On the nonexistence of some open immersions

1.   Introduction

The topology involved in this paper is the Zariski topology. The closed sets of affine space $\mathbf{A}_{k}^{n}$ are those of the form $V(S) = \{x \in \mathbf{A}_{k}^{n} \mid f(x) = 0, \forall f \in S\}$ where $S$ is any set of polynomials in $n$ variables over $k$. A variety is an integral separated scheme of finite type over an algebraically closed field.

It is well-known that $GL_{m/k}$ is an open subset of $\mathbf{A}^{m^{2}}_{k}$, where $k$ is a field. A natural question is whether there exists an open immersion $\phi : SL_{m/k} \hookrightarrow \mathbf{A}^{m^{2}-1}_{k}$. If $k$ is algebraically closed, it is easy to show that such an open immersion does not exist, by combining two classical results on algebraic groups. Indeed, if such an immersion exists, then the complement of its image is the zero locus of some polynomial function $f$. The restriction of this function to $SL_{m/k}$ is a regular invertible function on this connected algebraic group, and hence a scalar multiple of a character by a result of Rosenlicht(see ^[8]). Since every character of $SL_{m}$ is trivial, $f$ is constant, and hence $SL_{m/k}$ is an affine space. But this contradicts the fact that every algebraic group with underlying variety an affine space is unipotent(see ^[6]). In this paper we study the nonexistence of some open immersions between affine varieties over a field $k$, and we will prove a theorem which can explain the nonexistence of $\phi$ in another way. In addition, there is a result which is similar in spirit: any injective endomorphism of an affine variety is also surjective(see ^[7]). The main theorem is Theorem 2.15.

Since an open subset of an irreducible affine scheme is dense, we can reduce to the case where the open immersion $\phi:X \hookrightarrow Y$ is dominant.We first prove the theorem when $k$ is a finite field and then we prove a more general result by reducing to this case. As a simple application, we have the following conclusions: open immersions $SL_{n/k} \hookrightarrow \mathbf{A}_{k}^{n^{2}-1}$, $Sp_{n/k} \hookrightarrow \mathbf{A}_{k}^{2n^{2}+n}$, $SO_{n/k} \hookrightarrow \mathbf{A}_{k}^{\frac{n^{2}-n}{2}}$, $PGL_{n/k} \hookrightarrow \mathbf{A}_{k}^{n^{2}-1}$ do not exist, and open immersions $SL_{n/k}\times_{k} \mathbf{G}_{a/k} \hookrightarrow \mathbf{A}_{k}^{n^{2}}$, $Sp_{n/k} \hookrightarrow SO_{2n+1/k}$, $Sp_{n/k} \hookrightarrow SL_{n/k}\times_{k} \mathbf{A}^{n^{2}+n+1}_{k}$ do not exist either, where $k$ is an arbitrary field.

2.   Results

2.1.   Result over a finite field

To prove the main result of this subsection, we require the Lang-Weil estimate and some other lemmas.

Let $X$ be an irreducible scheme defined over a finite field $\mathbf{F}_{q}$. Assume that $X$ is embedded into a projective space of fixed dimension $n$, dim$(X) = r$, deg$(X) = d$, then we have universal estimates for #$X(\mathbf{F}_{{q}^{k}})$, in terms of $r$, $d$, and $q^{k}$. More precisely, we show the following:

LEMMA 2.1 ([Lang-Weil]). Given nonnegative integers $n$, $d$ and $r$, with $d > 0$, there is a positive constant $A(n, d, r)$, such that for every finite field $k = \mathbf{F}_{q}$, and every absolutely irreducible subscheme $X\subseteq \mathbf{P}_{k}^{n}$ of dimension $r$ and degree $d$, we have

Proof. See ^[11].

LEMMA 2.2. Given $n$, $d$ and $r$ as in Lemma 2.1, there is a positive constant $A_{1}(n, d, r)$, such that for every finite field $k = \mathbf{F}_{q}$, and every subscheme $X\subseteq \mathbf{P}_{k}^{n}$ of pure dimension $r$ and degree $\le d$, we have

Proof. See Lemma 1 in ^[11].

If X is allowed to have components of smaller dimension, we still have

COROLLARY 2.3. If $X$ is an $r$-dimensional scheme over $k = \mathbf{F}_{q}$, then there is $c_{X} > 0$ such that $\#X(\mathbf{F}_{q^{e}})\le c_{X}q^{er}$ for every $e\ge 1$.

Proof. The proof is omitted.

COROLLARY 2.4. If $X$ is an $r$-dimensional absolutely irreducible scheme over $\mathbf{F}_{q}$, then there is $c_{X} > 0$ such that $|\#X(\mathbf{F}_{q^{e}})-q^{er}|\le c_{X}q^{e(r-\frac{1}{2})}$ for every $e\ge 1$.

Proof. The proof is omitted.

LEMMA 2.5. Let $A$ be an integrally closed noetherian domain. Then

where the intersection is taken over all prime ideals of height 1.

Proof. See Theorem 38 in ^[1].

LEMMA 2.6. For any noetherian, integral, normal affine scheme $X$ = Spec $A$, and any nonempty closed subscheme $Z$ of $X$ with codimension at least 2, $X-Z$ is not an affine scheme.

Proof. If $X-Z$ is affine, then $i: X-Z \rightarrow X$ is a morphism of affine schemes, hence $i$ is totally determined by $i^{\#}: \Gamma(X, \mathcal{O}_{X})\rightarrow \Gamma(X-Z, \mathcal{O}_{X-Z}) = \Gamma(X-Z, \mathcal{O}_{X})$. Since $X$ is integral, $i^{\#}$ is injective.

Moreover, since $Z$ does not contain any codimension 1 point of $X$, so for any $f \in \Gamma(X-Z, \mathcal{O}_{X})$, $f$ is regular at all codimension 1 points.Hence by Lemma 2.5, $f$ is regular on $X$ bexause $A$ is integrally closed. As a result, $i^{\#}$ is surjective.

In sum, $i^{\#}$ is both injective and surjective, so it is actualy an isomorphism. Therefore $i$ is an isomorphism, which means $Z$ is empty and we get a contradiction.

COROLLARY 2.7. If $X$ is a noetherian, integral, separated scheme $X$, $U$ is an affine open subset, then the complement of $U$ has codimension 1 in $X$.

Proof. Replace $X$ by its normalization, and $U$ by its preimage in$X$. Then the codimension of the complement of $U$ doesn't change, and so we reduce to the normal case. Intersecting $U$ with the members of an open affine cover of $X$, we reduce to the case when $X$ is affine.($X$ is separated, so the intersection of two open affines is open affine.)Then Lemma 2.6 applies.

We can now prove the main result of this subsection.

THEOREM 2.8. Assume X and Y are affine $n$-dimensional schemes over $\mathbf{F}_{q}$, $\#X(\mathbf{F}_{q^{m}}) \ne \#Y(\mathbf{F}_{q^{m}})$, if $\left|\#X(\mathbf{F}_{q^{m}}) - \#Y(\mathbf{F}_{q^{m}})\right| = o(q^{m(n-1)})\quad (m \rightarrow \infty \quad m \in \mathbb{Z})$, then there does not exist an open immersion $X \hookrightarrow Y$.

$o(q^{m(n-1)})$ represents a polynomial of $q$, denoted by $f$, such that $\mathop{\text{lim}}\limits_{m \rightarrow \infty} \frac{f(q^{m})}{q^{m(n-1)}} = 0$.

Proof. Suppose that there exists an open immersion $X \hookrightarrow Y$. Let $Y = \text{Spec}R$, $V = Y \setminus X$ which is associated with an induced scheme structure $R/I$ that is reduced, $I \subset R$ is the defining ideal of $V$.

We denote by $V_{\overline{\mathbf{F}_{q}}}$ the scheme $V\times _{\text{Spec}\mathbf{F}_{q}}\text{Spec}\overline{\mathbf{F}_{q}}$.

Let $X_{1}, X_{2}, \ldots, X_{l}$ be the irreducible components of $V_{\overline{\mathbf{F}_{q}}}$ of maximal dimension $r$, there is a finite extention $\mathbf{F}_{q^{e}}$ of $\mathbf{F}_{q}$ such that for some closed subscheme $V_{i}$ of $V_{\mathbf{F}_{q^{e}}}$, we have $V_{i}\times _{\text{Spec}\mathbf{F}_{q^{e}}}\text{Spec}\overline{\mathbf{F}_{q}} = X_{i}$, $1 \le i \le l$. This implies that each $V_{i}$ is absolutely irreducible. Note that the dimension of any other irreducible component of $V_{\mathbf{F}_{q^{e}}}$ is smaller than $r$, and $\text{dim}(V_{i}\bigcap V_{j}) < r$ when $i\ne j$.

Combining Corollary 2.3 with Corollary 2.4, we have

That is

Then we have

According to Corollary 2.7, $r = n-1$, we obtain

On the other hand,

this implies

This contradiction shows that there does not exist an open immersion $X$ $\hookrightarrow$ $Y$.

2.2.   Result over an arbitrary field

We need the following four lemmas to prove the main theorem.

LEMMA 2.9. Let $A\subseteq B$ be two integral domains. If $B$ is faithfullly flat over $A$ and $\text{qf}$($A$) = $\text{qf}$($B$). Then $A = B$.

qf($A$) means the quotient field of $A$, and qf($B$) is similar.

Proof. Take $x\in B$ with $x = b/a$ ($a, b\in A)$. B is faithfully flat over A, so it follows that $b = ax\in aB\cap A = aA$ (cf.[^[1], (4.C)]). Hence $x = b/a\in A$. Therefore $A = B$.

LEMMA 2.10. Let $A$ be an integral domain with quotient field $K$ and let $B$ be an extension of A. $B$ is a finitely generated $A$-algebra. Then the canonical morphism $f:\mathit{\text{Spec}}(B) \to \mathit{\text{Spec}}(A)$ is an open immersion $\Longleftrightarrow$ $B$ is a birational extension of $A$ in $K$ and $B$ is flat over $A$.

Proof. $\Rightarrow$ $B$ is flat over $A$ if and only if $B_{P}$ is flat over $A_{p}$ for every $P\in B$ ($p = P\cap A$). Since $f$ is an open immersion, it follows that $B_{P}\cong A_{p}$, and $B_{P}$ is flat $A_{p}$. Furthermore $f$ is birational, this implies that $\text{qf} (B)$ = $K$.

$\Leftarrow$ Any flat morphism that is locally of finite type is open. Next we will show that this map is injective. Let $P, P'\in \text{Spec}(B)$ with $P\cap A = P'\cap A: = p$, then $A_{p} \rightarrow B_{P}$ is flat. As a flat extension of rings satisfies Going-Down Theorem(cf.[^[1], (5.D)]), $\text{Spec}(B_{P}) \rightarrow \text{Spec}(A_{p})$ is surjective, it follows that $B_{P}$ is faithfully flat over $A_{P}$(cf.[^[1], (4.D)]). Hence $A_{p} = B_{P}$ by Lemma 2.9. Similary, $A_{p} = B_{P'} = B_{P}$. Hence $P = P'$. So $\text{Spec}(B)$ is homeomorphic to an open subset $U$ of $\text{Spec}(A)$. Since open sets of the form $D(g)$ form a base for the topology of $\text{Spec}(A)$, take $B_{g} \subset U$, it is enough to prove $A_{g} \cong B_{g}$. We have a surjective homomorphism $B_{g}\otimes_{A_{g}}B_{g} \overset{\phi}{\twoheadrightarrow} B_{g}$, since $B_{g}\otimes_{A_{g}}B_{g}\subset B_{g}\otimes_{A_{g}}K = K$, $\phi$ is also injective, namely $\phi$ is an isomorphism. As $B_{g}$ is faithfully flat over $A_{g}$, we conclude that $A_{g} \cong B_{g}$.

LEMMA 2.11. Let $A\subset B\subset C$ be rings. Suppose that $A$ is Noetherian, that $C$ is finitely generated as an $A$-algebra and that $C$ is either (1) finitely generated as a $B$-module or (2) integral over $B$. Then $B$ is finitely generated as an $A$-algebra.

Proof. See Proposition 7.8 in ^[5].

LEMMA 2.12. Let $k$ be a finitely generated $\mathbb{Z}$-algebra. If $k$ is a field, then $k$ is finite.

Proof. We have a homomorphism $\mathbb{Z}\overset{f}{ \to }k$, if ker($f$) = $p\mathbb{Z}$($p$ is a prime), then $k$ is a finitely generated $\mathbf{F}_{p}$-algebra, so $k$ is a finite algebraic extension of $\mathbf{F}_{p}$(cf.[^[5], (7.9)]), and $k$ is a finite field.

If ker($f$) = (0), we have $\mathbb{Z}\subset \mathbb{Q}\subset k$. Since $k$ is a finitely generated $\mathbb{Z}$-algebra, it is a finitely generated $\mathbb{Q}$-algebra, similarly, $k$ is a finite algebraic extension of $\mathbb{Q}$, hence $k$ is a finitely generated $\mathbb{Q}$-module, by Lemma 2.11, $\mathbb{Q}$ is a finitely generated $\mathbb{Z}$-algebra, a contradiction. Let $\mathbb{Q} = \mathbb{Z}[c_{1}, \dots, c_{s}], \quad c_{i} = \frac{a_{i}}{b_{i}}, \quad a_{i}, b_{i}\in \mathbb{Z}, 1\le i\le s$, take $p$ such that $p$ and $b_{i}$ are coprime, then $\frac{1}{p}\notin \mathbb{Z}[c_{1}, \dots, c_{s}]$, this contradiction shows that $k$ is a finite field.

Suppose $S_{0}$ is a scheme, and $\mathscr{A}_{\lambda}$ are commutative quasi-coherent $\mathcal{O}_{S_{0}}$-algebras, then $\mathscr{A} = \mathop{\text{lim}}\limits_{ \to } \mathscr{A}_{\lambda}$ is a quasi-coherent $\mathcal{O}_{S_{0}}$-algebra. Denote by $S_{\lambda}$(resp.$S$) the spectrum of the $\mathcal{O}_{S_{0}}$-algebra $\mathscr{A}_{\lambda}$(resp.$\mathscr{A}$), and let $u_{\lambda\mu}:S_{\mu} \rightarrow S_{\lambda}$(for $\lambda \le \mu$) and $u_{\lambda}:S\rightarrow S_{\lambda}$ be respectively the $S_{0}$-morphisms corresponding to homomorphisms $\varphi_{\mu\lambda}:\Gamma(S_{0}, \mathscr{A}_{\lambda}) \rightarrow \Gamma(S_{0}, \mathscr{A}_{\mu})$ and $\varphi_{\lambda}:\Gamma(S_{0}, \mathscr{A}_{\lambda}) \rightarrow \Gamma(S_{0}, \mathscr{A})$; it is clear that ($S_{\lambda}$, $u_{\lambda\mu}$) is a projective system in the category of $S_{0}$-schemes.

Given two $S_{\alpha}$-schemes $X_{\alpha}$, $Y_{\alpha}$, we define two projective systems of ($X_{\lambda}$, $v_{\lambda\mu}$) and ($Y_{\lambda}$, $w_{\lambda\mu}$) by setting $X_{\lambda} = X_{\alpha}\times_{S_{\alpha}}S_{\lambda}$, $Y_{\lambda} = Y_{\alpha}\times_{S_{\alpha}}S_{\lambda}$, $v_{\lambda\mu} = \text{id}_{X_{\alpha}}\times u_{\lambda\mu}$, $w_{\lambda\mu} = \text{id}_{Y_{\alpha}}\times u_{\lambda\mu}$(for $\alpha \le \lambda \le \mu$), whose projective limits are respectively $X = X_{\alpha}\times_{S_{\alpha}}S$, $Y = Y_{\alpha}\times_{S_{\alpha}}S$, the canonical morphisms $v_{\lambda}:X\rightarrow X_{\lambda}$ and $w_{\lambda}:Y\rightarrow Y_{\lambda}$ are respectively equal to $\text{id}_{X_{\alpha}}\times u_{\lambda}$ and $\text{id}_{Y_{\alpha}}\times u_{\lambda}$. We denote by $f_{\lambda}$ the morphism $X_{\lambda} \rightarrow Y_{\lambda}$, and for $\alpha \le \lambda \le \mu$, we have $f_{\mu} = f_{\lambda}\times \text{id}_{S_{\mu}}:X_{\lambda}\times_{S_{\lambda}} S_{\mu} \rightarrow Y_{\lambda}\times S_{\mu}$, $f = f_{\lambda}\times \text{id}_{S}:X_{\lambda}\times_{S_{\lambda}} S \rightarrow Y_{\lambda} \times_{S_{\lambda}}S$.

LEMMA 2.13. Suppose $S_{0}$ is quasi-compact, $X_{\alpha}$ and $Y_{\alpha}$ are of finite presentation over $S_{\alpha}$, let $f_{\alpha}:X_{\alpha} \rightarrow Y_{\alpha}$ be an $S_{\alpha}-morphism$.Then f is an open immersion if and only if there exists $\lambda \ge \alpha$ such that $f_{\lambda}$ is an open immersion(in which case $f_{\mu}$ is also an open immersion for $\mu \ge \lambda$).

Proof. See Theorem 8.10.5 in ^[13].

LEMMA 2.14. Let $X$ and $Y$ be affine $n$-dimensional integral schemes of finite type over a field $k$. If we have an open immersion $X \overset{f}{\hookrightarrow} Y/k$, then there exists a finitely generated $\mathbb{Z}$-subalgebra $R$ of $k$, affine scheme $\mathscr{X}$, $\mathscr{Y}$ of finite type over $S = \mathit{\text{Spec}}R$, and an open immersion $\mathscr{X} \overset{f_{S}}{\hookrightarrow} \mathscr{Y}/S$, such that $\mathscr{X}\times_{S} \mathit{\text{Spec}}k \cong X$, $\mathscr{Y}\times_{S} \mathit{\text{Spec}}k \cong Y$, and $f_{S}\times_{S} \text{id}_{k} = f$.

Proof. In order to use Lemma 2.13, we should find a field $k^{'}$, two affine schemes $X^{'}$, $Y^{'}$ that are defined over $k^{'}$, an open immersion $f^{'}: X^{'} \rightarrow Y^{'}$, also defined over $k^{'}$, such that $k^{'}$ is the quotient field of a finitely generated $\mathbb{Z}$-subalgebra $R^{'}\subset k$, and $X = X^{'}\times_{\text{Spec}k^{'}} \text{Spec}k$, $Y = Y^{'}\times_{\text{Spec}k^{'}} \text{Spec}k$, $f = f^{'}\times \text{id}_{\text{Spec}k}:X^{'}\times_{\text{Spec}k^{'}} \text{Spec}k \rightarrow Y^{'}\times_{\text{Spec}k^{'}} \text{Spec}k$, because $k^{'} = \mathop{\mathop{\text{lim}}\limits_{ \to }}\limits_{s\in R^{'}, s\ne 0}R^{'}_{s}$.

Write $[\underline{x}] = [x_{1}, \dots, x_{l}]$, $[\underline{y}] = [y_{1}, \dots, y_{m}]$, $X = \text{Spec}k[\underline{x}]/I$, $Y = \text{Spec}k[\underline{y}]/J$, where $I = (\mathtt{f}_{1}, \dots, \mathtt{f}_{s})k[\underline{x}]$ and $J = (\mathtt{g}_{1}, \dots, \mathtt{g}_{t})k[\underline{y}]$ are respectively the defining ideals of $X$ and $Y$.

