Let $A$ be a noetherian local ring with residue field $k$, one can consider $\operatorname{Ext}^i(k,k)$ for every natural number $i$. If it is zero for large $i$, then $A$ is regular and the converse is also true. Then, for which rings the dimension of $\operatorname{Ext}^i(k,k)$ is bounded with respect to $i$ ?

Example: $\mathbb Z /p^n$

Non-example: $k[x,y]/(x^2,y^2)$

What about polynomial growth?


2 Answers 2


The commutative noetherian rings such that the Betti numbers of $k$ eventually grow polynomially are precisely the complete intersections. This is a theorem of Gulliksen, see Theorem 2.3 here .

The degree of the polynomial giving the growth (which is called the complexity if you add 1) determines the codimension of the complete intersection. In particular, if the Betti numbers are eventually constant the ring is a hypersurface.

Both of your examples are complete intersections, as the theorem says, with $\mathbb{Z}/p^n\mathbb{Z}$ a hypersurface and $k[x,y]/(x^2,y^2)$ of codimension $2$.

  • $\begingroup$ Thank you, is there a general bound such as exponential bound for any noetherian local ring? $\endgroup$
    – sawdada
    Oct 31, 2018 at 17:37

To add to Greg's answer and to address zzy's question in the comment: there is an exponential bound. First, the Betti numbers are unchanged by completion, and we can therefore write $R$ as a quotient of a regular local ring $S$ in a minimal way. Then Serre proved long ago that there is a term-wise inequality of power series: $$\sum \dim Ext^i_R(k,k)t^i \leq \frac{p(t)}{q(t)} $$ Where $p,q$ are polynomial whose coefficients depends on the minimal resolution of $R$ as a module over $S$. For details, google "Golod rings" (those where the equality is achieved).

To see an easy example of exponential behavior, let $S = k[[x_1,...,x_d]]$ and $R=S/m^2$ where $m$ is the maximal ideal of $S$. Then the first syzygy of $k$ is $m_R= m_R/m_R^2 \cong k^{d}$. So the i-th Betti number is $d^i$. By the way, when $d\leq 2$, and quotient of $S$ is Golod unless if they are complete intersections!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .