Linear independence of powers

Steven V Sam; Andrew Snowden

arXiv:1907.02659·math.AC·July 8, 2019

Linear independence of powers

Steven V Sam, Andrew Snowden

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TL;DR

This paper proves that for r elements in an integral domain over an algebraically closed field, if any two are linearly independent, then their e-th powers are also linearly independent for some e between 1 and r!.

Contribution

It establishes a bound on the exponent e ensuring the linear independence of powers of elements in an integral domain, extending previous understanding.

Findings

01

Existence of an integer e between 1 and r! for which powers are linearly independent.

02

Generalization of linear independence properties to powers of elements.

03

Provides a new bound related to the factorial of the number of elements.

Abstract

Given r elements in an integral domain over an algebraically closed field such that any two are linearly independent, we show that there is an integer e between 1 and r! such the eth powers of these elements are linearly independent.

Equations2

f_{i_{1}} (x_{1}) \dots f_{i_{s}} (y) \neq = f_{j_{1}} (x_{1}) \dots f_{j_{s}} (y) .

f_{i_{1}} (x_{1}) \dots f_{i_{s}} (y) \neq = f_{j_{1}} (x_{1}) \dots f_{j_{s}} (y) .

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TopicsAdvanced Differential Equations and Dynamical Systems · Mathematical Dynamics and Fractals · Polynomial and algebraic computation

Full text

Linear independence of powers

Steven V Sam

Department of Mathematics, University of California, San Diego, CA

[email protected] http://math.ucsd.edu/~ssam/ and

Andrew Snowden

Department of Mathematics, University of Michigan, Ann Arbor, MI

[email protected] http://www-personal.umich.edu/~asnowden/

(Date: July 4, 2019)

SS was partially supported by NSF DMS-1500069, DMS-1651327, and a Sloan Fellowship. AS was supported by NSF DMS-1453893.

Fix an algebraically closed field $\mathbf{k}$ . We prove the following result:

Theorem 1.

Let $R$ be an integral $\mathbf{k}$ -algebra, and let $f_{1},\ldots,f_{r}\in R$ be non-zero elements such that $f_{i}/f_{j}\not\in\mathbf{k}$ for all $i\neq j$ . Then there exists an integer $1\leq e\leq r!$ such that $f_{1}^{e},\ldots,f_{r}^{e}$ are $\mathbf{k}$ -linearly independent.

Remark 2.

We were motivated by [KTB, Conjecture 16]. This conjecture takes $R$ to be a polynomial ring over the real numbers, and asks for a bound $E$ depending on the number of variables and $r$ such that $f_{1}^{e},\dots,f_{r}^{e}$ are linearly independent whenever $e\geq E$ . Our methods do not seem able to obtain this result. We note that such a bound does not exist for general domains: consider the rings $R=\mathbf{k}[x_{1},\dots,x_{d}]/(x_{1}^{s}+\cdots+x_{d}^{s})$ with $f_{i}=x_{i}$ , for example. ∎

We may as well replace $R$ with the subalgebra generated by the $f_{i}$ ’s. Thus, in what follows, we assume that $R$ is finitely generated. Thus $X=\operatorname{Spec}(R)$ is an integral scheme of finite type over $\mathbf{k}$ . If $R=\mathbf{k}$ then the theorem is clear, so in what follows we assume $\dim(X)\geq 1$ . The following is the key lemma:

Lemma 3.

Let $1\leq s\leq r$ be given. There exist $\mathbf{k}$ -points $x_{1},\ldots,x_{s}$ of $X$ such that the following two conditions hold:

(a)

$f_{i}(x_{j})\neq 0$ * for all $1\leq i\leq r$ and $1\leq j\leq s$ .* 2. (b)

Given $(i_{1},\ldots,i_{s})\neq(j_{1},\ldots,j_{s})$ in $[r]^{s}$ , we have $f_{i_{1}}(x_{1})\cdots f_{i_{s}}(x_{s})\neq f_{j_{1}}(x_{1})\cdots f_{j_{s}}(x_{s})$ .

Proof.

Let $Y$ be the open subvariety of $X$ where all the $f_{i}$ ’s are non-zero. We proceed by induction on $s$ . The result is tautologically true for $s=0$ . Suppose now that the result has been proven for $s-1$ . Let $x_{1},\ldots,x_{s-1}$ be the $\mathbf{k}$ -points witness this; note that these points all belong to $Y$ . We now produce $x_{s}$ .

For $i_{\bullet}\neq j_{\bullet}\in[r]^{s}$ , let $U_{i_{\bullet},j_{\bullet}}$ be the locus of points $y\in Y$ such that

[TABLE]

This is an open set. We claim that it is non-empty. There are two cases.