If we have an open immersion $X \overset{f}{\hookrightarrow} Y/k$, by Lemma 2.10, we have a flat and birational k-morphism $f^{\#}: k[\underline{y}]/J \rightarrow k[\underline{x}]/I$. So we have an isomorphism $\text{qf}\{k[\underline{x}]/I\} \overset{g}{\rightarrow} \text{qf}\{k[\underline{y}]/J\}$.

Let $g_{0}$ be the restriction to $k[\underline{x}]/I$ of $g$.

Set $\varphi = g_{0}\circ \pi$, $\pi :k[\underline{y}]\twoheadrightarrow k[\underline{x}]/I$.

We have a homomorphism

$\bar{f_{i}}, \bar{g_{i}}\in k[\underline{y}]/J$, $\bar{f_{i}}, \bar{g_{i}}$ have no non-unit common factor, $1\le i\le l$.

Set $R_{0}$ = {the subring of $k$ generated by identity 1, all the coefficients of $\mathtt{f}_{1}, \dots, \mathtt{f}_{s}$, $\mathtt{g}_{1}, \dots, \mathtt{g}_{t}$}.

Set $R_{1}$ = {the subring of $k$ generated by $S_{0}$, all the coefficients of $\bar{f_{i}}$, $\bar{g_{i}}$ and their inverses}.

It is noted that $R_{1}$ is a finitely generated $\mathbb{Z}$-algebra.

Consider the restriction to $R_{1}[\underline{x}]$ of $\varphi$

$R_{1}[\underline{x}]$ is noetherian, so $\text{ker}\varphi_{1}$ is finitely generated.

Let $b_{1}, b_{2}, \dots, b_{\ell}$ be the generators of $\text{ker}\varphi_{1}$

Set $R_{2}$ = {the subring of $k$ generated by $R_{1}$, all the coefficients of $P_{ij}$ and their inverses. $1\le i \le \ell$, $1 \le j \le s$}

$R_{2}$ is still a finitely generated $\mathbb{Z}$-algebra.

Consider the restriction to $R_{2}[\underline{x}]$ of $\varphi$

Clearly we have $\text{ker}\varphi_{2}\supset (\mathtt{f}_{1}, \dots, \mathtt{f}_{s})R_{2}[\underline{x}]$.

We denote by $\text{IM}(\varphi_{1})$ the image of $\varphi_{1}$ in $\text{qf}\{k[\underline{y}]/J\}$, it is a subring of $\text{qf}\{k[\underline{y}]/J\}$ generated by $R_{1}$ and $\varphi_{1}(x_{i})$, so we have an exact sequence

$\text{IM}(\varphi_{1})\otimes_{R_{1}}R_{2}$ is the subring of $\text{qf}\{k[\underline{y}]/J\}$ generated by $R_{2}$ and $\varphi_{1}(x_{i})$, so $\text{ker}\varphi_{2}$ is the image of $(b_{1}, \dots, b_{\ell})R_{1}[\underline{x}]\otimes_{R_{1}}R_{2}$ in $R_{1}[\underline{x}]\otimes_{R_{1}}R_{2}\cong R_{2}[\underline{x}]$, namely $(b_{1}, \dots, b_{\ell})R_{2}[\underline{x}]$.

By (2.8) we have

$(\mathtt{f}_{1}, \dots, \mathtt{f}_{s})R_{2}[\underline{x}]\supset (b_{1}, \dots, b_{\ell})R_{2}[\underline{x}] = \text{ker}\varphi_{2}$.

Hence $\text{ker}\varphi_{2} = (\mathtt{f}_{1}, \dots, \mathtt{f}_{s})R_{2}[\underline{x}]$.

On the other hand, we have a homomorphism of rings:

$f^{\#}: k[\underline{y}]/J \rightarrow k[\underline{x}]/I$

Set $\psi = f^{\#}\circ \rho$, $\rho : k[\underline{y}] \twoheadrightarrow k[\underline{y}]/J$.

We have a homomorphism

$\bar{f^{'}_{i}}, \bar{g^{'}_{i}}\in k[\underline{x}]/I$, $\bar{f^{'}_{i}}, \bar{g^{'}_{i}}$ have no non-unit common factor, $1\le i\le m$.

Set $R_{3}$ = {the subring of $k$ generated by $R_{2}$ and all the coefficients of $\bar{f^{'}_{i}}$, $\bar{g^{'}_{i}}$ and their inverses}.

Consider the restriction to $R_{3}[\underline{y}]$ of $\psi$

Let $c_{1}, c_{2}, \dots, c_{\ell^{'}}$ be the generators of $\text{ker}\psi_{1}$

$Q_{ij}\in k[\underline{y}]$, $1\le i\le \ell^{'}$, $1\le j\le t$.

