Affine space over non algebraically closed field 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.
 A: 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 $k$?
A: 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. 
A: 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 $  
Edit
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.
