Gaussian Rings and Flat Dimension

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:

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

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