Let $R$ be a commutative ring with unity and let $S$ be a multiplicatively closed subset of $R$ such that $S$ contains no zero divisor . So the canonical map $f : R \to S^{-1}R$ is invective , hence w.l.o.g. , let us assume $R$ is a subring of $S^{-1}R$ . Now if for every $f :R \to R , \exists \hat f (x) \in (S^{-1}R)[x]$ such that $f(r)=\hat f (r) , \forall r \in R$ , then is it true that $R$ is a field ? 


Comparing cardinality , I can easily show that under the condition I have given , $R$ is finite , so equivalently I am asking :


Is $R$ an integral domain ? 


[I can show that $R$ is a field only for the case $S=\{1\}$ ; but that's trivial . ]



Functions on rings and polynomials with coefficients in a certain kind of localisation
Yes (assuming $R$ is nonzero).


Suppose $a,b\in R\setminus\{0\}$ satisfy $ab=0$. Suppose further that $f=\frac{r\_n}{s\_n}x^n+\cdots+\frac{r\_1}{s\_1}x+\frac{r\_0}{s\_0}\in (S^{-1}R)[x]$ satisfies $f(0)=1$ and $f(a)=f(b)=0$. In particular $r\_0/s\_0=1$. Then $0=f(a)f(b)=f(a)+f(b)-1=-1$, because all terms in the product containing both an $a$ and a $b$ will be $0$.


