Content of a polynomial

Solution 1:

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.

Solution 2:

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".