Set $R_{4}$ = {the subring of $k$ generated by $R_{3}$ and all the coefficients of $Q_{ij}$ and their inverses. $1\le i \le \ell^{'}$, $1\le j \le t$}

Let $\psi_{2}$ be the restriction to $R_{4}[\underline{y}]$ of $\psi$. Similarly, $R_{4}$ is a finitely generated $\mathbb{Z}$-algebra and $\text{ker}\psi_{2} = (\mathtt{g}_{1}, \dots, \mathtt{g}_{t})R_{4}[\underline{y}]$.

The restriction to $R_{4}[\underline{y}]/J$ of $f^{\#}$

is birational according to the construction of $R_{4}$.

Write $k_{4} = \text{qf}(R_{4})$. Obviously $k_{4} \subset k$, the restriction to $k_{4}[\underline{y}]/J$ of $f^{\#}$

is still birational.

We have a commutative diagram

$i_{1}$ is flat and $f^{\#}$ is flat, so $f\circ i_{1} = i_{2}\circ f^{\#}_{k_{4}}$ is flat. As $i_{2}$ is faithfully flat, $f^{\#}_{k_{4}}$ is flat.

According Lemma 2.10, $f_{k_{4}}$ is an open immersion.

Let $k^{'} = k_{4}$, $R^{'} = R_{4}$, $X^{'} = \text{Spec}R^{'}[\underline{x}]/I$, $Y^{'} = \text{Spec}R^{'}[\underline{y}]/J$, $f^{'} = f_{k_{4}}$, then Lemma 2.13 applies.

THEOREM 2.15. Assume $R$ is an integral domain which is a finitely generated $\mathbb{Z}$-algebra, $S = \text{Spec}R$, $\xi \in S$ is the generic point. $X$ and $Y$ are affine integral $S$-schemes of finite type, such that $X_{\xi}$ and $Y_{\xi}$ are $n$-dimensional affine schemes. If for a sufficiently general closed point $t \in S$, $\#X_{t}(\kappa(t)^{m}) \ne \#Y_{t}(\kappa(t)^{m})$, and $\left|\#X_{t}(\kappa(t)^{m}) - \#Y_{t}(\kappa(t)^{m})\right| = o(|\kappa(t)|^{m(n-1)})\quad (m \rightarrow \infty \quad m \in \mathbb{Z})$, then there does not exist an open immersion $X_{\xi} \hookrightarrow Y_{\xi}$("sufficiently general" is made precise in the proof).

Proof. $\kappa(t))$ is the residue field of the point $t$, because $t$ is a closed point of $S$, $\kappa(t))$ is a finitely generated $\mathbb{Z}$-algebra. According to Lemma 2.12, $\kappa(t))$ is a finite field. Set $\kappa(t) = \mathbf{F}_{q}$, $\kappa(t)^{m} = \mathbf{F}_{q^{m}}$. So $\#X_{t}(\kappa(t)^{m}) = \#(X \times_{S} \kappa(t)^{m})$ makes sense, so does $\#Y_{t}(\kappa(t)^{m})$.

Write $[\underline{x}] = [x_{1}, \dots, x_{l}]$, $[\underline{y}] = [y_{1}, \dots, y_{m}]$, $X = \text{Spec}R[\underline{x}]/I$, $Y = \text{Spec}R[\underline{y}]/J$, where $I = (\mathtt{f}_{1}, \dots, \mathtt{f}_{s})R[\underline{x}]$ and $J = (\mathtt{g}_{1}, \dots, \mathtt{g}_{t})R[\underline{y}]$ are respectively the defining ideals of $X$ and $Y$.

Suppose there is an open immersion $X_{\xi} \overset{f_{\xi}}{\hookrightarrow} Y_{\xi}/\kappa({\xi})$, by Lemma 2.14, there exists a finitely generated $\mathbb{Z}$-subalgebra $R^{'}$ of $\kappa({\xi})$, an open immersion

such that $\text{Spec}R^{'}[\underline{x}]/I\times_{\text{Spec}R^{'}} \text{Spec}\kappa({\xi}) \cong X_{\xi}$, $\text{Spec}R^{'}[\underline{y}]/J\times_{\text{Spec}R^{'}} \text{Spec}\kappa({\xi}) \cong Y_{\xi}$, and $f_{R^{'}}\times_{\text{Spec}R^{'}} \text{id}_{\kappa(\xi)} = f_{\xi}$.

Set $R^{''}$ = {the subring of $\kappa(\xi)$ generated by $R$ and $R^{'}$}, then $\text{Spec}R^{''}[\underline{x}]/I\times_{\text{Spec}R^{'}} \text{Spec}\kappa({\xi}) \cong X_{\xi}$, $\text{Spec}R^{''}[\underline{y}]/J\times_{\text{Spec}R^{'}} \text{Spec}\kappa({\xi}) \cong Y_{\xi}$, $\text{Spec}R^{''}[\underline{x}]/I \overset{f_{R^{''}}}{\hookrightarrow} \text{Spec}R^{'}[\underline{y}]/J$ is an open immersion, and $f_{R^{''}}\times_{\text{Spec}R^{''}} \text{id}_{\kappa(\xi)} = f_{\xi}$.

