TITLE: Noetherian local ring and the growth of $\dim_k \operatorname{Ext}^i(k,k)$
QUESTION [12 upvotes]: 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?

REPLY [11 votes]: 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!