What's the difference between $\mathbb{A}^n$ and $\mathbb{A}^{n+1}$?

If we would like imitate the case of the natural topology on the real affine spaces: I shall work in the scheme setting and I shall use the Čech and Grothendieck cohomologies!


I start from $\mathbb{A}^1_{\mathbb{R}}\equiv\mathbb{A}^1$; let $X_1=\mathbb{A}^1\setminus\{(x_1)\}$, via Rabinowitsch's trick one can prove that $X_1$ is isomorphic to hyperbola $H=V(xy-1)\subset\mathbb{A}^2$, in other words $X_1$ is an affine variety.

I recall the Serre's Criterion for the affiness of a scheme (see [B], theorem 7.7.8).

Let $S$ be a quasi-compact scheme. $S$ is affine if and only if for any quasi-coherent $\mathcal{O}_S$-module $\mathcal{F}$ one has $H^1(S,\mathcal{F})=0$.

In particular, without change the name, $H^1\left(X_1,\mathcal{O}_{X_1}\right)=0$.

Again, I recall that for any scheme $S$ and for any $\mathcal{O}_S$-module $\mathcal{F}$, $H^1(S,\mathcal{F})$ is in bijection with $\check{H}^1(S,\mathcal{F})$; therefore, I start to study $\check{H}^1\left(X_n,\mathcal{O}_{X_n}\right)$ for any $n\geq2$, where $X_n=\mathbb{A}^n_{\mathbb{R}}\setminus\{(x_1,\dots,x_n)\}$.

First all, the $X_n$'s are not affine, for $n\geq2$; because they are not the spectrum of $\mathcal{O}_{X_n}(X_n)$!

Indeed, let $\mathcal{U}=\{D(x_k)\subset\mathbb{A}^n\}_{k\in\{1,\dots,n\}}$ an affine open covering of $X_n$, where $D(x_k)=Spec\mathbb{R}[x_1,\dots,x_n]_{x_k}$; considering the Čech cocomplex: \begin{equation} 0\to C^0\left(\mathcal{U},\mathcal{O}_{X_n}\right)\stackrel{d_0}{\longrightarrow}C^1\left(\mathcal{U},\mathcal{O}_{X_n}\right)\stackrel{d_1}{\longrightarrow}\dots \end{equation} where:

  1. $C^q\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\displaystyle\prod_{1\leq i_1<\dots<i_{q+1}\leq n}\mathcal{O}_{X_n}\left(D\left(x_{i_1}\dots x_{i_{q+1}}\right)\right)$,

  2. $\left(d^qf\right)_{i_1\dots i_{q+2}}=\displaystyle\sum_{j=1}^{q+2}(-1)^{j+1}f_{i_1\dots\widehat{i_j}\dots i_{q+2}\displaystyle|D\left(x_{i_1}\dots x_{i_{q+2}}\right)}$.

Trivially: \begin{gather*} \mathcal{O}_{X_n}(X_n)\cong\check{H}^0\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\ker d^0=\\ =\left\{(f_1,\dots,f_n)\in\mathcal{O}_{X_n}\left(D\left(x_1\right)\right)\times\dots\times\mathcal{O}_{X_n}\left(D\left(x_n\right)\right)\mid\\ \forall h\neq k\in\{1,\dots,n\},\,f_{h\displaystyle|D(x_hx_k)}=f_{k\displaystyle|D(x_hx_k)}\right\}=\bigcap_{i=1}^n\mathcal{O}_{X_n}(D(x_i))=\mathbb{R}[x_1,\dots,x_n]. \end{gather*} At this point, one can state that $X_1$ is not isomorphic to $X_n$ for $n\geq2$; that is, $\mathbb{A}^1$ can not be isomorphic to $\mathbb{A}^n$ for $n\geq2$.

In consequence to Serre's criterion: \begin{equation*} \forall n\geq2,\,H^1\left(X_n,\mathcal{O}_{X_n}\right)\neq0; \end{equation*} then what are $H^1\left(X_n,\mathcal{O}_{X_n}\right)$'s? The answer is a little bit complicated...

