# Betti sequence of finite dimensional commutative algebras

Given a finite dimensional commutative local $$K$$-algebra $$A$$ for a field $$K$$. Associated to $$A$$ is its dimension $$d_A$$ and the Betti-sequence $$c_i=dim(Ext_A^i(S,S))$$ where $$S$$ is the unique simple $$A$$-module.

Question: Are there only finitely many such $$K$$-algebras $$A$$ with a fixed dimension $$d_A$$ and associated sequence $$c_i$$?

If not, then can one characterise for which sequences there are only finitely many for a fixed dimension? Is there an example of such a sequence that is not constant?

For example for $$c_i=1$$ constant and $$d_A=n$$, the unique ones are $$A=K[x]/(x^n)$$ as we saw in a previous thread Commutative algebras with modules of small complexity .

I will first show that the general answer is no (in fact one should expect the opposite even in the graded case) by the Pigeonhole principle. Then we shall construct some concrete examples via an old result of Tate.

Let $$(A,k)$$ be a local artin algebra. We write $$A=R/I$$ as a (minimal) quotient of a regular local ring $$(R,k)$$. Note that $$\dim R$$ is bounded above by $$d_A-1$$.

Claim 1: Fixing $$d_A$$, then the set of possible sequences $$\{c_i\}_{i\geq 0}$$ is countable.

Proof: this follows from an inequality proved by Serre: $$\sum c_it^i \ll \frac{(1+t)^{\dim R}}{1-\sum_{i>0}\dim_k Tor_i^R(k,A) t^{i+1}}$$ (where $$\ll$$ means that the LHS is term by term dominated by the RHS). For references, look up "Golod rings", this is such a popular subject even I had a paper about it.

Now, to finish the proof, note that the denominator of RHS is a polynomial of degree $$\dim R$$, and each $$\dim_k Tor_i^R(k,A)$$ is bounded above by $$d_A\dim_k Tor_i^R(k,k)$$ as $$A$$ has a filtration by $$d_A$$ copies of $$k$$ and $$\dim Tor_i(k,-)$$ is subadditive on short exact sequences. Finally, $$\dim_k Tor_i^R(k,k) = \binom{\dim R}{i}$$.

Claim 2: Fixing $$d_A$$, and assume $$k$$ is algebraically closed, the set of isoclasses of $$A$$, even assuming $$A$$ standard graded, can be uncountable.

Proof: this comes from standard facts that the set of isoclasses of $$A$$ with a fixed Hilbert function are parametrized by Hilbert schemes.

Putting together Claim 1 and 2 shows that there are sequences with uncountably many $$A$$s.

Example: For a concrete example, a result by Tate (mentioned in Eisenbud's paper on complete intersection) implies that when $$A$$ is a complete intersection, the $$\{c_i\}$$ only depends on $$\dim R$$, so even two general quadrics in $$k[x,y]$$ would give infinitely many examples.

• Thanks, another question might be what about countable fields such as $k= \mathbb{Q}$ and whether there might exist a non-constant sequence $(c_i)$ where only finitely many algebras exist even for uncountable fields. – Mare Nov 14 at 15:51
• Even when $k$ is small, one can get examples of infinitely many local $A$ by looking at complete intersections. Say $I=(x^2+f, y^2+g)$ for $f,g$ of high degrees. – Hailong Dao Nov 14 at 16:11
• Also, note that it is easy to give sequence $(c_i)$ with no $A$. For instance, we know that it has to be eventually non-decreasing. Or, if it is eventually constant, then $A$ is a hypersurface and the sequence has to be constant from the beginning. So, a lot of constraints would have to be added to avoid trivial examples. – Hailong Dao Nov 14 at 16:58