Content of a polynomial 
Define the content of a polynomial (over an arbitrary commutative ring $A$) to be the ideal generated by its coefficients, denoted $c(f)$. I want to show that
  $$c(fg) = c(f)c(g).$$
(I was told this is true.) 

What I was able to show was that $c(fg) \subseteq c(f)c(g)$ (this is obvious), and that their radicals are the same. My reasoning for the latter was as follows: let $f = a_0 + \dotsb + a_nx^n,$ $g = b_0 + \cdots + a_mx^m,$ and consider the matrix
$$
\begin{pmatrix}
 a_0 b_0 & \cdots & a_0b_m\\
\vdots & \ddots & \vdots\\
a_nb_0 & \cdots & a_nb_m
\end{pmatrix}
.$$
Then $c(f)c(g)$ is generated by the entries of this matrix, while $c(fg)$ is generated by the sums along the diagonals. Let $\mathfrak p$ be a prime ideal of $A$ not containing $c(f)$ or $c(g)$, and let $i,\,j$ be minimal such that $a_i,\, b_j \notin \mathfrak p$.  Then all the terms in the generator of $c(fg)$ corresponding to the coefficient of $x^{i+j}$ are in $\mathfrak p$ except for $a_ib_j$, so that $c(fg) \not\subset\mathfrak p$. It follows that $c(fg) \subseteq \mathfrak p \Longleftrightarrow c(f)c(g) \subseteq\mathfrak p$, and hence $\sqrt{c(fg)} = \sqrt{c(f)c(g)}$.

Can I go all the way and show that in fact $c(fg) = c(f) c(g)$?

 A: I will give another example where $c(fg)\neq c(f)c(g)$. This example is taken from an exercise given in Kaplansky's book "Commutative Rings". 
Let's consider the integral domain $\Bbb Z[2i]=\{a+2bi: a,b\in \Bbb Z\}$. We take the polynomials $f=2+2ix$ and $g=2-2ix$. It's easy to see that $fg=8x^2$, so $c(fg)=(8)$. On the other hand, $c(f)=c(g)=(2,2i)$, so $c(f)c(g)=(2,2i)^2$. Finally, we note that $(2+2i)^2\notin (8)$ because $(2+2i)^2=8i$ and if $8i\in (8)$, then we would have that $8i=8(a+2bi)$, which leads to $8=16b$, contradiction. Hence, $c(fg)\neq c(f)c(g)$.

Another way to see why $\Bbb Z[2i]$ is not a Gaussian domain it's because this domain is not integrally closed (just take $1+i\in \Bbb Q[i]$ and note that $1+i$ is a root of the polynomial $x^2-2i$), whereas that a Gaussian domain is necessarily integrally closed (this follows from the fact that a domain is Gaussian iff is Prufer* and it's well-known that Prufer domains are integrally closed, see e.g. here).
(*) A proof of this result can be found in chapter IV of Gilmer's book "Multiplicative Ideal Theory".
A: The displayed equation is not true in general.  For example, let $A=k[t,u]$, $k$ a field and $t, u$ indeterminates.  Let $f=t+ux$ and $g=t-ux$.  Then $c(fg) = c(t^2 - u^2 x^2) = (t^2, u^2)$, but $c(f)c(g) = (t,u)^2 = (t^2, tu, u^2)$, a strictly larger ideal of $A$.
A ring $A$ for which the displayed equation does hold for all $f,g \in A[x]$ is sometimes called a Gaussian ring.  It is closely related to the condition of being a Prüfer domain, and indeed holds whenever $A$ is Prüfer.
