On the number of universal algebraic geometries

“$\Leftarrow$”: We assume that for all $f,g\in\mathcal{C}^{[n]}$ with $f|_{X}=g|_{X}$, we have $f(\mathbfsl{a})=g(\mathbfsl{a})$. We show that $\mathbfsl{a}\in\text{V}_{\mathcal{C}}(X)$. To this end, we show that for each $B\in\operatorname{Alg}_{n}\mathcal{C}$ with $X\subseteq B$, we have $\mathbfsl{a}\in B$. Since $B$ is algebraic with respect to $\mathcal{C}$, there exists a system of $\mathcal{C}$-equations $\{p_{i}\approx q_{i}\mid i\in I\}$ such that $B=\{\mathbfsl{x}\in A^{n}\mid\forall i\in I\colon p_{i}(\mathbfsl{x})=q_{i}(\mathbfsl{x})\}$. Since $X\subseteq B$, we have $p_{i}|_{X}=q_{i}|_{X}$ for all $i\in I$. Therefore, the assumption yields $p_{i}(\mathbfsl{a})=q_{i}(\mathbfsl{a})$ for all $i\in I$, and thus $\mathbfsl{a}\in B$.

For each $d\in\mathbb{N}\setminus\{1\}$, let ${\mathbf{A}}_{d}$ and $f_{d}$ be as in the statement of Lemma 3.1. For $d\in\mathbb{N}\setminus\{1\}$, for $m\in\mathbb{N}$ and for $X\subseteq\mathbb{Z}_{np^{2}}^{m}$, we use $\text{V}_{d}(X)$ as a shorthand for $\text{V}_{\operatorname{Pol}({\mathbf{A}}_{d})}(X)$. Let $l,i\in\mathbb{N}$ with $l>i\geq 2$. We claim that $\operatorname{Pol}({\mathbf{A}}_{i})\not\sim_{\text{alg}}\operatorname{Pol}({\mathbf{A}}_{l})$. For proving this, we let

and show that $\text{V}_{i}(Q)\neq\text{V}_{l}(Q)$; as a consequence $\operatorname{Pol}({\mathbf{A}}_{i})\not\sim_{\text{alg}}\operatorname{Pol}({\mathbf{A}}_{l})$.

First we prove that $(1,\dots,1)\notin\text{V}_{l}(Q)$. To this end, we consider the term equation $f_{l}(x_{1},\dots,x_{l})\approx 0$. We have that $f_{l}(1,\dots,1)=np$ and $f_{l}|_{Q}=0$. Hence Lemma 2.1 yields $(1,\dots,1)\not\in\text{V}_{l}(Q)$.

We now prove that $\text{V}_{i}(Q)=\mathbb{Z}_{np^{2}}^{l}$. To this end, let us consider an equation of the form $t_{1}(x_{1},\dots,x_{l})\approx t_{2}(x_{1},\dots,x_{l})$ with $t_{1},t_{2}\in\operatorname{Pol}_{l}({\mathbf{A}}_{i})$. This equation is equivalent to $(t_{1}-t_{2})\,(x_{1},\dots,x_{l})\approx 0$. Hence it suffices to consider equations of the form $t(x_{1},\dots,x_{l})\approx 0$. If $t(x_{1},\dots,x_{l})\approx 0$ is satisfied by all elements of $Q$, then the function $t$ is absorbing and has arity greater than $i$. Lemma 3.1 implies that $t$ is the constant zero function. Therefore, the equation $t(x_{1},\dots,x_{l})\approx 0$ is satisfied by all elements in $\mathbb{Z}_{np^{2}}^{l}$. Hence Lemma 2.1 yields $(1,\dots 1)\in\text{V}_{i}(Q)$ and therefore $\text{V}_{i}(Q)\neq\text{V}_{l}(Q)$.

Let $K$ be the set of all unary constant functions on $A^{\oplus}$. We will show that all projections on $A^{\oplus}$ satisfy (4.1), and that for all $k,n\in\mathbb{N}$, for all $n$-ary

and for all $k$-ary $t_{1},\ldots,t_{n}$ that satisfy (4.1), also $s(t_{1},\ldots,t_{n})$ satisfies (4.1). From this, we conclude that every function in $\mathcal{C}^{\oplus}$ satisfies (4.1).

