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Re-edited: Let $k$ be a field, not necessarily algebraically closed. Then what is the relation of the affine space $\mathbb{A}^n(k)$ with $k^n$ or $\bar{k}^n$?

Note: I am quite confused about what an affine space is, so i am asking this question, hoping to gain more insight about what an affine space is.

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The notation $\text{Spec } k[x_1, ... x_n]$ is actually quite ambiguous; it can mean at least four things (two of which are equivalent but not obviously equivalent) depending on how much algebraic geometry you know. Please be more specific. – Qiaochu Yuan Oct 9 '12 at 20:56
@QiaochuYuan: I honestly don't know what to write. I was prompted by Mariano to give the definition of the affine space, however i am not sure what this definition is, because i have seen at least two of them. Essentially, my question asks what is the relation of the affine space with $k^n$. – Manos Oct 9 '12 at 21:00
$k^n$ is the set of morphisms from $\text{Spec } k$ to $\text{Spec } k[x_1, ... x_n]$ (when these have been defined appropriately), while $\bar{k}^n$ is the set of morphisms from $\text{Spec } \bar{k}$ to $\text{Spec } k[x_1, ... x_n]$. What formalism you want to use to understand affine space over a non-algebraically closed field depends on what you want to use it for. The definition in my answer is suitable for talking about varieties over $\mathbb{Q}$ in the context of doing number theory, for example. – Qiaochu Yuan Oct 9 '12 at 21:02
up vote 7 down vote accepted

As a complement to the excellent answers already given by Mariano and Qiaochu, let me give a description and two illustrations of the closed points of $\mathbb A^n_k$.

First take an algebraic closure $\bar k$ of $k$.
The closed points of $\mathbb A^n_k$ are the maximal ideals of the polynomial ring $k[x_1, ..., x_n]$.
Now, given an $n$-tuple $a=(a_1, ..., a_n)\in \bar k^n$, we can associate to it the set $\mathfrak m_a$ of polynomials $P\in k[x_1, ..., x_n]$ vanishing at $a$ (i.e. $P(a_1, ..., a_n)=0$).
Then $\mathfrak m_a$ is a maximal ideal of $k[x_1, ..., x_n]$ and the pleasant surprise is that we obtain all the maximal ideals by this procedure.
Moreover, we know for which $n$-tuples $a,b\in \bar k^n$ we'll have $\mathfrak m_a=\mathfrak m_b$: that will be the case if and only if there exists an automorphism $\sigma\in Aut_k(\bar k)$ of $\bar k$ fixing $k$ such that $\sigma (a)=b$ (meaning $\sigma (a_i)=b_i$ for $i=1, ...,n$)

Let me illustrate with the smallest example: the affine line $\mathbb A^1_{\mathbb F_2}$ over the field $\mathbb F_2$ with $2$ elements.
Its closed points are given by the elements $ a\in \overline {\mathbb F_2}$ of an algebraic closure of $\mathbb F_2$ and two such elements give the same maximal ideal iff they have the same minimal polynomial. (Equivalently the closed points are given by the irreducible polynomials in $\mathbb F_2[X]$).
So despite apperances $\mathbb A^1_{\mathbb F_2}$ has infinitely many closed points!
(And, for the record, one non closed point: its so-called generic point)

As another illustration I'll describe the closed points of $\mathbb A^2_{\mathbb R}=Spec(\mathbb R[x,y])$ with the help of the group $Aut_{\mathbb R}( \mathbb C)=\lbrace I, \sigma\rbrace$ generated by complex conjugation $\sigma:z\mapsto \bar z$.
$\bullet$ First we have the classical maximal ideals $\langle x-r,y-s \rangle (r,s\in \mathbb R)$ corresponding to $(r,s)\in \mathbb R^2$ $ \bullet \bullet $ The other closed points correspond to the less classical maximal ideals $\mathfrak m_{(z,w)}$ associated to $(z,w)\in \mathbb C^2 \setminus \mathbb R^2$.
If $z=a+bi, w=c+di$ a little calculation shows that $$m_{(z,w)}=m_{(\bar z,\bar w)}=\langle dx-by-ad+bc, (x-a)^2+b^2, (y-c)^2+d^2\rangle \subset \mathbb R[x,y]$$ (One of the last two generators is redundant: if, say, $d\neq0$ keep the last one and dismiss the second. If both $b,d\neq0$, arbitrarily dismiss one of the last two generators)
For example $m_{(1,2-3i)}=\langle -3x+3,(x-1)^2+0^2,(y-2)^2+3^2 \rangle =\langle -x+1,y^2-4y+13 \rangle $

Although there are a gazillion books describing the Hilbert Nullstellensatz, the description I give at the beginning of the maximal ideals of $k[x_1, ..., x_n]$ is not so easy to find in the literature.
As always Bourbaki comes to the rescue: Commutative Algebra, Chapter V, §3.4, Proposition 2, page 351.

