Divisors of zero in polynomial ring I have the following theorem: 

McCoy: Let $R$ be a commutative ring with identity. If $f=\sum_{i=0}^na_iX^i$ is a zero divisor in $R[X]$, then there exists a nonzero $c$ in $R$ such that $cf=0$.

Ok. But if we consider the field $\mathbb{Z}_7$ and the polynomials $f=X^6-\hat{1}, g=X$ in $\mathbb{Z}_7[X]$, we see that $f$ is nonzero and non constant and $fg=X^7-X$, which is $\hat{0}$ by Fermat's Little Theorem (thus $f$ is a zero divisor). It is easy to see that there is no $c\in\mathbb{Z}_7, c\neq \hat{0}$ such that $cf=0$. What is wrong here? 
 A: What is wrong is this: there is a difference between evaluating to zero at all points (as does $X^7-X$), and being zero. 
It is only over infinite fields that evaluating to zero at all points implies that the polynomial itself is zero. Your polynomial is not equal to zero, because $X$ and $X^7$ are different in $\mathbb Z_7[X]$. In fact, one of the definitions of $k[X]$ is the free algebra generated by the symbol $X$, meaning that $X$ satisfies no relations whatsover (except those required of it for $k[X]$ to still be a ring, like $X^a X^b = X^{a+b}$).
A: As noted above, this is a confusion between the polynomial and the function defined by the polynomial.
In particular, if $i:R_1\to R_2$ is an inclusion of rings (or any homomorphism, really), under you definition, there is no homomorphism between $R_1[x]\to R_2[x]$. That's because a polynomial that evaluates to zero on all of $R_1$ does not necessarily evaluate to zero on all of $R_2$.
For example, given the inclusion $\mathbb F_7\to \mathbb F_{49}$, it is not true that in $\mathbb F_{49}$ that $x^7-x$ always evaluates to zero.
So that means that there would be no inclusion $\mathbb F_7[x]\to\mathbb F_{49}[x]$.
If you must think of polynomials over a ring $R$ as evaluation functions, think of them as corresponding to the collection of evaluation functions on all $S$ given $R\to S$.