First, suppose that $i_{s}=j_{s}$ . Then $(i_{1},\ldots,i_{s-1})\neq(j_{1},\ldots,j_{s-1})$ , and so $f_{i_{1}}(x_{1})\cdots f_{i_{s-1}}(x_{s_{1}})\neq f_{j_{1}}(x_{1})\cdots f_{j_{s-1}}(x_{s-1})$ by assumption. Thus the two sides of (4) are different multiples of $f_{i_{s}}(y)=f_{j_{s}}(y)$ , and so $U_{i_{\bullet},j_{\bullet}}=Y$ .

Second, suppose that $i_{s}\neq j_{s}$ . Then $f_{j_{s}}$ and $f_{i_{s}}$ are not scalar multiples of each other, by assumption, and so the two sides of (4) are unequal functions of $y$ . Thus the claim follows.

Now let $U$ be the intersection of all the sets $U_{i_{\bullet},j_{\bullet}}$ . This is a non-empty open set. We can take $x_{s}$ to be any $\mathbf{k}$ -point of it. ∎

We also require the following simple lemma:

Lemma 5.

Let $\alpha_{1},\ldots,\alpha_{t}$ be distinct non-zero elements of $\mathbf{k}$ , and let $\beta_{1},\ldots,\beta_{t}$ be elements of $\mathbf{k}$ that are not all zero. Then there exists $1\leq j\leq t$ such that $\sum_{i=1}^{t}\beta_{i}\alpha_{i}^{j}\neq 0$ .

Proof.

Let $A$ be the $t\times t$ matrix with entries $A_{i,j}=\alpha_{i}^{j}$ and let $B$ be the column vector with entries $B_{i}=\beta_{i}$ . The determinant of $A$ is non-zero by the Vandermonde identity, and so $AB\neq 0$ . Since the $j$ th row of $AB$ is $\sum_{i=1}^{t}\beta_{i}\alpha_{i}^{j}$ , the result follows. ∎

We can now prove the main result:

Proof of Theorem 1.

Let $x_{1},\ldots,x_{r}\in X$ be the points produced by Lemma 3 with $s=r$ . For $\sigma\in S_{r}$ , let $c_{\sigma}=f_{\sigma(1)}(x_{1})\cdots f_{\sigma(r)}(x_{r})$ . The $c_{\sigma}$ are distinct non-zero elements of $\mathbf{k}$ . Let $1\leq e\leq r!$ be such that $\sum_{\sigma\in S_{r}}\operatorname{sgn}(\sigma)c_{\sigma}^{e}\neq 0$ , which exists by Lemma 5. For $1\leq i\leq r$ , let $v_{i}$ be the vector $(f_{i}^{e}(x_{1}),\ldots,f_{i}^{e}(x_{r}))$ . These vectors are linearly independent, as the determinant of the matrix with columns $v_{1},\ldots,v_{r}$ is $\sum_{\sigma\in S_{r}}\operatorname{sgn}(\sigma)c_{r}$ . It follows that the $f_{1}^{e},\ldots,f_{r}^{e}$ are linearly independent, as a dependency would give one between the $v_{i}$ ’s. ∎

Remark 6.

Suppose $\mathbf{k}$ is not algebraically closed. Theorem 1 remains true if we assume that $R$ is geometrically integral, i.e., that $\overline{\mathbf{k}}\otimes_{\mathbf{k}}R$ is integral. However, it is not true if we simply assume $R$ is integral. Indeed, if $R$ is a finite extension field of $\mathbf{k}$ and $r>[R:\mathbf{k}]$ then $f_{1}^{e},\ldots,f_{r}^{e}$ are linearly dependent for all $e$ , since any set of $r$ elements of $R$ is linearly dependent. ∎

Remark 7.

The upper bound of $r!$ in Theorem 1 is not optimal: for $r=3$ , we can take $1\leq e\leq 2$ in characteristic not 2, and $e\in\{1,3\}$ in characteristic 2. It is an interesting problem to determine the optimal upper bound on $e$ . ∎

Bibliography1

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[KTB] Joe Kileel, Matthew Trager, Joan Bruna, On the expressive power of deep polynomial neural networks, ar Xiv:1905.12207 v 1 .

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Abstract

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Linear independence of powers

Theorem 1**.**

Remark 2**.**

Lemma 3**.**

Proof.

Lemma 5**.**

Proof.

Proof of Theorem 1.

Remark 6**.**

Remark 7**.**

Theorem 1.

Remark 2.

Lemma 3.

Lemma 5.

Remark 6.

Remark 7.