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Why is the affine space $\mathbb{A}^{2}$ not isomorphic to $\mathbb{A}^{2}$ minus the origin?

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    $\begingroup$ Lots of reasons. One is that the latter isn't the spectrum of a ring. $\endgroup$
    – user14972
    Mar 21, 2012 at 7:03

4 Answers 4

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It is enough to show that $X=\mathbb A^2_k\setminus \lbrace 0\rbrace$ is not affine since $\mathbb A^2_k \:$ is affine.

First proof of non-affineness
The key point is that the restriction map $\Gamma(\mathbb A^2_k,\mathcal O_{\mathbb A^2_k})=k[T_1,T_2] \to \Gamma(X,\mathcal O_X )$ is bijective.
This is the analogue of the Hartogs phenomenon in several complex variables and in algebraic geometry results from the fact that for a normal ring $A$ we have $A=\cap_{\mathfrak p} A_{\mathfrak p}$, where the intersection is over primes of height $1$.

Now if $X$ were affine we would have the canonical isomorphism of schemes
$X\stackrel {\cong}{\to} Spec(\Gamma(X,\mathcal O_X ))=Spec (k[T_1,T_2])=\mathbb A^2_k$
which is false since the origin of $\mathbb A^2_k$ is not in $X$.

Edit: Second proof of non-affineness
Consider the open covering $\mathcal U$ of $X$ by the two open affine subsets $U_1=D(T_1)$ and $U_2=D(T_2)$, i.e. those open subsets determined by $T_1\neq0$ and $T_2\neq0$.
The covering is a Leray covering for $\mathcal O$ (acyclic opens with acyclic intersection), since $U_1, U_2$ and $U_1\cap U_2$ are affine.
Hence by Leray's theorem we have $H^1(X,\mathcal O)=\check {H}^1(\mathcal U,\mathcal O)$ and thus $H^1(X,\mathcal O)$ is the cohomology of the complex $\Gamma(U_1,\mathcal O)\times \Gamma(U_2,\mathcal O)\to \Gamma(U_1\cap U_2,\mathcal O)\to 0$ where the non-trivial map is $$k[T_1,T_1^{-1},T_2]\times k[T_1,T_2,T_2^{-1}] \to k[T_1,T_1^{-1},T_2,T_2^{-1}]:(f,g)\mapsto g-f$$

Thus $H^1(X,\mathcal O)=\oplus _{i,j\lt 0} \;\; k \cdot T^{i}T^{j}$, an infinite-dimensional $k$-vector space, in sharp contrast to Serre's theorem stating that positive dimensional cohomology groups of coherent sheaves on affine schemes are zero.

Other Edit: Third proof of non-affineness
If $k=\mathbb C $ let us show that $X$ is not affine by proving that its underlying holomorphic manifold $X_{hol}$ is not Stein.
Indeed suppose it were and consider the discrete closed subset $D=\lbrace (1/n,0): n=1,2,3,...\rbrace\subset X$.
Since $D$ is a $0$-dimensional submanifold the restriction map $\Gamma(X_{hol}, \mathcal O_{X_{hol}})\to \Gamma(D, \mathcal O_D)$ would be surjective.
On the other hand, by Hartogs's theorem the restriction map $\Gamma(\mathbb C^2,\mathcal O_{\mathbb C^2}) \to \Gamma(X_{hol},\mathcal O_{X_{hol}}) $ is also surjective so that by composition we would conclude if $X_{hol}$ were Stein to the surjectivity of the restriction map $$ \Gamma(\mathbb C^2,\mathcal O_{\mathbb C^2}) \to \Gamma(D, \mathcal O_D): f\mapsto f_0=f\mid D $$

But this is clearly false because a holomorphic function $f_0:D\to \mathbb C$ is an arbitrary function, and if $f_0$ is not bounded [ for example $f_0(1/n,0)=n$ ] it cannot be the restriction of a holomorphic function $f:\mathbb C^2\to \mathbb C$.

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    $\begingroup$ You give wonderful explanations! $\endgroup$ Mar 21, 2012 at 7:50
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    $\begingroup$ @Georges Elencwajg: Thanks. I'm actually trying to show that $\mathbb{A}^{2}\setminus \{(0,0)\}$ is not affine so it suffices to show (because the ring of regular functions coincides with the coordinate ring in the affine case) that there is no isomorphism between $\mathbb{A}^{2}$ and $\mathbb{A}^{2} \setminus \{(0,0)\}$ without an affine argument. (Otherwise the argument is circular), do you know another argument? $\endgroup$
    – user6495
    Mar 21, 2012 at 7:55
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    $\begingroup$ Thanks a lot to @Zhen for his illuminating comment. Although the proof I posted is logically sufficient, I have posted a second proof, for the sheer pleasure of computing a cohomology group (but that second proof is a bit more advanced). $\endgroup$ Mar 21, 2012 at 8:32
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    $\begingroup$ Hey Georges +1--nice answer as usual. It occurred to me the other day that one can essentially capture the cohomological argument without ever having to mention cohomology. Namely, if $X=\text{Spec}(A)$ is an affine scheme and $x,y\in\mathcal{O}_X(X)$ are such that $D(x)\cup D(y)=X$ then, necessarily, $(x,y)=A$--so $x$ and $y$ are comaximal. That said, one can canonically identify $\mathcal{O}_{\mathbb{A}^2-\{0\}}(\mathbb{A}^2-\{0\})$ with $k[x,y]$. Then, $x,y$ are global sections of $\mathbb{A}^2-\{0\}$ such that $D(x)\cup D(y)$ cover the whole space, but which are not comaximal. Of course, $\endgroup$ Nov 24, 2016 at 10:05
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    $\begingroup$ this is really, secretly, the point/idea of the cohomological argument since precisely the reason that $\frac{1}{xy}$ is not a coboundary in $\check{H}^1(\{D(x),D(y)\},\mathcal{O}_{\mathbb{A}^2-\{0\}})$ is because $x$ and $y$ are not comaximal (i.e. 1 is not in $(x,y)$). It just, to me, seems nice that one can somehow give the cohomological argument without ever saying cohomology. Of course, this is (essentially) the point of the proof that cohomology vanishes for affine schemes though--so nothing really new here. :) Thanks again for the nice answer! (I especially appreciated the last part!). $\endgroup$ Nov 24, 2016 at 10:07
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I'll call your punctured plane $X$, as Georges does. As he says, the key is to prove a Hartogs lemma for the inclusion $X \to \mathbf A^2$. What follows is (I think) a special case of the result in commutative algebra that he mentions.

