Compatible ring structure on ring-valued set functions

[Kreizhn]

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.

 

[Kreizhn]

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

 

[micromass]

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

Yes, that is correct!