An exercise asks me to describe this ring:
Let $R$ be a ring and $X$ a non-empty set. Consider $A=R^X$ with sum and multiplication defined by:
$\forall \,\,f,g\in A, \forall x\in X$: $\,\,\,\,(f+g)(x)=f(x)+g(x)$; $\,\,\,\,\,(f\cdot g)(x)=f(x)\cdot g(x)$.
Now $(A,+,\cdot)$ is a commutative ring. Identities are $c_0(x)=0_X$ and $c_1(x)=1_X, \forall x\in X$ and, supposing to know the invertible elements of $X$, those of $A$ are the set of applications which don't assume zero as a value.
About of zero divisors of $A$:
considering an application $f \in A, \,\,f\neq c_0$ such that $f(x)=0$ for some $x\in X$ and $f(\bar x)\neq 0, \,\,\, \forall \bar x \in X$ with $\bar x \neq x$. I can assume that exists an application $g \in A$ such that: $$ g(y) = \begin{cases} 0 &\text{if}\;\;\; y \neq x \\ 0 \neq k \in X &\text{if}\;\;\; y = x \end{cases}$$ so $f,g \neq c_0$ and $f \cdot g = c_0$. Hence can I say that the set of zero divisors of $A$ is composed by all the applications which assume $0$ as a value?
Let $D$ be that set, is $D$ an ideal of $A$? I am not sure that I can suppose $\forall f,g \in D$ that $(f-g)\in D$ where for $f \cdot g$ is trivial.
Lastly, agreed upon $X=\{0,1\}$, show an isomorphism between $R^X$ and $R \times R$:
Can I consider $\lambda_a :X \to R$ such that $\lambda_a (x) = a \cdot x$, $\,\, \forall x \in X$, $\forall a\in R\,\,\,\,$ and $\,\,\,\, \phi : R^X \to R \times R$ such that $\phi(\lambda_a) = (a \cdot x,a \cdot x)$? Or do I have a loss of informations?
1) The zero divisors of $R^X$ are those functions which either vanishes somewhere or admits a zero divisor as value.
2) $D$ is not stable under addition. Take for example $$f(x)=\begin{cases} 1 \; \text{if } x\in A \\ 0 \; \text{otherwise} \end{cases} \qquad \text{and} \qquad g(x)=\begin{cases} 1 \; \text{if } x\in \bar{A} \\ 0 \; \text{otherwise} \end{cases},$$ for some $\emptyset \subsetneq A \subsetneq X.$ Then $f,g\in D$ while $f+g\notin D.$
3) A function in $f\in R^X$ is uniquely determined by its values on $\{0,1\}$. So if $f(0)=x$ and $f(1)=y$ you can define $\phi : R^X \longrightarrow R \times R, \, \phi(f)=(x,y) .$ This is a ring homomorphism from the definition of $R^X$, it has inverse $\psi : R \times R \longrightarrow R^X, (x,y)\mapsto (f: f(0)=x, f(1)=y)$.