Let $R$ be an infinite, characteristic zero, commutative ring. I can furthermore suppose it is reduced and indecomposable (no nontrivial nilpotents and idempotents).
My question is whether there is a nonzero polynomial $f\in R[x]$ which is identically zero on $R$, when $R$ has no nontrivial nilpotents or idempotents.
Note: it is easy to show that there are polynomials with infinitely many roots, let $R=\mathbb Z[s]/(2s)$ and consider $f\in R[x]$ given by $f(x)=sx^2+sx$. All integers are roots of $f$.
But my question is whether we can have $f$ vanishing on all of $R$, not just on an infinite subset.
On the other hand, if we further mod out by $s^2$, turning $f$ (I believe) identically vanishing, we create a nilpotent element.
A technique that I tried is trying and produce a Vandermonde matrix $V$ associated to the elements of $a_1,...,a_k$ of $R$ that be a nonzero element, so that $V$ multiplied by the matrix of coefficients of the canonical basis $e_i$ of polynomials of degree up to $k$, with the its $i$-th element replaced by the coefficients of $f$, $f_i$'s, would have two proportional columns and yield, $\det V\,f_i=0$ and therefore, if I can manage to make $\det V$ regular, I will get $f_i=0$.
But I believe we may have reduced rings where all elements are zerodivisors, so I am currently trying to modify this Vandermonde argument, by using the very coefficients of $f$ as $a_i$'s and create a nice Vandermonde lattice.
No. The existence of more roots than the degree of the polynomial necessarily implies the existence of zero divisors: Let f(x) have degree n, and let $r_1$, $r_2$, ..., $r_n$ be zeros of f(x). Then f(x) factors thus: (x - $r_1$)(x - $r_2$)...(x - $r_n$) [at least in some extension field of R]. If $a$ is another zero [i.e one different from $r_1$, $r_2$, ..., $r_n$], then in R, ($a$ - $r_1$)($a$ - $r_2$)...($a$ - $r_n$) = 0, whereas none of the parenthesized factors is zero.
Or at least, I THOUGHT characteristic zero precluded zero-divisors.