# Numbers of solutions equal on every finite commutative ring

Let $X,Y$ be two schemes finite type over $\Bbb Z$, assume $\#X(A)=\#Y(A)$ for every finite commutative ring $A$, then

1. must these two schemes be isomorphic ?
2. What invariants of schemes coincide on $X,Y$ (dimension, cohomology and so on)?
3. If one is smooth, must the other one be?
4. Is the condition above equivalent to $\#X(A)=\#Y(A)$ for all but finitely many finite commutative ring $A$?

The motivation for this problem is from Weil conjecture or Igusa zeta-function, which either count points over finite fields or quotient rings like $\Bbb Z/p^n\Bbb Z$ to get the information of underlying scheme.

Firstly, as every finite ring is Artinian we can only consider finite local rings. If $X$ is a equidimensional smooth scheme finite type over $\Bbb Z$, then by criterion of formally etaleness we know that $\#X(A)=\#X(A/I)(\#I)^{\text {dim} X}$ for every ring $A$ with square-zero ideal $I$ i.e $I^2=0$ if both sides are finite. (Working locally, we can assume $X \overset{\text{etale}}\rightarrow \Bbb A_\Bbb Z ^{\text {dim} X} \rightarrow \text{Spec} \Bbb Z$, then $X(A)=X(A/I)\times_{\Bbb A_\Bbb Z ^{\text {dim} X}(A/I)} \Bbb A_\Bbb Z ^{\text {dim} X}(A)$ as sets and $\Bbb A_\Bbb Z ^{\text {dim} X}(A) \rightarrow \Bbb A_\Bbb Z ^{\text {dim} X}(A/I)$ is surjective with each fiber has $(\#I)^{\text {dim} X}$ elements).

Hence in above case we have $\#X(A)=\#X(A/m)\prod_{i=1}^{\infty}(\#m^i/m^{i+1})^{\text {dim}X}$ for any finite local ring $(A,m)$, so knowing information over finite fields is enough to decide information over every finite rings in the smooth case. So for equidimensional smooth proper schemes over $\text{Spec} \Bbb Z$ , the condition is equivalent to their Hasse-Weil function equal.

Secondly, two elliptic curves over $\Bbb F_q$ have same zeta-function if and only if they are isogenous. So two isogenous non-isomorphic elliptic curves over $Spec \Bbb Z$ may serve as a counter example, but there is no elliptic curves over $\Bbb Q$ with good reduction everywhere. (Probably the same holds for abelian varieties). In some sense, smooth proper schemes over $\text{Spec} \Bbb Z$ are rare (those connected etale ones are trivial by Minkowski's theorem), so I think we need to consider singular ones in general.

On the other hand, as in Classification of finite commutative rings shows finite rings are rich, we already have lots of finite rings like $\mathbb F_p[x,y] / \langle x^2, xy, y^2\rangle$. Furthermore, there is an estimate for the number of commutative rings of order $≤N$. It is

$exp[\frac{2}{27} \frac{log(N)^3}{(log 2)^2} \; +O(log(N)^{\frac {8}{3}})] \quad for N\to \infty$

by Bjorn Poonen, and many related interesting theorems is in the article

So in the non-smooth case the condition over finite rings may not be totally determined by finite fields, there may be new phenomenons. This recent article explains some relationships of rational singularity and growth of $\#X(\Bbb Z/p^n\Bbb Z)$.

At last, are there some examples of computation of $\Bbb \#X(A)$ in the non-smooth case? I worked out some cases for $X=\text{Spec} \Bbb Z[x,y]/(y^2-x^3)$ but not completely.

Edit: I am mostly interested in the case $X$ is proper or $X \rightarrow \text{Spec} \Bbb Z$ is surjective, as there are counterexamples using isogenous elliptic curves for the first problem in below's comments. The title contains the words "commutative" because we can also define points over finite rings in general (like $M_n(\Bbb F_q)$) for $X$ finite type over $\Bbb Z$ and ask the same question, which I have no idea to deal with even in the smooth case at present.

• If $E$ and $E'$ are isogenous elliptic curves over $\mathbb Q$ with minimal Weierstraß models $\mathcal E$, $\mathcal E'$, and $U \subseteq \operatorname{Spec} \mathbb Z$ is the locus where $\mathcal E$ and $\mathcal E'$ have good reduction, doesn't what you say imply that $X = \mathcal E|_U$ and $Y = \mathcal E'|_U$ are a counterexample to your first question? (Properness is not needed for your argument in the smooth case.) Nov 14, 2017 at 6:43
• @R.vanDobbendeBruyn Oh, thanks for that observation. We can ignore bad points by base change to the good parts, but it is somehow artificial to just ignore them. Also even in this case the two curves are isogenous, so I wonder whether there will be a counter example with the proper condition or not the empty scheme over every point of $Spec \Bbb Z$. Nov 14, 2017 at 7:04

No, that is not true. Let $B$ be a finite type scheme over $\text{Spec}\ \mathbb{Z},$ for instance, $B=\mathbb{P}^1_{\text{Spec}\ \mathbb{Z}}.$ Let $X,$ respectively $Y,$ be the relative Proj of the symmetric algebra of a locally free sheaf $E,$ resp. $F,$ on $B.$ If the rank of $E$ equals the rank of $F,$ then $X$ and $Y$ have the same number of points over every finite ring. Yet there is no reason that $X$ should equal $Y,$ e.g., for Hirzebruch surfaces the minimal self-intersection number of an irreducible curve on the surface is an invariant that distinguishes Hirzebruch surfaces.
• Another example is $X = \mathbb{P}^1_{\mathbb{Z}}$ and $Y = \mathbb{A}^1_{\mathbb{Z}} \sqcup \mathrm{Spec}(\mathbb{Z})$. Though your examples are proper so it even better. This answers 1)&2) but 3)&4) are left.
• @js21. I am not certain that those are examples. The OP wants to consider all finite rings, not just finite fields. So that gives the OP access to, for instance, the finite-field point-counting functions of arc spaces over the scheme. In your examples, I believe this distinguishes the two examples: for a finite field $\mathbb{F}_q$, for the finite ring $\mathbb{F}_q[\epsilon]/\langle \epsilon^2 \rangle,$ your first scheme has $q^2-1$ points, yet your second scheme has $q^2-q+1$ points. Nov 14, 2017 at 11:26
• Oh, you are right (though I am counting $q(q+1)$ points for $X$).
• @js21. You are correct. I was counting only closed immersions from $\text{Spec}\ \mathbb{F}_q[\epsilon]/\langle \epsilon^2 \rangle$ (those are the ones parameterized by arc spaces). The correct point counts are, respectively, $q(q+1)$ and $q^2+1.$ Nov 14, 2017 at 12:14
• Thanks a lot, by your answer I realize that if $X$ is covered by $U_i$, then $\#X(A)=\sum \#U_i(A)- \sum \#U_i \cap U_j(A)+\dots$ for every finite local ring $A$ as $\text{Spec} A$ is just a point so we can count points locally. But locally isomorphic does not imply global isomorphic , so we get the counterexample. Nov 14, 2017 at 12:31