Clearly, all the projections satisfy (4.1).

Let $k,n\in\mathbb{N}$, let $t_{1},\ldots,t_{n}$ be $k$-ary functions on $A^{\oplus}$ that satisfy (4.1), and let $s$ be an $n$-ary function on $A^{\oplus}$ from $\{f^{\oplus}\mid f\in\mathcal{C}\}\cup\{\cdot\}\cup K$. We will show that $s(t_{1},\ldots,t_{n})$ satisfies (4.1).

We first consider the case that $s=f^{\oplus}$ with $f\in\mathcal{C}^{[n]}$. If there exists $i\in[{n}]$ such that $t_{i}|_{A^{k}}=0$, then by the definition of $f^{\oplus}$, $f^{\oplus}(t_{1},\dots,t_{n})|_{A^{k}}=0$, and hence $s(t_{1},\ldots,t_{n})$ satisfies (4.1). If for all $i\in[{n}]$, there exists $\hat{t_{i}}\in\mathcal{C}^{[k]}$ such that $t_{i}|_{A^{k}}=\hat{t_{i}}$, then by the definition of $f^{\oplus}$ we have

Therefore $s(t_{1},\ldots,t_{n})$ satisfies (4.1).

Now we consider the case that $s$ is the function $\cdot$ (and $n=2$). Let $t_{1},t_{2}$ be $k$-ary functions on $A^{\oplus}$ satisfying (4.1). We show that $t_{1}\cdot t_{2}$ satisfies (4.1). Since $t_{1},t_{2}$ satisfy (4.1), we have to consider the following cases.

Case 1: $t_{1}|_{A^{k}}=t_{2}|_{A^{k}}=0$: In this case, the definition of $\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=1$. Since $\mathcal{C}$ is constantive, $t_{1}\cdot t_{2}|_{A^{k}}\in\mathcal{C}$, and thus $t_{1}\cdot t_{2}$ satisfies (4.1).

Case 2: $t_{1}|_{A^{k}}=0$ and there exists $\hat{t_{2}}\in\mathcal{C}^{[k]}$ such that $t_{2}|_{A^{k}}=\hat{t_{2}}$: In this case, the definition of $\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=\hat{t_{2}}$. This implies that $t_{1}\cdot t_{2}|_{A^{k}}\in\mathcal{C}$, and thus $t_{1}\cdot t_{2}$ satisfies (4.1).

Case 3: $t_{2}|_{A^{k}}=0$ and there exists $\hat{t_{1}}\in\mathcal{C}^{[k]}$ such that $t_{1}|_{A^{k}}=\hat{t_{1}}$: In this case, the definition of $\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=\hat{t_{1}}$. This implies that $t_{1}\cdot t_{2}|_{A^{k}}\in\mathcal{C}$, and thus $t_{1}\cdot t_{2}$ satisfies (4.1).

Case 4: There exist $\hat{g_{1}},\hat{g_{2}}\in\mathcal{C}^{[k]}$ such that $g_{1}|_{A^{k}}=\hat{g_{1}}$ and $g_{2}|_{A^{k}}=\hat{g_{2}}$: In this case, the definition of $\cdot$ yields $t_{1}\cdot t_{2}|_{A^{k}}=0$, thus $t_{1}\cdot t_{2}$ satisfies (4.1).

Without loss of generality, we assume that there exists $B\subseteq A^{k}$ that is algebraic with respect to $\mathcal{C}_{1}$ but not with respect to $\mathcal{C}_{2}$. Note that then $B\neq A^{k}$. Since $\mathcal{C}_{2}$ is constantive, $B$ is not empty. Then there exists $\mathbfsl{a}\in A^{k}\setminus B$ such that $\mathbfsl{a}\in\text{V}_{\mathcal{C}_{2}}(B)$.