$\text{Spec}R^{''}\subset \text{Spec}R$ is a dense open subset, for a sufficiently general closed point $t \in \text{Spec}R$, we have $t \in \text{Spec}R^{''}$.

Let $P \subset R^{''}$ be the maximal ideal corresponding to $t \in S$, we have

The property of being an open immersion is stable under base change, so ${\bar{f}_{R^{''}}}$ is an open immersion.

$R^{''}/P$ is a finitely generated $\mathbb{Z}$-algebra which is a field, according to 2.12, it is a finite field, set $R^{''}/P\cong \mathbf{F}_{q}$.

Namely there exists an open immersion $X_{t}/\kappa(t)\hookrightarrow Y_{t}/\kappa(t)$.

$\left|\#X_{t}(\kappa(t)^{m}) - \#Y_{t}(\kappa(t)^{m})\right| = o(|\kappa(t)|^{m(n-1)}) (m \rightarrow \infty \quad m \in \mathbb{Z})$ implies

by Theorem 2.8, this can not happen. This contradiction shows that there does not exist an open immersion $X_{\xi} \hookrightarrow Y_{\xi}$.

REMARK 2.16. If it happens that $\#X_{t}/\kappa(t) > \#Y_{t}/\kappa(t)$, there does not exist an open immersion $X \hookrightarrow Y$.

REMARK 2.17. Denote $\left|\#X_{t}(\kappa(t)^{m}) - \#Y_{t}(\kappa(t)^{m})\right|$ by $f_{m}$, if $\text{deg}f_{m} \ne m(\text{dim}X-1) (\forall m \in \mathbb{Z})$, there does not exist an open immersion $X \hookrightarrow Y$.

REMARK 2.18. Given $X$ and $Y$ as in Theorem 2.15, if $Y$ is quasi-projective, there does not exist an open immersion $X \hookrightarrow Y$. Regard $X$ as an affine open subscheme of $Y$, for any affine open subscheme $U \subset Y$, $U \cap X$ is still affine because $Y$ is separated over $S$. Then we come to the conclusion.

COROLLARY 2.19. Let $X$ and $Y$ be affine $n$-dimensional integral schemes of finite type over a field $k$. If there exists a finitely generated $\mathbb{Z}$-subalgebra $R$ of $k$ and affine schemes $\mathscr{X}$, $\mathscr{Y}$ of finite type over $S = \mathit{\text{Spec}}R$, such that $\mathscr{X}\times_{S} \mathit{\text{Spec}}k\cong X$, $\mathscr{Y}\times_{S} \mathit{\text{Spec}}k\cong Y$, and for a sufficiently general closed point $t \in S$, $\#\mathscr{X}_{t}(\kappa(t)^{m}) \ne \#\mathscr{Y}_{t}(\kappa(t)^{m})$ and $\left|\#\mathscr{X}_{t}(\kappa(t)^{m}) - \#\mathscr{Y}_{t}(\kappa(t)^{m})\right| = o(|\kappa(t)|^{m(n-1)})\quad (m \rightarrow \infty \quad m \in \mathbb{Z})$, then there does not exist an open immersion $X \overset{f}{\hookrightarrow} Y/k$.

Proof. Write $[\underline{x}] = [x_{1}, \dots, x_{l}]$, $[\underline{y}] = [y_{1}, \dots, y_{m}]$, $X = \text{Spec}k[\underline{x}]/I$, $Y = \text{Spec}k[\underline{y}]/J$, where $I = (\mathtt{f}_{1}, \dots, \mathtt{f}_{s})k[\underline{x}]$ and $J = (\mathtt{g}_{1}, \dots, \mathtt{g}_{t})k[\underline{y}]$ are respectively the defining ideals of $X$ and $Y$.

Assume there is an open immersion $X \overset{f}{\hookrightarrow} Y/k$, by Lemma 2.14, there exists a finitely generated $\mathbb{Z}$-subalgebra $R^{'}$ of $k$, an open immersion

such that $\text{Spec}R^{'}[\underline{x}]/I\times_{\text{Spec}R^{'}} \text{Spec}k \cong X$, $\text{Spec}R^{'}[\underline{y}]/J\times_{\text{Spec}R^{'}} \text{Spec}k \cong Y$, and $f_{R^{'}}\times_{\text{Spec}R^{'}} \text{id}_{k} = f$.

Set $R^{''}$ = {the subring of $k$ generated by $R$ and $R^{'}$}, we have $\text{Spec}R^{''}[\underline{x}]/I\times_{\text{Spec}R^{'}} \text{Spec}k \cong X$, $\text{Spec}R^{''}[\underline{y}]/J\times_{\text{Spec}R^{'}} \text{Spec}k \cong Y$, $\text{Spec}R^{''}[\underline{x}]/I \overset{f_{R^{''}}}{\hookrightarrow} \text{Spec}R^{'}[\underline{y}]/J$ is an open immersion, and $f_{R^{''}}\times_{\text{Spec}R^{''}} \text{id}_{\kappa(\xi)} = f_{\xi}$.

