Skip to main content
1 of 2
Hailong Dao
  • 30.6k
  • 5
  • 102
  • 188

To add to Greg's answer and to address zzy's question in the comment: there is an exponential bound. First, the Betti number 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 $d^i$.

Hailong Dao
  • 30.6k
  • 5
  • 102
  • 188