7
$\begingroup$

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.)

$\endgroup$
6
  • 3
    $\begingroup$ 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. $\endgroup$
    – YCor
    May 30, 2019 at 17:36
  • 2
    $\begingroup$ 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) $\endgroup$
    – YCor
    May 30, 2019 at 17:45
  • 1
    $\begingroup$ 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? $\endgroup$
    – Kevin
    May 30, 2019 at 17:55
  • 1
    $\begingroup$ ("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. $\endgroup$
    – YCor
    May 30, 2019 at 18:20
  • 4
    $\begingroup$ 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. $\endgroup$
    – YCor
    May 30, 2019 at 20:06

1 Answer 1

5
$\begingroup$

$\def\CC{\mathbb{C}}\def\fm{\mathfrak{m}}\def\PP{\mathbb{P}}$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.


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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.