Let $R$ be any commutative ring. The **content** of a polynomial $f\in R[x]$ is by definition the two-sided ideal in $R$ generated by the coefficients of $f$. If $f,g\in R[x]$, then $c(fg)\subseteq c(f)c(g)$, because each coefficient of $fg$ is a linear combination of elements of $c(f)c(g)$. Sometimes, however, this inclusion is strict. For example, if $k$ is a field of characteristic two, and $R = k[u,v]$ then $f = u + vX$ satisfies $c(f^2)\subset c(f)^2$, where the inclusion is strict. Indeed, $f^2 = u^2 + v^2X^2$ so $c(f^2) = (u^2,v^2)$, whereas $c(f)c(f) = (u^2,v^2,uv)$. A ring $R$ in which $c(fg) = c(f)c(g)$ for all $f,g\in R[x]$ is called **Gaussian**. We have just seen that $k[u,v]$ is *not* Gaussian, and in fact, we didn’t even have to specify that $k$ is characteristic two. What about a polynomial ring $k[u]$ over a field $k$ in one variable? Since $k$ is a field, $k[u]$ is a principal ideal domain (PID), and PIDs are always Gaussian. These observations can be clarified by looking at the concept of weak dimension.

## Weak Dimension

Recall that the weak dimension of any ring $R$ is the supremum over the length of all flat resolutions of all left (or right) $R$-modules. Typically the notation ${\rm w.dim}(R)$ is used to denote the weak dimension of $R$.

For example, if $k$ is a field then ${\rm w.dim}(k) = 0$. What about $k[u]$? The module $k \cong k[u]/u$ is not flat since $k\otimes_{k[u]}$ doesn’t preserve injections (can you find an example?), but since $k[u]$ is a principal ideal domain, every $k[u]$-module has a length-one projective resolution and hence a length-one flat resolution. Hence ${\rm w.dim}(k[u]) = 1$.

The concept of weak dimension is useful in the world of Gaussian rings. Here is one such result:

**Theorem (Glaz [1]).**For a commutative ring $R$, the following are equivalent:

- ${\rm w.dim}(R)\leq 1$.
- $R$ is a Gaussian ring and all principal ideals of $R$ are flat.
- $R$ is Gaussian and reduced.

In particular, a domain $R$ is Gaussian if and only if ${\rm w.dim}(R) \leq 1$. For example, $k[u]$ is Gaussian whenever $k$ is a field, since ${\rm w.dim}(k[u]) = 1$. On the other hand, $k[u,v]$ is not Gaussian, since ${\rm w.dim}(k[u,v]) = 2$ and $k[u,v]$ is a domain. That ${\rm w.dim}(k[u,v]) = 2$ follows because $k[u,v]$ has global dimension two and the concepts of global dimension and weak dimension coincide for Noetherian rings.

**Question.**Can you find an explicit example of two polynomials $f,g\in k[u,v][X]$ for an arbitrary field $k$ such that $c(fg)\subset c(f)c(g)$ (strict inclusion)?

Glaz’s theorem allows us to slickly avoid finding an explicit example to conclude that $k[u,v]$ is not Gaussian. The theorem also shows having all principal ideals are flat as in $k[u,v]$ is not enough to guarantee that ${\rm w.dim}(R)\leq 1$, or in other words, it is not enough to guarantee that all ideals are flat. I feel this isn’t too surprising though, since any principal ideal in a domain is always free and hence flat.

## The Relation Between Gaussian Rings and Weak Dimension is Mysterious

A commutative ring $R$ is called **coherent** if every finitely generated ideal is finitely presented. Glaz proved in [1] that every coherent Gaussian ring has weak dimension in the set $\{ 0,1,\infty\}$. On the other hand, whether *every* Gaussian ring has weak dimension in $\{0,1,\infty\}$ is an open question!

[1] Glaz, Sarah. The weak dimensions of Gaussian rings. Proc. Amer. Math. Soc. 133 (2005), no. 9, 2507-2513