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.
 A: Suppose M has at least two distinct elements : $m_1$ and $m_2$.
Then, consider $f :M\to R$ such that $f(m_1)=1$ and $f(m)= 0$ otherwise; and $g:M\to R$ such that $g(m_2)=1$ and $g(m) =0$ otherwise.
Then, as functions (that is, as elements of $F$), $f$ and $g$ are non zero but $fg=0$.
So $F$ can't be a field.
A: 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. 
A: Hint  If $\mid M\mid\ge2$, then it is easy to construct two nonzero functions whose product is zero.  Thus $\mathcal F$ is not an integral domain. 
If $M=\emptyset$, then $\mathcal F$ has only one element. 
If $R$ is not a field,  take a nonunit $0\neq r\in R$.  Consider the function $f(p)= r$, where $M=\{p\}$.  Then $0\neq f\in \mathcal F$ and $f$ isn't a unit. 
