affine scheme over a ring R I red an article and encountered some concepts from algebraic geometry. Let $R=\mathbb{Q}[\alpha_1,\ldots,\alpha_5]$ be a polynomial ring in the variables $\alpha_i$. Define $f(x,y)\in R[x,y]$ by
$$f(x,y)=y^2+\alpha_1 xy+\alpha3 y-x^3-\alpha_2 x^2-\alpha_4 x-\alpha_5.$$
We now consider the affine scheme $\mathcal{E}:f(x,y)=0$ over $R$. What does this mean? What is the definition of the n-fold fibered product of $\mathcal{E}$?
Edit:
I have a concrete question: 
If $\mathcal{E}=Spec\  R[X,Y]/(f(X,Y))$, how does the field of rational functions on $\mathcal{E}$ look like? I am using Geometry of schemes (Eisenbud, Harris), but I do not find any explanation about this nor the definition. And the same question for $\mathcal{E}^2=Spec\ \Big(   R[X,Y]/(f(X,Y))\otimes_R R[X,Y]/(f(X,Y))\Big)$
 A: There is much, much material behind the situation you describe. Here is one aspect of it.   
The scheme $E$ described bt $f(x,y)=0$ is a subscheme of the affine plane over $R$, namely $E=V(I)\subset \mathbb A^2_R$. Its ring of regular functions is $A=R[X,Y]/(f)$, so that you may write  $E=Spec(A)$.
The interesting point is that  you have a morphism $f: E\to Spec(R)=\mathbb A^5_{\mathbb Q}$.
So in reality, you are studying a family of  affine curves, one for each $s\in \mathbb A^5_{\mathbb Q}$.
And believe me: $\mathbb A^5_{\mathbb Q}$, five-dimensional space over the rationals, is really, really big and complicated!
For example, whenever you choose five rational numbers $q_1,...,q_5\in \mathbb Q$, you get a so-called "rational point" point $q=\lt \alpha_1-q_1,..., \alpha_5-q_5\gt\in Spec (R)=\mathbb A^5_{\mathbb Q}$ whose fiber with respect to  $f$ is the affine curve $E_q \subset \mathbb A^2_{\mathbb Q}$
given by the  equation $$f(x,y)=y^2+q_1 xy+q_3 y-x^3-q_2 x^2-q_4 x-q_5.$$
At the other extreme, if you take the generic point $\eta=(0)\in  Spec (R)=\mathbb A^5_{\mathbb Q}$ its fiber will be the generic curve $E_\eta\subset \mathbb A^2_{ \mathbb{Q}(\alpha_1,\ldots,\alpha_5)}$ given ( a little confusingly!) by the original equation 
$$f(x,y)=y^2+\alpha_1 xy+\alpha_3 y-x^3-\alpha_2 x^2-\alpha_4 x-\alpha_5.$$
And, as mentioned before, there are many, many other points in $\mathbb A^5_{\mathbb Q}$, for example   $\lt\alpha_2 ^3-\alpha_2 +1, \alpha_4^{2011}-17 \gt \;\; \in Spec(R) \quad $  (and it will be worse next year...) 
This illustrates Grothendieck's philosophy that you should not study a scheme like $E$  per se, but rather as a family of schemes: here the family is the set of fibers of $f$, parametrized by $\mathbb A^5_{\mathbb Q}$.     
Edit
Nadori has now completely changed  his question in an  an edit and his new question is : what is the field of  functions of $E$ ?.
The affine scheme $E$  corresponds to the ring $A=R[X,Y]/(f)$.
Since the polynomial $f$   contains the isolated term $-\alpha_5$, the ring $A$ is isomorphic to $\mathbb Q[\alpha_1, \alpha_2,\alpha_3, \alpha_4; X,Y] $ ,  so that $E\simeq \mathbb A^6_{\mathbb Q}$ and the required function field is $\mathbb Q(\alpha_1, \alpha_2,\alpha_3, \alpha_4; X,Y)$, a purely transcendental extension of $\mathbb Q$.  
A: If $R$ is an algebraically closed field, then the affine scheme you wrote down is an affine patch of an elliptic curve. To get a feeling for these babies you should start with studying them over the complex numbers.
As a set an affine variety  $X$ is the set of prime ideals in $R$. This is not very interesting on its own. But if you define a closed set of $X$ to be of the form $Z(I) = \{x \in X : s(x) =0 \ \textrm{for all} \ s\in I\}$ you get an interesting topology: the Zariski topology. (Here $s(x)$ is the image of $s$ in $k(x)$.) So the closed sets of $X$ are precisely the algebraic sets. 
If $\epsilon$ is an affine variety, say $\textrm{Spec} \ S$, where $S$ is an $R$-algebra, then the $n$-fold fibre product $\epsilon^n:= \textrm{Spec} (S\otimes_R S \otimes_R \ldots \otimes_R S)$. This might give you a way to understand a bit better what's going on.
