We know that for a finite field $F$, every function on $F$ can be given by a polynomial with coefficients in $F$. What about the converse?.. i.e. if $R$ is a ring (not necessarily commutative or with unity) such that every function on $R$ can be given by a polynomial with coefficients in $R$, can we say $R$ is a finite field ? We show that the answer is yes, and that in fact it is enough to only require that all bijections be given by polynomials. If we allow our rings to have unity, we show that the property that all characteristic functions can be given by polynomials actually characterizes finite fields and if we moreover allow our rings to be commutative, then to characterize finite fields, it is enough that some special characteristic function be given by a polynomial (with coefficients even in some extension ring). Motivated by this, we determine all commutative rings with unity which admits a characteristic function which can be given by some polynomial with coefficients in the ring. read more

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Commutative Algebra