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

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.)

• For $d=2$ one can get $5$ (rather than 6). Let $V$ be a 3-dimensional space with an isotropic nondegenerate symmetric bilinear form $b$, over the field $K$. Choose in $V$ a family $(x_i)$ of isotropic vectors with $b(x_i,x_j)$ nonzero for all $i\neq j$ (not hard to check they exist, as many as the cardinal of the field, using that $\dim(V)\ge 3$). Write $A=K\oplus V\oplus K$, where the left-hand $K$ is generated by the unit, the multiplication on $V$ is $b$ valued in the right hand $K$, and the right-hand $K$ has multiplication zero with everybody.
– YCor
May 30, 2019 at 17:36
• PS for $d=2$ 5 is optimal (at least for semilocal algebras) in view of the short classification of local algebras of dimension $\le 4$ (as is recalled in www-math.mit.edu/~poonen/papers/dimension6.pdf for instance)
– YCor
May 30, 2019 at 17:45
• Great, thank you for that reference! The bound of 5 for $d=2$ can also be seen from the fact that the dimension of the space of partial derivatives of $f$ is 5 (my bound of $d \cdot \binom{\lceil 3d/2 \rceil}{\lfloor d/2 \rfloor}$ is sloppy). Are there any keywords I should look up for the general problem? May 30, 2019 at 17:55
• ("Semilocal" is an empty condition in my previous comment.) No I just checked; I have no idea of any particular useful keyword. Useless here, but it can be checked (over $\mathbf{C}$) that if these's such a countable family, then there's one of continuum cardinal.
– YCor
May 30, 2019 at 18:20
• A lower bound is $2^d$. Indeed, if $x_1,\dots,x_d$ satisfy the condition, The products $x_J=\prod_{j\in J}x_J$ for $J\subset\{1,\dots,d\}$ form a linearly free family. Indeed, if we have a nontrivial relation $w$, then $w$ is not a multiple of $x_1\dots x_d$, choose a monomial of minimal degree $\delta$ appearing in $w$, and choose $i$ not appearing in this monomial; then $x_iw$ gives a nontrivial relation with larger $\delta$, and hence choosing $\delta$ maximal yields a contradiction.
– YCor
May 30, 2019 at 20:06

$$\def\CC{\mathbb{C}}\def\fm{\mathfrak{m}}\def\PP{\mathbb{P}}$$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 $$k, then quotienting $$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.

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.

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.

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.