Let $f$ be a regular function on $X$. Then $X - Z(x)$ is affine, isomorphic to $Z(xz - 1) \subset \mathbf A^3$, and hence the restriction of $f$ to this open set of agrees with $g(x, y)/x^n$ for some $g \in k[x, y]$ and $n \geq 0$. Similarly, $f$ restricted to $X - Z(y)$ looks like $h(x, y)/y^m$.

Now, regular functions on $\mathbf A^2$ correspond exactly to elements of $k[x, y]$, and you have two such elements \[ y^mg(x, y) \quad \text{and} \quad x^nh(x, y) \] which agree on the dense subset $\mathbf A^2 - Z(xy)$. Can you argue that, after putting some conditions on $g, h, n, m$, we must have $n = m = 0$ and $f = g$?

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One of the two is an affine variety while the other isn't.

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    $\begingroup$ This gets downvotes from time to time. I find it extremely sad, to he honest, and quite characteristic of the site. $\endgroup$ Feb 18, 2022 at 4:55
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Other people have answered this question in a satisfactory way. I will just add that I figured out an elementary proof that $\mathbb{A}^2$ is not isomorphic to $X := \mathbb{A}^2 - \{(0,0)\}$ when the field $K$ has characteristic 0, which it would have to be if $X$ were affine $\big($because $\mathcal{O}(X) \cong K[x,y]\big)$.

Suppose $\phi:\mathbb{A}^2 \rightarrow X$ is an isomorphism. Then one can think of $\phi$ as an injective morphism $\mathbb{A}^2 \rightarrow \mathbb{A}^2$ whose image is $X$. Any morphism $\mathbb{A}^2 \rightarrow \mathbb{A}^2$ is at each coordinate a polynomial, i.e. $\phi(a,b) = [f(a,b), g(a,b)]$ where $f,g$ are polynomials. (This takes a bit of work to show but it's not too hard).

Think of $f,g$ as polynomials in $y$ with coefficients that are polynomials in $x$. So $f = h_n(x)y^n+\cdots+h_0(x)$ and similar for $g$. Then one can use a parallel algorithm to the Euclidean algorithm to get a polynomial $S$ in the ideal of $f,g$ that only contains the variable $x$. Note that $S$ may be constant. We call $S$ the solution polynomial.

Lemma: There exists a $b \in K$ such that $(a,b)$ is a solution to $f=0,g=0$ iff $a$ is a root of the solution polynomial $S$.

pf: The variety of $S$ contains $V(f,g)$. Since the x-coordinate of any element of $V(S)$ must be a root, $(a,b) \in V(f,g)$ must imply that $a$ is a root of $S$. Going the other way, if $a$ is a root of $S$, then running the parallel Euclidean algorithm on $f(a,y),g(a,y)$ (now just the Euclidean algorithm) will give us $S(a) = 0$. Thus the gcd of $f(a,y),g(a,y)$ is a nonconstant polynomial so they share a root, $b$. Thus $(a,b)$ a solution to $f=0,g=0$.

Now, if $\phi(a,b) = (c,d)$ then $(a,b)$ is in $V(f-c,g-d)$ thus $a$ is a root of the solution polynomial for $f-c$ and $g-d$. In fact, let's introduce two new variables $w,z$. The solution polynomial of $f-w,g-z$ is now a three variable polynoimal which I will call the global solution polynomial, denoted $S(w,z)$. Similar to before we can think of $S(w,z)$ as a polynomial in $x$ with coefficients that are polynomials in $w$ and $z$. So $S(w,z) = c_m(w,z)x^m + \cdots + c_0(w,z)$.

Since $\phi$ doesn't map to $(0,0)$, $S(0,0)$ must be a nonzero constant polynomial. So in particular, $c_i(0,0) = 0$ for all $i > 0$ and $c_0(0,0) \neq 0$. Thus, $V(c_1)\backslash V(c_0)$ is nonempty and open, so it is dense and contains another point $(w_0,z_0)$. If $S(w_0,z_0)$ is constant, it is nonzero, so $(w_0,z_0)$ is not in the image and $\phi$ is not surjective. If $S(w_0,z_0)$ is not constant then it must have multiple distinct roots, because in characteristic 0 any polynomial of the form $d(x-a)^r$ cannot have linear coefficient 0 and constant term nonzero. Thus multiple points map to $(w_0,z_0)$ so $\phi$ is not injective. Either way this is a contradiction to $\phi$ being an isomorphism, so no isomorphism exists.

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