [/itex] is injective ... can someone help me with this task?Note that I suspect one would proceed as follows:
Suppose we have [itex] f, g \in Hom_R (R,X) [/itex] such that:
[itex] \theta (f) = f(1_R) [/itex] and [itex] \theta (g) = f(1_R) [/itex]
Now we have, of course, by definition of g, that [itex] \theta (g) = g(1_R) [/itex]
So [itex] f(1_R) = g(1_R) [/itex] ... but how do we proceed from here to show that f = g?
Hope someone can help.
Peter