It is noted that $\text{Spec}R^{''} \subset \text{Spec}R$, then we come to the conclusion according to Theorem 2.15.

3.   Examples

EXAMPLE 3.1. Since

$\begin{aligned} & \left|\#SL_{n/\mathbf{F}_{q}}(\mathbf{F}_{{q}^{m}}) - \#\mathbf{A}_{\mathbf{F}_{q}}^{n^{2}-1}(\mathbf{F}_{{q}^{m}})\right| \\ & = q^{mn^{2}-m}-(q^{mn}-1)\cdots(q^{mn}-q^{mn-2m})q^{mn-m}\\ & = q^{m(n^{2}-3)}+{\text{lower-degree terms}}+\ldots \end{aligned}$

We have ${\text{dim}}SL_{n/\mathbf{F}_{q}} = {\text{dim}}\mathbf{A}_{\mathbf{F}_{q}}^{n^{2}-1} = n^{2} - 1$, so for an arbitrary field $k$, there does not exist an open immersion $SL_{n/k} \hookrightarrow \mathbf{A}_{k}^{n^{2}-1}$.

EXAMPLE 3.2. Since

$\begin{aligned} & \left|\#Sp_{n/\mathbf{F}_{q}}(\mathbf{F}_{{q}^{m}}) - \#\mathbf{A}_{\mathbf{F}_{q}}^{2n^{2}+n}(\mathbf{F}_{{q}^{m}})\right|\\ & = q^{m(2n^{2}+n-2)} +{\text{lower-degree terms}}+\ldots\\ \end{aligned}$

We have ${\text{dim}}Sp_{n/\mathbf{F}_{q}} = \mathbf{A}_{\mathbf{F}_{q}}^{2n^{2}+n} = 2n^{2} + n$, so for an arbitrary field $k$, there does not exist an open immersion $Sp_{n/k} \hookrightarrow \mathbf{A}_{k}^{2n^{2}+n}$.

EXAMPLE 3.3. Since

We have ${\text{dim}}SO_{n/\mathbf{F}_{q}} = \frac{n^{2}-n}{2}$, for $n > 2$, so for an arbitrary field $k$, there does not exist an open immersion $SO_{n/k} \hookrightarrow \mathbf{A}_{k}^{\frac{n^{2}-n}{2}}$.

EXAMPLE 3.4. Since

We have ${\text{dim}}PGL_{n/\mathbf{F}_{q}} = n^{2} -1$, so there does not exist an open immersion $PGL_{n/k} \hookrightarrow \mathbf{A}_{k}^{n^{2}-1}$ for an arbitrary field $k$.

EXAMPLE 3.5. Since

$\begin{aligned} & \left|\#SL_{n/\mathbf{F}_{q}}\times \mathbf{G}_{a/\mathbf{F}_{q}}(\mathbf{F}_{{q}^{m}}) - \#\mathbf{A}_{\mathbf{F}_{q}}^{n^{2}-1}(\mathbf{F}_{{q}^{m}})\right|\\ & = q^{mn^{2}-m}-(q^{mn}-1)\cdots(q^{mn}-q^{mn-2m})q^{mn}\\ & = q^{m(n^{2}-2)}+{\text{lower-degree terms}}+\ldots \end{aligned}$

We have ${\text{dim}}SL_{n/\mathbf{F}_{q}}\times \mathbf{G}_{a/\mathbf{F}_{q}} = {\text{dim}}\mathbf{A}_{\mathbf{F}_{q}}^{n^{2}} = n^{2}$, so for an arbitrary field $k$, there does not exist an open immersion $SL_{n/k}\times_{k} \mathbf{G}_{a/k} \hookrightarrow \mathbf{A}_{k}^{n^{2}}$, and Example 1 is a corollary of this.

EXAMPLE 3.6. We have ${\text{dim}}Sp_{n/\mathbf{F}_{q}} = {\text{dim}}SO_{2n+1/\mathbf{F}_{q}} = 2n^{2}+n$, $\#SO_{2n+1/\mathbf{F}_{q}} > \#Sp_{n/\mathbf{F}_{q}}$ for $n > 2$, so when $n > 2$, there does not exist an open immersion $Sp_{n/k} \hookrightarrow SO_{2n+1/k}$ for an arbitrary field $k$.

EXAMPLE 3.7. There does not exist an open immersion $Sp_{n/k} \hookrightarrow SL_{n/k}\times_{k} \mathbf{A}^{n^{2}+n+1}_{k}$ for any field $k$.

Acknowledgments

The author thanks Professor KeZheng Li for his guide and useful discussions with him. The author thanks Professor Fei Xu and Professor Lei Fu.

Conflict of interest

The author declares that there are no conflicts of interest.