Then I change strategy: by construction $\forall n\geq1,q\geq n,\,C^q\left(\mathcal{U},\mathcal{O}_{X_n}\right)=0$ and therefore $\forall n\geq1,q\geq n,\,\check{H}^q\left(\mathcal{U},\mathcal{O}_{X_n}\right)=0$; in particular: \begin{equation*} \forall n\geq1,\,\check{H}^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\ker d^{n-1}_{\displaystyle/\operatorname{Im}d^{n-2}}=\mathcal{O}_{X_n}\left(D(x_1\dots x_n)\right)_{\displaystyle/\operatorname{Im}d^{n-2}}. \end{equation*}

I recall a Leray's theorem (see [B], theorem 7.7.5)

Let $\mathcal{U}$ be an open covering of a scheme $X$ and $\mathcal{F}$ be an $\mathcal{O}_X$-module. Assume $H^q(U,\mathcal{F})=0$ for any $q\geq1$ and for any finite intersection $U$ of sets in $\mathcal{U}$; then the canonical map $\check{H}^q(\mathcal{U},\mathcal{F})\to\check{H}^q(X,\mathcal{F})$ is bijective for all $q\geq0$.

and the Vanishig Cohomology Theorem for the Affine Schemes (see [B], proposition 7.6.4 and corollary 7.7.7)

Let $S$ be an affine scheme over a ring $R$. Then for any quasi-coherent $\mathcal{O}_S$-module $\mathcal{F}$, one has $\forall q\geq1,\,H^q(S,\mathcal{F})=0$.

By the previous theorem, $\mathcal{U}$ satisfies the hypothesis of Leray's theorem and then \begin{equation*} \forall n\geq1,\,\check{H}^{n-1}\left(X_n,\mathcal{O}_{X_n}\right)\cong\check{H}^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right)=\mathcal{O}_{X_n}\left(D\left(x_1\dots x_n\right)\right)_{\displaystyle/\operatorname{Im}d^{n-2}} \end{equation*} where $\mathcal{O}_{X_n}\left(D\left(x_1\dots x_n\right)\right)=\mathbb{R}[x_1,\dots,x_n]_{x_1\dots x_n}$.

By construction: \begin{gather*} C^{n-2}\left(\mathcal{U},\mathcal{O}_{X_n}\right)\ni\left(\frac{f_1}{x_2^{\alpha_{1,2}}\dots x_n^{\alpha_{1,n}}},\frac{f_2}{x_1^{\alpha_{2,1}}x_3^{\alpha_{2,3}}\dots x_n^{\alpha_{2,n}}},\dots,\frac{f_n}{x_1^{\alpha_{n,1}}\dots x_{n-1}^{\alpha_{n,n-1}}}\right)\mapsto\sum_{j=1}^n(-1)^{j+1}\frac{f_jx_j}{x_1^{\alpha_{j,1}}\dots x_j\dots x_n^{\alpha_{j,n}}}\in C^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right); \end{gather*} in other words: \begin{equation*} a,b\in\mathbb{R}[x_1,\dots,x_n]_{x_1\dots x_n},\,[a]=[b]\in\check{H}^{n-1}\left(\mathcal{U},\mathcal{O}_{X_n}\right)\iff a-b=\sum_{\underline{i}\in(\mathbb{N}_0)^n\setminus\{\underline{0}^n\}}\frac{r_{\underline{i}}}{x^{\underline{i}}} \end{equation*} where:

  1. $r_{\underline{i}}\in\mathbb{R}$,

  2. for any $\underline{i}=(i_1,\dots,i_n)\in(\mathbb{N}_0)^n,\,x^{\underline{i}}=x_1^{i_1}\dots x_n^{i_n}$;

then: \begin{equation*} \forall n\geq2,\,\check{H}^{n-1}\left(X_n,\mathcal{O}_{X_n}\right)=\bigoplus_{\underline{i}\in(\mathbb{N}_0)^n\setminus\{\underline{0}^n\}}x^{-\underline{i}}\mathbb{R}. \end{equation*}

Let $n>m\geq2$, then $X_n$ is not isomorphic to $X_m$ because, by the previous reasoning, $\check{H}^{n-1}\left(X_n,\mathcal{O}_{X_n}\right)\neq\check{H}^{n-1}\left(X_m,\mathcal{O}_{X_m}\right)=0$; in consequence $\mathbb{A}^m$ can not be isomorphic to $\mathbb{A}^n$.


Bibliography

[B] Bosch S. - Algebraic Geometry and Commutative Algebra (2013), Springer Verlag.