The content of a polynomial vs the ideal of its values Let $f(x) = \sum_i a_i x^i$ be a degree $d$ polynomial over some ring $A$. Define the content of $f$ to be the ideal:
$$c(f) = (a_0,\dots,a_d).$$
One can ask for the relation of the above ideal to the ideal:
$$v(f) = (f(a): a \in A.)$$
It is clear that $v(f) \subset c(f).$ It is also well known that the inclusion can be proper. Take $A = \Bbb Z$ and $f = x^2+x$. Then $c(f) = (1)$ while $v(f) = (2)$. Is there a "nice" description of rings where $c(f) = v(f)$?
 A: As you correctly observed, the condition which is necessary and sufficient is that the cardinality of every residue field of $A$ at maximal ideals is greater than $n=\deg f$. Your example gives the necessity. 
To prove sufficiency, first, we may go modulo $\nu(f)$ and then we want to show that $c(f)=0$. Thus, we may assume $f(a)=0$ for all $a\in A$. Notice that the constant term of $f$ must therefore be zero, since it is $f(0)$. So, we may write $f(x)=xg(x)$ with $\deg g=n-1$. We show that $c(f)=0$ when we localize at any maximal ideal which will prove what we need. Given any maximal ideal $\mathfrak{m}$, by our hypothesis, we can find $t_1,\ldots, t_n\in A$, so that their images in  $A/\mathfrak{m}$ are non-zero and distinct. Then, $g(t_i)=0$ in $A_{\mathfrak{m}}$ for all $i$. Using Van der Monde determinant, you can see that all the coefficients of $g$ must be zero in $A_{\mathfrak{m}}$.
A: This is not a solution but some ideas:
First, a necessary condition: If $A$ has a maximal ideal $\mathfrak m$ such that it's residue field $k$ has finite order $q$, then the polynomial $x^q-x$ has $c(f) = 1$ and $v(f) \subset \mathfrak m$ and therefore a necessary condition is that all residue fields are of infinite size.
Second, an example where we have equality: Take $k$ to be any infinite field, say $k = \mathbb C$ and consider $A = k[t]$. The maximal ideals are all of the form $(x-a), a \in k$ and let $f(x) = \sum_i a_i(t)x^i$. Since $A$ is a UFD, we can assume $c(f)$ is $A$ by factoring out common factors.
Then, assume for contradiction that $v(f) \subset (x-\alpha)$. Then, in the residue field $k(a)=k$, we have $\sum_i a_i(\alpha)x^i = 0$ for all $x\in k$ which is impossibly since $k$ is infinite. Therefore $v(f) = A$.
