Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $k$ is an algebraically closed field, and let $\mathbb{A}^2$ denote the affine $2$-space $k^2$.

An affine scheme is defined to be a locally ringed space $(X, \mathcal{O}_X )$ which is isomorphic (as locally ringed spaces) to the spectrum of some ring.

Then is $\mathbb{A}^2$ an affine scheme? Is it isomorphic (as locally ringed spaces) to the spectrum of the polynomial ring $k[x,y]$?

Let $\mathbb{A}^2_k = \operatorname{Spec}k[x,y]$, and $\phi: \mathbb{A}^2 \rightarrow \mathbb{A}^2_k, (a,b) \mapsto \langle x-a, y-b \rangle$, then $\phi$ injects $\mathbb{A}^2$ to the set of maximal ideals of $k[x,y]$, i.e., the set of closed points of $\mathbb{A}^2_k$. I see that $0$, as a prime ideal of $k[x,y]$, is also a point in $\mathbb{A}^2_k$, but it does not lie in $\phi(\mathbb{A}^2)$. The Krull dimension of $k[x,y]$ is $2$, so beside $0$ and maximal ideals, there is still another kind of prime ideals in $k[x,y]$. In fact, any irreducible polynomial $f(x,y) \in k[x,y]$ generates a prime ideal $\langle f(x,y) \rangle$. In a word, $\phi$ is not surjective. Can it still be the map of topological spaces in the isomorphism of ringed spaces? The definition for such isomorphism says that the map of topological spaces should be a homeomorphism. Can a non-surjective map be a homeomorphism? Or, is there any other way to define the map?

Thank you very much.

Edited: I neglected the condition of $k$ being algebraically closed, and I thought the denotation $\mathbb{A}^2$ is generally known. I add these in the front. Sorry for the misunderstanding :)

share|improve this question
There's a difference betweean \mathrm{Spec}k[x,y] and \operatorname{Spec}k[x,y]. I changed it. Look: $\mathrm{Spec}k[x,y]$, $\operatorname{Spec}k[x,y]$. \operatorname does not only prevent italicization, but also provides proper spacing and in some cases affects formatting more generally (for example: $\displaystyle\sup_{a\in A} f(x)$). –  Michael Hardy Nov 27 '11 at 19:59
@Shinya: What is $\mathbb{A}^2$ in the first place? The definition of the scheme $\mathbb{A}^2_k$ is, as you say, $\operatorname{Spec} k[x, y]$. –  Zhen Lin Nov 27 '11 at 20:10
And $\mathbb A^2$, which I guess is $k^2$, consists in the rational points of $\mathbb A^2_k$. –  user18119 Nov 27 '11 at 20:23
@Michael Hardy: Thank you for the editing and explanation. I didn't know about that. –  ShinyaSakai Nov 28 '11 at 12:12
@Zhen Lin: Sorry, I thought it was commonly used, so I didn't explain it. Qil is right. $\mathbb{A}^2$ is the affine space $k^2$, where $k$ is the base field. –  ShinyaSakai Nov 28 '11 at 12:14
add comment

2 Answers

up vote 12 down vote accepted

I assume by $\mathbb{A}^2$ you mean the "classical" affine plane over $k$. This can't be homeomorphic to the underlying topological space of the scheme $\mathbb{A}_k^2=\mathrm{Spec}(k[x,y])$ because the latter has a non-closed "generic" point for each irreducible closed subset. When $k$ is algebraically closed (in which case all the maximal ideals of $k[x,y]$ are of the form $(x-a,y-b)$ for $a,b\in k$), the topological space $\mathbb{A}^2$ is homeomorphic to the set of closed points of $\mathbb{A}_k^2$. Basically the scheme $\mathbb{A}_k^2$ is obtained from $\mathbb{A}^2$ by throwing in a generic point for each irreducible closed subset. A detailed explanation of this (making things precise, at least for $k$ algebraically closed) can be found in the second chapter of Hartshorne. Specifically, see Proposition 2.6, where a fully faithful functor from "classical varieties" over $k$ to $k$-schemes is constructed.

share|improve this answer
As QiL points out in his comment, when $k$ is arbitrary, $\mathbb{A}^2$ can be identified with the set of $k$-rational points of $\mathbb{A}_k^2$, i.e., the set of $k$-scheme morphisms $\mathrm{Spec}(k)\rightarrow\mathbb{A}_k^2$, which are the same as $k$-algebra maps $k[x,y]\rightarrow k$. The latter objects are just pairs of elements of $k$ by the universal property of $k[x,y]$. –  Keenan Kidwell Nov 27 '11 at 20:28
Thank you very much. Now $\mathbb{A}^2$ is homeomorphic to the the set of closed points of $\mathbb{A}^2_k$, rather than $\mathbb{A}^2$ itself. So, the affine space $\mathbb{A}^2$ is not an affine scheme? –  ShinyaSakai Nov 28 '11 at 14:20
The underlying topological space of a scheme is "sober," which means that every irreducible closed subset has a unique generic point. The space $\mathbb{A}^2$ doesn't have this property, so it cannot be the underlying topological space of a scheme. –  Keenan Kidwell Nov 28 '11 at 15:26
Thanks a lot! Now I understand! –  ShinyaSakai Nov 28 '11 at 16:15
add comment

Let us define $\mathbb A^2_k=Spec (k[x,y])$. Your notation $\mathbb A^2$ is not standard in algebraic geometry and I think you just mean $k^2$, which I will use henceforth.

There is indeed a set theoretic injective map $\phi:k^2\to \mathbb A^2_k: (a,b)\mapsto \langle x-a,x-b\rangle $, just as you wrote.
The image of $\phi $ consists of the $k$-rational points of $\mathbb A^2_k$: these are the points corresponding to ideals ${\mathfrak m}\subset A$ such that the natural morphism of $k$-algebras $k\to A/{\mathfrak m}$ is surjective.
These rational points are closed ( in other words these ${\mathfrak m}$ are maximal) but, unless $k$ is algebraically closed, there are other closed points in $\mathbb A^2_k$.
For example in $\mathbb A^2_\mathbb Q$, the maximal ideal $M=\langle x^2-2,y\rangle \subset k[x,y]$ yields a closed point which is not rational (according to a theorem proved 25 centuries ago, albeit not in the language of scheme theory).

To sum up, the image of $\phi $ is never surjective and misses many points of $\mathbb A^2_k$:
i) The closed but not rational points (if $k$ is not algebraically closed).
ii) The height one points corresponding to nonzero prime ideals which are not maximal .
iii) The generic point corresponding to the zero ideal.

share|improve this answer
+1 for "25 centuries ago." –  Keenan Kidwell Nov 27 '11 at 20:36
Thanks @Keenan. And +1 to you for your sense of fair-play toward the "competition" (and of course for the quality of your answer!). –  Georges Elencwajg Nov 27 '11 at 21:07
Thank you very much. Now I am more sure of the non-surjectivity of $\phi$. Maybe there is no such homeomorphism... –  ShinyaSakai Nov 28 '11 at 14:16
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.