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Thank you for your amazing answer. Where can i find proofs for the statements you made in the first part of your answer? – Manos Oct 10 '12 at 13:30
Glad you liked the answer, Manos. I have added a reference in an Edit. – Georges Elencwajg Oct 10 '12 at 18:20

If affine space means to you «the spectrum of $k[x_1,\dots,x_n]$» then it is not true that its points are in a (sensible) bijection with $n$-tuples of scalars, even in the case where the field is algebraically closed. This statement is only true if you look at the maximal spectrum of that polynomial ring in the algebraically closed field.

Before someone gives you the answer to your question: can you find a maximal point in the affine line which does not correspond to a point of $k$?

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I am not sure what affine space means to me, at this point. – Manos Oct 9 '12 at 19:55
I ask this question so that i can extract insight from the answer. – Manos Oct 9 '12 at 20:04
Well... Ideally, you should edit the question to explain what you understand by «affine space». – Mariano Suárez-Alvarez Oct 9 '12 at 20:20
@Manos: to start with: do you know what the (maximal) spectrum of a ring is? – Nils Matthes Oct 9 '12 at 20:21
He defines affine space in page 1 like that... and later gives a different definition, which is not equivalent; under the definition given in page 1 of that book, affine space is the set of tuples with entries in the field whatever the field. For your question to make sense at all, you have to fix one definition! – Mariano Suárez-Alvarez Oct 9 '12 at 20:32

Even when $k$ is algebraically closed, $\mathbb{A}^n/k$ is not just the set of $n$-tuples over $k$. There is more data than this, namely the data of what counts as a regular function on $\mathbb{A}_n/k$, which is a polynomial function with coefficients in $k$.

When $k$ is not algebraically closed, there are various levels of sophistication from which to address the question of what exactly $\mathbb{A}^n/k$ is. One perspective is that $\mathbb{A}^n/k$ is

  • the set of $n$-tuples over the algebraic closure $\bar{k}$, together with
  • the action of $\text{Aut}(\bar{k}/k)$, together with
  • the data of what counts as a regular function on $\mathbb{A}^n/k$, which is still a polynomial function with coefficients in $k$.

A more sophisticated perspective is to use the language of schemes (although you will only need to understand affine schemes, and in particular you do not need to know what a sheaf is). In this language, the category of affine schemes over $k$ is anti-equivalent to the category of $k$-algebras, and $\mathbb{A}^n/k$ is the affine scheme corresponding to $k[x_1, ... x_n]$ under this equivalence.

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Dear Qiaochu, it would be better not to mention Galois groups in this context since the extension $\bar k/k$ is not Galois (unless $k$ is perfect). – Georges Elencwajg Oct 9 '12 at 20:36
I do not think that more sophistication is going to be of help here! I am consistently surprised at how you seem to think introducing categories into everything is helpful, Qiaochu :-) – Mariano Suárez-Alvarez Oct 9 '12 at 20:40
@Mariano: well, introducing categories always makes things make more sense to me. How can you understand what X is until you understand what morphisms between X'es are supposed to be? The data of a bare set of points, or even of a set of points together with the Zariski topology, captures very little of what people mean when they say "algebraic variety," and to me the categorical description is the cleanest way to capture what people actually mean when they say "algebraic variety" because it tells you what morphisms are in a coordinate-free way. – Qiaochu Yuan Oct 9 '12 at 20:57
@Mariano: in particular, to me an absolutely crucial point in understanding this situation is that when $k$ is not algebraically closed then the functor sending a $k$-variety to its set of $k$-points is not faithful and so one should look for a better one. The fact that this explanation requires more sophistication to appreciate is to me a property of the situation Manos is trying to understand and not a property of my approach to understanding it. – Qiaochu Yuan Oct 9 '12 at 21:06
In standard notation, isn't $\mathbb{A}^n (k)$ the set of $k$-points in $\mathbb{A}^n$, i.e. the set of homomorphisms $\mathbb{Z}[x_1, \ldots, x_n] \to k$? Perhaps it would be better to talk about the scheme $\mathbb{A}^n_k$. – Zhen Lin Oct 9 '12 at 21:14

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