I'm interested in the following strange question: for some $d > 1$, what is the minimum dimension of a commutative $\mathbb{C}$-algebra containing infinitely many elements that square to zero, but where the product of any $d$ of these elements is nonzero?

The best upper bound I know is $d\cdot \binom{\lceil 3d/2 \rceil}{\lfloor d/2 \rfloor} < d\cdot 2.6^d$, and can be obtained as follows. Let $f = \sum_{i_1 < i_2 < \cdots < i_d} \prod_{1\le j < k \le d}(i_j - i_k)^2 x_{i_1} \cdots x_{i_d}$ be an element of $S = \mathbb{C}[[x_1, x_2, \ldots]]$, the ring of formal power series in infinitely many variables. Let $T = \mathbb{C}[[\partial_{x_1} , \partial_{x_2}, \ldots]]$ be the ring of partial differential operators/dual power series in infinitely many variables. $T$ acts on $S$ via differentiation, which I denote by $\circ$. Now let $f^\perp = \{g \in T : g \circ f = 0\}$ be the "apolar ideal" to $f$. I claim that the ring $R = T/f^\perp$ has the desired properties.

First, it can be shown that that $\dim \text{span}\{g \circ f : g \in T\} \le d\cdot \binom{\lceil 3d/2 \rceil}{\lfloor d/2 \rfloor}$, and hence the stated bound holds for $\dim R$. Now since each variable appearing in $f$ has degree at most $1$, we have that $\partial^2_{x_i} \circ f = 0$ for all $i$. Furthermore, for all $i_1<\cdots < i_d$ we have $\partial_{x_{i_1}} \cdots \partial_{x_{i_d}} \circ f = \prod_{j,k}(i_j - i_k)^2 \neq 0$. This shows that the images of $\partial_{x_1}, \partial_{x_2}, \ldots$ under the quotient map have the desired properties in $T$.

I would appreciate any pointers to concepts or related work that could be useful here.

**EDIT**

As YCor points out, there is a lower bound of $2^d$: if $x_1, \ldots, x_d$ are elements of such an algebra with the desired properties, then the products $\{x_S : S \subseteq [d]\}$ must be linearly independent (if $\sum_{S \subseteq [d]} \alpha_S x_S = 0$ is a nontrivial relation, then letting $U \subseteq [d]$ be such that $\alpha_U \neq 0$ and $U$ is minimal with respect to set inclusion among the sets in the support of this relation, multiplying by $x_{[d]-U}$ we find that $x_{[d]}= 0$, a contradiction.)

Also, the stated upper bound actually holds for the following more general family of algebras: let $(a_i)_{i \in \mathbb{N}}$ be a sequence of distinct elements in $\mathbb{C}$, let $f_a = \sum_{i_1 < i_2 < \cdots < i_d} \prod_{1\le j < k \le d}(a_{i_j} - a_{i_k})^2 x_{i_1} \cdots x_{i_d}$, and take $R =T/f_a^\perp$. (Even more generally, one can take a matrix $A \in \mathbb{C}^{d \times \mathbb{N}}$ with nonvanishing $d \times d$ minors, and take $f = \sum_{S \subset \mathbb{N}, |S| = d} \det(A_S)^2 x_S$ and $R = T/f^\perp$, although the dimension bound, while finite, ends up being worse.)

1more comment