Let $B_{0}:=B\cup((A^{\oplus})^{k}\setminus A^{k})$. We claim that

To prove (4.2), let $\mathbfsl{c}\notin B_{0}$. By Lemma 2.1 it suffices to show that there are two functions $p_{1},p_{2}\in\mathcal{C}_{1}^{\oplus}$ such that $p_{1}|_{B_{0}}=p_{2}|_{B_{0}}$ and $p_{1}(\mathbfsl{c})\neq p_{2}(\mathbfsl{c})$. Since $\mathbfsl{c}\in A^{k}\setminus B$ and $B$ is algebraic with respect to $\mathcal{C}_{1}$, Lemma 2.1 implies that there are $f_{1},f_{2}\in\mathcal{C}_{1}^{[k]}$ such that $f_{1}|_{B}=f_{2}|_{B}$ and $f_{1}(\mathbfsl{c})\neq f_{2}(\mathbfsl{c})$. We set $p_{1}=f_{1}^{\oplus}$ and $p_{2}=f_{2}^{\oplus}$. Then $p_{1}|_{B_{0}}=p_{2}|_{B_{0}}$ and $p_{1}(\mathbfsl{c})\neq p_{2}(\mathbfsl{c})$, which concludes the proof of (4.2).

We now claim that

Suppose that $\mathbfsl{a}\notin\text{V}_{\mathcal{C}_{2}^{\oplus}}(B_{0})$. Then Lemma 2.1 implies that there are $f_{1},f_{2}\in(\mathcal{C}_{2}^{\oplus})^{[k]}$ such that $f_{1}|_{B_{0}}=f_{2}|_{B_{0}}$ and $f_{1}(\mathbfsl{a})\neq f_{2}(\mathbfsl{a})$. Since $B\subseteq B_{0}$ we infer that $f_{1}|_{B}=f_{2}|_{B}$. From Lemma 4.2 and $B\neq\emptyset$, we deduce that $f_{1}|_{B}=f_{2}|_{B}\neq 0$. Moreover, by Lemma 4.1, there are $\hat{f_{1}},\hat{f_{2}}\in\mathcal{C}_{2}^{[k]}$ such that $f_{1}|_{A^{k}}=\hat{f_{1}}$ and $f_{2}|_{A^{k}}=\hat{f_{2}}$. Therefore, $\hat{f_{1}}|_{B}=\hat{f_{2}}|_{B}$ and $\hat{f_{1}}(\mathbfsl{a})\neq\hat{f_{2}}(\mathbfsl{a})$. Hence by Lemma 2.1, $\mathbfsl{a}\notin\text{V}_{\mathcal{C}_{2}}(B)$, contradicting the choice of $\mathbfsl{a}$.

(1) Let $n\in\mathbb{N}$, let $f\in F_{n}$ with $\text{essArity}(f)\leq 1$ and let $\mathbfsl{x}\in Z^{n}$. We prove that $f(\mathbfsl{x})=0$. If $\text{essArity}(f)\leq 1$, then there exists $i\in[{n}]$ such that for all $j\in[{n}]\setminus\{i\}$ the function $f$ does not depend on its $j$-th argument. Let us define $f^{\prime}\in Z^{Z}$ by $f^{\prime}(y)=f(y,\dots,y)$ for all $y\in Z$. We prove that $f^{\prime}=0$. To this end, let $a\in Z$. Since $(a,\dots,a)\notin D_{n}$, we have $f^{\prime}(a)=f(a,\dots,a)=0$. Moreover, since for all $j\neq i$, $f$ does not depend on its $j$-th argument, $f(\mathbfsl{x})=f(x_{i},\dots,x_{i})=f^{\prime}(x_{i})=0$. This concludes the proof of (1).

(2) Let $n\in\mathbb{N}$ and let $f\in F_{n}$. If $f$ is not constantly zero, then there exists $\mathbfsl{a}\in D_{n}$ such that $f(\mathbfsl{a})=1$. Changing any one of the coordinates of $\mathbfsl{a}$ to $0$, the value of $f$ changes from $1$ to $0$. Hence $f$ depends on all of its arguments and (1) implies that it cannot be essentially unary. This concludes the proof of (2).

