- #1
Kreizhn
- 714
- 1
Homework Statement
Let R be a ring and S be any set. Let R^S be the set of set-functions S \to R. Endow R^S with a ring structure such that if S is a singleton, then R^S is just a copy of R.
The Attempt at a Solution
It seems to me that the obvious (and perhaps only?) way to endow R^S with a ring structure that is compatible with R is to define for \alpha, \beta \in R^s
(\alpha + \beta)(s) = \alpha(s) + \beta(s), \quad (\alpha \cdot \beta)(s) = \alpha(s)\beta(s)
And I have checked that this makes R^S a ring with additive and multiplicative identities given by the constant functions
0(s) = 0_R, \quad 1(s) = 1_R, \forall s \in S
So all that remains to show is that when S is a singleton, then R^S \cong R. Intuitively, I think I know how to do this, but I'm having trouble formalizing. It seems to me that the easiest way to do this is to give the mapping \phi: R \to R^S where \phi(r) is the constant function taking all elements of S to r. Namely,
[\phi(r)](s) = r, \forall s \in S.
Now I want to show that this is an isomorphism. The preservation of the ring structure is simple and follows from definition of the ring structure on R^S so all that remains is to show that \phi is bijective. It is easily injective, since if r_1 \neq r_2 in R then certainly \phi(r_1) \neq \phi(r_2). My problem is showing surjectivity.
I know that I can use set-magic to show that |R^S | = |R| when |S| = 1. I also know that if I can show that phi has a right-inverse, it will be surjective. I'm just stuck here and not sure how to proceed. Is this just vacuously true? That is, since the domain of the function \alpha: S \to R is a singleton, do all functions just look like constant functions and I can claim I'm done? This just seems a little shaky, so I want to make it more sound.