Exercise in abstract algebra Assume that $M$ is a set, and that $R$ is a ring. Let $\cal{F}$ be the collection all functions $M\rightarrow{R}$. Prove that the following two statements are equivalent.


*

*$\cal{F}$ is a field.

*$M$ is a singleton (i.e. consisting one element), and $R$ is a field.
The implication 2. $\implies$ 1. is straightforward. If $M=\{m\}$, one identifies any particular element $r\in{R}$ with the function $f(m) = r$, and checks that $\cal{F}$ satisfies the field axioms. But I am stuck on the implication 1. $\implies$ 2.

If $R$ is not a field and $M$ is a singleton, then clearly $\mathcal{F}\cong R$, as you've already shown. 
If $M$ is not a singleton, then either it's the empty set (in which case, $\mathcal{F}$ is the trivial ring, which is not a field), or it contains two distinct elements $a$ and $b$. Thus, there exist $f,g\in\mathcal{F}$ such that $f(a) = g(b) = 1$, and all other images are zero. Then clearly, $f$ and $g$ are non-zero, but $fg = 0$, hence $\mathcal{F}$ is not a field. 
