Compatible ring structure on ring-valued set functions

[Kreizhn]

Homework Statement

Let R be a ring and S be any set. Let [itex] R^S [/itex] be the set of set-functions [itex] S \to R [/itex]. Endow [itex] R^S [/itex] with a ring structure such that if S is a singleton, then [itex] R^S [/itex] is just a copy of R.

The Attempt at a Solution

It seems to me that the obvious (and perhaps only?) way to endow [itex] R^S [/itex] with a ring structure that is compatible with R is to define for [itex] \alpha, \beta \in R^s [/itex]
[tex] (\alpha + \beta)(s) = \alpha(s) + \beta(s), \quad (\alpha \cdot \beta)(s) = \alpha(s)\beta(s) [/tex]
And I have checked that this makes [itex] R^S [/itex] a ring with additive and multiplicative identities given by the constant functions
[tex] 0(s) = 0_R, \quad 1(s) = 1_R, \forall s \in S [/tex]

So all that remains to show is that when S is a singleton, then [itex] R^S \cong R [/itex]. 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 [itex] \phi: R \to R^S [/itex] where [itex] \phi(r) [/itex] is the constant function taking all elements of S to r. Namely,
[tex] [\phi(r)](s) = r, \forall s \in S. [/tex]
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 [itex] R^S [/itex] so all that remains is to show that [itex] \phi [/itex] is bijective. It is easily injective, since if [itex] r_1 \neq r_2 [/itex] in R then certainly [itex] \phi(r_1) \neq \phi(r_2) [/itex]. My problem is showing surjectivity.

I know that I can use set-magic to show that [itex] |R^S | = |R| [/itex] when [itex] |S| = 1 [/itex]. 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 [itex] \alpha: S \to R [/itex] 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.

 

[Kreizhn]

Wait, I might have figured it out. Let [itex] \alpha \in R^S [/itex] with S = {s}. Then [itex] \phi [/itex] is surjective since [itex] \phi(\alpha(s)) = \alpha [/itex], yes?

 

[micromass]

Wait, I might have figured it out. Let [itex] \alpha \in R^S [/itex] with S = {s}. Then [itex] \phi [/itex] is surjective since [itex] \phi(\alpha(s)) = \alpha [/itex], yes?

Yes, that is correct!