# Noetherian local ring and the growth of $\dim_k \operatorname{Ext}^i(k,k)$

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

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 .
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 $$2$$.
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!