Finite dimensional commutative algebras containing infinitely many nilpotents whose $d$-way products are nonzero

Question

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.)

Answer 1

$\def\CC{\mathbb{C}}\def\fm{\mathfrak{m}}\def\PP{\mathbb{P}}$Here are some ideas, but not a solution. First, I will reduce the problem to the sort of examples the OP is already considering. I will then make some connections to secant varieties. I will prove the optimum value for $d=3$ to be $12$, given by the OP's construction. Finally, I will report on some numeric investigations of the OP's construction. Calling the OP's ring $R^d$, my data suggests that $$\dim R^d_k = \begin{cases} \binom{2d-k}{k} & 0 \leq k \leq d/2 \\ \binom{d+k}{d-k} & d/2 \leq k \leq d \end{cases}.$$

We assume throughout $d \geq 2$.

In this section, our goal is to reduce to the case that $R$ is a graded Gorenstein ring, with the $x_i \in R_1$ and with socle in degree $d$.

Let $V$ denote the vector space spanned by the $x_i$ and let $\fm$ be the ideal they generate. Let $X$ be the Zariski closure of $\{ x_i \}$ in $\PP(V)$ and let $CX$ be the cone on $X$ in $V$.

Reduction 1 We can assume that the $x_i$ generate $R$ as a $\CC$-algebra.

Proof Passing to the subalgebra they generate reduces the dimension of $R$ and preserves the hypotheses. $\square$

Having made this reduction, I claim that $\fm$ is a maximal ideal and that $R$ is a local ring. Proof: Since $\fm$ is nilpotent and $R \neq 0$, we have $R \neq \fm$ and $R/\fm$ is nonzero. But, since the $x_i$ generate $R$, the dimension of $R/\fm$ is at most $1$. We conclude that $R/\fm \cong \CC$, so $\fm$ is maximal. Also, since this maximal ideal is nilpotent, we deduce that $R$ is local.

Reduction 2 We may assume that $X$ is irreducible.

Proof If $X$ has multiple irreducible components, one of them must contain infinitely many $x_i$.

Reduction 3 We may assume that $\fm^{d+1}=0$.

Proof Note that the condition on $d$-fold products ensures that $\fm^d \neq 0$ so, by Nakayama, we have $\fm^d/\fm^{d+1} \neq 0$. Replace $R$ by the ring $R/\fm^{d+1}$. We will now construct a new infinite subset of $CX$ with the given conditions. We note that, for any $x \in CX$, we have $x^2=0$.

Since $CX$ spans $V$, which generates $\fm$, the set of $d$-fold products of elements of $CX$ spans $\fm^d/\fm^{d+1}$. In particular, there are some $x_1$, $x_2$, \dots, $x_d$ in $CX$ with $x_1 x_2 \cdots x_d \not \in \fm^{d+1}$. Now, suppose inductively that we have constructed $x_1$, $x_2$, ..., $x_N$ in $CX$, for $N \geq d$, so that every $d$-fold product of $x_i$ is nonzero in $\fm^d/\fm^{d+1}$. For every $x_{i_1}$, ..., $x_{i_{d-1}}$, there exists an $y \in CX$ such that $x_{i_1} \cdots x_{i_{d-1}} y \neq 0$ modulo $\fm^{d+1}$ (namely, any $x_j$ distinct from $x_{i_1}$, ..., $x_{i_{d-1}}$). The condition that $x_{i_1} \cdots x_{i_{d-1}} y \neq 0$ is an open condition on $y$, and we already reduced to the case that $X$ is irreducible. The intersection of finitely many nonempty opens on an irreducible variety is nonempty, so we can find some $x_{N+1}$ in $CX$ so that all the $x_{i_1} \cdots x_{i_{d-1}} x_{N+1}$ are simultaneously nonzero modulo $\fm^{d+1}$. This concludes the inductive construction. $\square$.

Reduction 4 We may assume that $R$ is graded.

Proof Let $\hat{R}$ be the associated graded of $R$ with respect to the filtration by powers of $\fm$. So we have $\hat{R}_1 \cong \fm/\fm^2 \cong V$, and we can thus think of $CX$ as a subset of $\hat{R}_1$. The condition that $x^2=0$ for $x \in CX$ implies the same claim in $\hat{R}$. Since $\fm^{d+1}=0$, we have $\hat{R}_d \cong \fm^d$, so our condition on $d$-fold products in $R$ implies the same in $\hat{R}$. $\square$.

Reduction 5 We may assume that $R_d$ has dimension $1$.

Proof If $\dim R_d>1$, then quotient $R$ by a generic element of $R_d$. If this element is chosen generically enough, it will not conicide with any of the countably many products $x_{i_1} \cdots x_{i_d}$. $\square$.

Reduction 6 We may assume that $R$ is Gorenstein.

Proof We have already assumed that $\dim R_d=1$ and $R_k=0$ for $k>d$. If $R$ has socle in degree $k<d$, then quotienting $R$ by this socle produces another ring with the same properties.$\square$

As the OP clearly knows, a $0$-dimensional graded Gorenstein ring with $R_1 = V$ and socle in dimension $d$ is equivalent to the data of a nonzero symmetric multilinear form $V^{\otimes d} \to \CC$ (up to rescaling). Since we are in characteristic $0$, we can also think of this as a nonzero degree $d$ polynomial function on $V$; call this function $F$. Then $R_k$ can be identified with the vector space spanned by all $k$-fold partial derivatives of $F$, and we have $\dim R_k \cong \dim R_{d-k}$.

