I'm working in the context of $R$-modules where $R$ is a commutative ring. I can see a similar statement is true for $\text{Hom}_R(R, R \oplus R)$ using the fact that $\text{Hom}_R(R, M) \simeq M$ for any $R$-modules with the isomorphism $f \mapsto f(1)$, but I don't know how to prove it the other way around.
Asked
Active
Viewed 92 times
0
-
[Comment changed to an answer.] – Suzet Dec 09 '22 at 05:28
-
@Suzet -- is there a reason you left that as a comment rather than an answer? – HallaSurvivor Dec 09 '22 at 05:28
-
1@HallaSurvivor True, let me move it to an actual answer! – Suzet Dec 09 '22 at 05:29
1 Answers
1
The two modules are isomorphic. If $M$ is an $R$-module, to $f \in \mathrm{Hom}_R(R\oplus R,M)$, for $i = 0,1$ you can define the maps $f_i:R\to M$ given by $f_0(x) = f(x,0)$ and $f_1(x) = f(0,x)$.
Then $f(x,y) = f(x,0) + f(0,y)$, and since each $f_i$ is an element of $\mathrm{Hom}_R(R,M)$, as you have pointed out, they are entirely determined by their image of $1$. To sum up, the assignment $$f \mapsto (f_0(1),f_1(1)) = (f(1,0),f(0,1))$$ is the isomorphism you are looking for.
Suzet
- 5,482
- 12
- 35