(3) Let $f\in F^{\prime}_{n}$ with $\text{essArity}(f)=1$. Seeking a contradiction, we suppose that $f$ is not a projection. Then $f$ belongs to $F$ up to inessential arguments. Thus, there exist $1\leq j_{1}<\dots<j_{k}\leq n$ and $\hat{f}\in F$ such that $f(x_{1},\dots,x_{n})=\hat{f}(x_{j_{1}},\dots,x_{j_{k}})$ for all $\mathbfsl{x}\in Z^{n}$. If $\text{essArity}(\hat{f})\leq 1$, then (1) yields $\hat{f}=0$, and so $f$ is the constant $0$-function, contradicting $\text{essArity}(f)=1$. If $\text{essArity}(\hat{f})>1$, then $\text{essArity}(f)>1$, contradicting $\text{essArity}(f)=1$. This concludes the proof of (3).

(4) Let us assume that the image of $\rho$ is $\{j_{1},\dots,j_{k}\}\subseteq[{n}]$, with $j_{1}<\dots<j_{k}$, and let us define

We first prove that $h$ is a functional relation. To this end, let $\mathbfsl{x},\mathbfsl{y}\in Z^{n}$ be such that $(x_{j_{1}},\dots,x_{j_{k}})=(y_{j_{1}},\dots,y_{j_{k}})$. We show that $g(x_{\rho(1)},\dots,x_{\rho(l)})=g(y_{\rho(1)},\dots,y_{\rho(l)})$. Clearly, if $(x_{j_{1}},\dots,x_{j_{k}})=(y_{j_{1}},\dots,y_{j_{k}})$, then $(x_{\rho(1)},\ldots,x_{\rho(l)})=(y_{\rho(1)},\ldots,y_{\rho(l)})$, and therefore $g(x_{\rho(1)},\dots,x_{\rho(l)})=g(y_{\rho(1)},\dots,y_{\rho(l)})$. This concludes the proof that $h$ is a functional relation. Now we prove that $h\in F_{k}$. Condition (5.1) is clearly satisfied since $g\in F_{l}$. We prove that $h$ satisfies (5.2). To this end, let $\mathbfsl{x}\in A^{n}$ with $g(x_{\rho(1)},\dots,x_{\rho(l)})=1$. We show that $(x_{j_{1}},\dots,x_{j_{k}})\in D_{k}$. Since $g\in F_{l}$ and $g(x_{\rho(1)},\dots,x_{\rho(l)})=1$, there exists $a\in[{l}]$ such that $x_{\rho(a)}=1$ and $x_{\rho(b)}=2$ for all $b\in[{l}]\setminus\{a\}$. Since $j_{1}<\dots<j_{k}$ we have $\lvert\{r\in[{k}]\colon j_{r}=\rho(a)\}\rvert\leq 1$. Moreover, since $\{j_{1},\dots,j_{k}\}$ is the image of $\rho$, we have $\lvert\{r\in[{k}]\colon j_{r}=\rho(a)\}\rvert\geq 1$. Thus, $\lvert\{r\in[{k}]\colon j_{r}=\rho(a)\}\rvert=1$. Hence there exists $i\in[{k}]$ such that $x_{j_{i}}=x_{\rho(a)}=1$. Then for each $r\in[{k}]\setminus\{i\}$ there exists $b\in[{l}]\setminus\{a\}$ such that $j_{r}=\rho(b)$, and then $x_{j_{r}}=x_{\rho(b)}=2$. Thus, $(x_{j_{1}},\dots,x_{j_{k}})\in D_{k}$, and so $h$ satisfies (5.2). This proves that $h\in F_{k}$. Since $j_{1}<\dots<j_{k}$ and $g_{\rho}(\mathbfsl{a})=h(a_{j_{1}},\dots,a_{j_{k}})$ for all $\mathbfsl{a}\in A^{n}$, $g_{\rho}$ belongs to $F$ up to inessential arguments. Hence $g_{\rho}\in F_{n}^{\prime}$. This concludes the proof of (4).