The question, then, is how to choose $F$ to make $\dim R_k$ small, while making sure we can still find an infinite set $\{ x_i \}$ as required. The following proposition gives me some insight into the second requirement. For $v \in V$, let $\partial_v$ denote differentiation by the constant vector field in direction $v$.

Proposition Let $F$ be a degree $d$ homogenous polynomial on $V$ and let $R$ be the corresponding Gorenstein ring. Let $CZ$ be the set of $v \in V$ for which $\partial_v^2 F=0$. Then $CZ$ is contained in the zero locus of $F$. If (1) $CZ$ is positive dimensional (2) $CZ$ spans $V$ and (3) $CZ$ is irreducible, then we can find an infinite subset $x_i$ of $CZ$ as required.

Of course, we write $Z$ for the projectivization of $CZ$ into $\PP(V)$.

Proof To see that $CZ$ is contained in the zero locus of $F$, note that $F(tv)$ must be of the form $c t^d$ for any $v \in V$. When $v \in CZ$, we have $\tfrac{d^2}{(dt)^2} c t^d=0$, so $c=0$.

Now, we turn to the main task of constructing an infinite subset of $CZ$ as required. For every $x \in CZ$, we have $x^2=0$ in $R$. Since $CZ$ spans $V$, the $d$-fold products of elements of $CZ$ span $R_d$ and, in particular, there must be some $x_1$, $x_2$, \dots, $x_d$ in $CZ$ with $x_1 x_2 \cdots x_d \neq 0$. Since $x^2$ is $0$ on $CZ$, these $x_i$ must be distinct. Now, as in Reduction 3, we inductively construct $\{ x_i \}$. $\square$

By the way, I can't find an example where $CZ$ is reducible and the claim fails. For example, this could happen if we had $Z = X_1 \cup X_2 \cup \cdots X_N$ for $N \geq d$ and the product of any two elements of the same $X_j$ are $0$. But I can't find a case of this.

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So we may take any degree $d$ homogenous polynomial $F$, compute the variety $Z$ and, if the conditions of this proposition hold, we can find the infinite set $\{ x_i \}$. In the next section, we discuss further relations between the geometry of $Z$ and of the hypoersurface $F$.

We saw above that we must have $X \subseteq Z \subseteq \{ F=0 \}$. In fact, much more is true:

Proposition The $(d-1)$-fold secant variety of $X$ is contained in $\{ F=0 \}$.

Proof Let $x_1$, ..., $x_{d-1}$ be in $X$. The restriction of $F$ to the plane spanned by the $x_i$ is determined by the $d$-fold partials $\partial_{x_{j_1}} \cdots \partial_{x_{j_d}}(F)$ (notice that these scalars, since $F$ has degree $d$). But every such $d$-fold partial repeats some $\partial_{x_j}$, and we have $\partial_x^2 F=0$ for $x \in X$, so all of these $d$-fold partials vanish. We deduce that $F$ is $0$ on the span of $x_1$, ..., $x_{d-1}$, so $F$ contains the $(d-1)$-fold secant variety of $X$.

This gives a possible strategy for finding good choices of $F$. First, guess an $X$. Compute the ideal of $\Sigma_{d-1}(X)$ in degree $d$. If $I(\Sigma_{d-1}(X))_d=0$, we lose. Otherwise, chose $F$ in $I(\Sigma_{d-1}(X))_d$ and hope that $\dim R$ is small.

Conveniently, I learned from Section 2.2 of Sidman and Vermiere that $I(\Sigma_{d-1}(X))_d$ is also the degree $d$ part of the $(d-1)$-fold symmetric power $I(X)^{(d-1)}$, and that is can be computed by an method called prolongation. In particular, if $I(X)_2=0$, then $I(\Sigma_{d-1}(X))_d$ vanishes, so we better chose $I(X)$ to have many quadratic generators (but not so many that $X$ becomes a finite set!). Beyond that, though, I have little intuiton for how to make $\dim R$ small.

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Finally, small $d$. We have $\dim X \geq 1$ and we assumed that $X$ spans $V$. We want $\Sigma_{d-1}(X)$ not to be all of $\PP(V)$. I am fairly sure this forces us to take $\dim V \geq 2d-1$. In particular, when $d=2$, the smallest possible $R$ is to take $\dim R_0 = \dim R_2 = 1$ and $\dim R_1 = \dim V = 3$. We can take $X$ to be any conic in $\PP(V)$ and $F$ to be its defining equation.

For $d=3$, we have $\dim R_0 = \dim R_3 = 1$ and $\dim R_1 = \dim R_2 = \dim V$, so minimizing $\dim R$ is the same as minimizing $\dim V$, and we noted above that the minimum is to take $\dim V =5$. So this proves the bound claimed above. Explicitly, we can think of $V$ as the vector space of $3 \times 3$ Toeplitz matrices, $X$ to be the rank $1$ Toeplitz matrices, and $F$ to be the determinant.

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Finaly, I tried to compute the dimension of the OP's ring. I didn't see how to get an closed formula, so I generated $200$ randomly chosen $k$-fold derviatives of the formal power series, extracted $200$ random coefficients from them and found the rank of the $200 \times 200$ matrix. (This is a slight simplification of what I actually did, more details if desired.) My data suggests that the degree $k$ part of the OP's ring has dimension $$\min \left( \binom{2d-k}{k}, \binom{d+k}{d-k} \right).$$ In particular, the $k=d/2$ term is $\binom{3d/2}{d/2}$, the OP's expected growth rate.