# Counting Divisors in $\mathbb{Z}^n$

Basically, I'm looking for ways to multiply elements of $\mathbb{R}^n$ that allow me to count divisors in $\mathbb{Z}^n$.

For every positive integer $n$, I'm looking for an algebra structure on $\mathbb{R}^n$ such that

1. Given $y,z \in \mathbb{Z}^n$ with $z$ non-zero, $x y = z$ has at most $c$ solutions $x \in \mathbb{Z}^n$, where $c$ is a constant independent of $y$ and $z$.
2. Given $z \in \mathbb{Z}^{n}$ with $z$ non-zero, $$\# \{ (x,y) \in \mathbb{Z}^{2n} : xy = z \} = \| z \|^{o(1)},$$ where $\| \cdot \|$ is the usual Euclidean norm on $\mathbb{R}^n$.

Can $\mathbb{R}^n$ be given such an algebra structure?

Examples: If $n = 2$, use $\mathbb{C}$. If $n=4$, maybe use the quaternions. It's not exactly clear what the divisor bound is for the quaternions.

I know that the only finite-dimensional real division algebras are the real numbers, the complex numbers, the quaternions, and the octonions.

• So it sounds like you're actually looking for ways to multiply elements of $\mathbb{Z}^n$? Where do the real numbers come into it? – Qiaochu Yuan Sep 13 '16 at 18:52
• You should say you want not an $n$-dimensional algebra, but a structure of algebra on $\mathbb{R}^n$ (otherwise the sequel is meaningless). And from your last sentence it seems that by algebra you mean any choice of bilinear law (not necessarily associative). – YCor Sep 13 '16 at 19:01
• Due to a weird sign-up thing, I am the original poster, Linden. I've put in a request to merge the accounts. @QiaochuYuan That might be true. I didn't want to require that Z^n be closed under the multiplication. – Linden Sep 13 '16 at 19:20
• Due to a weird sign-up thing, I am the original poster, Linden. I've put in a request to merge the accounts. @YCor Isn't an algebra a vector space with a multiplication? Every finite dimensional vector space over R is isomorphic to R^n. So is there really a difference between an n-dimensional algebra over R and a structure of algebra on R^n? Are you getting at the fact that the addition operation might not be the usual one if I just ask for an algebra over R? – Linden Sep 13 '16 at 19:24
• @RitterSport: yes, the difference is that it doesn't make sense to ask what "$\mathbb{Z}^n$" is inside an $n$-dimensional $\mathbb{R}$-algebra. The additional information you get from "algebra structure on $\mathbb{R}^n$" over "$n$-dimensional $\mathbb{R}$-algebra" is a distinguished choice of basis. – Qiaochu Yuan Sep 13 '16 at 19:30

I assume that conditions 1) and 2) are applies to nonzero $z$ only, otherwise each of them ia absurd.

Let me describe the construction more formally. Let $\circ$ be a componentwise multiplicaion on $\mathbb R^n$, i.e., $(x_1,\dots,x_n)\circ(y_1,\dots,y_n)=(x_1y_1,\dots,x_ny_n)$.

Choose a no-degenerate matrix $S$ such that each its column consists of $n$ algebraic numbers linearly independent over $\mathbb Q$. This ensures that for every nonzero $x\in\mathbb Z^n$ all the components of $xS$ are nonzero.

Now define $$x*y=e\bigl((xS)\circ(yS)\bigr)S^{-1},$$ where $e$ is the base of natural logarithms. We claim that $*$ is a desired operation. One may easily see that the algebras $(\mathbb R,+,*)$ and $(\mathbb R,+,\circ)$ are isomorphic, the isomorphism is provided by $x\mapsto exS$. Thus $*$ is associative and distributive over addition.

Next, for every nonzero $x,y\in\mathbb Z^n$, the product $(xS)\circ(yS)$ consists of nonzero algebraic numbers, so $\bigl((xS)\circ(yS)\bigr)S^{-1}$ is a nonzero vector with algebraic components. Thus $x*y$ contains at least one transcendent component and thus does not lie in $\mathbb Z^n$. Therefore, the conditions 1) and 2) are trivially satisfied (for nonzero $z$).

• Can you give an example of this construction? I find it a bit hard to follow at the moment. – Greg Martin Sep 13 '16 at 21:00
• @GregMartin: OK, I've tried. Is it more clear now? – Ilya Bogdanov Sep 13 '16 at 21:27
• I at least see the intent of the construction now, which is helpful! But I still would have liked to see $e$ specified, and I'm not yet ready to bet money on the assertion that no zero-divisors ever occur. – Greg Martin Sep 14 '16 at 0:09
• $e$ is the usual $e=\lim_{n\to\infty}\bigl(1+\frac1n\bigr)^n$; we need only its transcendence. – Ilya Bogdanov Sep 14 '16 at 7:51
• Now I've reformulated the whole construction more formally. – Ilya Bogdanov Sep 14 '16 at 13:38

I don't get every detail of Ilya's answer but it seems to do the job with the feature that for non-zero $x,y \in \mathbb{Z}^n$ one never has $xy \in \mathbb{Z}^n$ so the numbers (of solutions) you wish to bound are always $0.$

Here is another answer and a possible variation you might like better (if it works).

Associate a vector $a=(a_0,a_1,\cdots,a_{n-1}) \in \mathbb{R}^n$ with the polynomial $f_{a}(t)=\sum_0^{n-1}a_it^{i}$ and define the multiplication as usual polynomial multiplication with $t^{n}=r \tag{*}$ for an irrational real $r.$

With this construction we have$f_x,f_y$ and $f_xf_y$ all in $\mathbb{Z}[t]$ exactly when the multiplication does not need $(*)$ So solving $f_xf_y=f_z$ for $f_x$ is done by polynomial division and has at most one solution. The number of pairs $f_x,f_y$ with $f_xf_y=f_z$ is bounded by $2^{n-1}$ times the number of divisors of the content of $f_z.$

If you instead use polynomial multiplication$\bmod g(t)=t^n-r$ or some other irreducible degree $n$ polynomial in $\mathbb{Z}[t],$ then you are dealing with some ring $\mathbb{Z}[\alpha].$ It then becomes a question of algebra and the particular choice of $g(t)$ how well conditions $1$ and $2$ hold up. Certainly if there are units which are not roots of unity in $\mathbb{C}$ then there are infinitely many units and $f_xf_y=f_z$ has infinitely many solutions if it has any.

• Is the number of divisors of the content of $f_z$ comparable to $\| z \|_{\infty} = \max_{i} |z_i| = \text{ height of } f_z$? – Linden Sep 23 '16 at 18:48
• I should have said: Is the content of $f_z$ comparable to $\| z \|_{\infty} = \max_i |z_i| =$ height of $f_z$? – Linden Sep 25 '16 at 15:37