# 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 $c(fg) = c(f) c(g)$?

• The equation is not true as shown below, but a weakened version holds: $c(fg)=A$ if and only if $c(f)=c(g)=A.$ Jun 6, 2013 at 5:36
• @RagibZaman What you say follows easily from the property proved by the OP, namely $\sqrt{c(fg)}=\sqrt{c(f)c(g)}$.
– user26857
Jun 6, 2013 at 6:16
• @YACP Ahh yes, I hadn't read that. Sorry. Jun 6, 2013 at 6:22
• A general result concerning the content of the product of two polynomials is the so-called Dedekind-Mertens-Lemma: for $f,g\in A[x]$ the equation $c(f)^n c(f)c(g)=c(f)^n c(fg)$ holds, where $n$ is the degree of $g$. Jun 6, 2013 at 7:12
• A nice, elementary (no prime ideals) and constructive proof of the weaker statement with the radicals is given in Banaschewski, /Polynomials and radical ideals/, 1996. Sep 5, 2016 at 14:01

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.

• I think it's worthwhile to notice that $\overline{c(fg)}=\overline{c(f)c(g)}$ since $tu$ is integral over $(t^2,u^2)$. Moreover, the equation holds for any two polynomials $f,g\in A[x]$, where $A$ is a commutative unitary ring. Jun 25, 2017 at 17:10

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".

• Isn't $fg=4+4x^2$ and thus, $c(fg)=(4)$? Sep 23, 2022 at 5:01