Let $R$ be a commutative ring with unity which is a subring of a commutative ring $S$ with the same unity as that of $R$ . Suppose for every function $f:R \to R , \exists \hat f(X) \in S[X]$ such that $f(r)=\hat f(r) , \forall r \in R$ ; then is $R$ a finite ring ?
Considering the function $f : R \to R$ given by $f(0)=0$ and $f(r)=1 , \forall r \neq 0$ , we see that every $0 \neq r \in R$ has an inverse in $S$ , hence in particular , $R$ is an integral domain . Also , if $S$ were an integral domain , then if $\hat f(X) \in S[X]$ is a polynomial for the function $f : R \to R$ given by $f(0)=0$ and $f(r)=1 , \forall r \neq 0$ , then we see that $\hat f(X) -1$ is a non-zero polynomial with coefficients in an integral domain and vanishing on $R \setminus \{0 \}$ , hence $R$ must be finite . But this works only when $S$ is an integral domain which it necessarily is not . That's about all I am able to say from my condition .
Please help . Thanks in advance
Yes, $R$ must be finite.
I claim more generally that if $R\subset S$ are rings such that every non-zero element of $R$ has an inverse in $S$, then a polynomial with coefficients in $S$ that vanishes on infinitely many elements of $R$ must vanish identically. To prove this, suppose $$ f(x)=\sum_{i=0}^d a_i x^i\in S[x] $$ vanishes on $r_0,\ldots,r_d\in R$. Let $M\in M_{d+1}(R)$ be the Vandermonde matrix whose $(i,j)$ entry for $0\leq i,j\leq d$ is $r_i^j$, and write $v\in S^{d+1}$ for the vector $[a_0,a_1,\ldots,a_d]^T$. The hypothesis $f(r_i)=0$ for all $i$ implies $Mv=\vec{0}$. The determinant of $M$ is a product of pairwise differences of the $r_i$, so the determinant is invertible in $S$. This means the matrix $M$ is invertible in $S$, so we must have $v=\vec{0}$, i.e. $f=0$.