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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 $\mathcal{E}^2=Spec\ \Big( R[X,Y]/(f(X,Y))\otimes_R R[X,Y]/(f(X,Y))\Big)$

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    Dear Georges, you're right. I'm sorry. I have changed it again.2011-12-13

2 Answers 2

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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 $\mathbb Q$.

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    Dear Georges, you are too modeste !2011-12-12
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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.

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    If $R$ is an algebraically closed field, then an affine variety over $R$ is the same as an affine scheme over Spec R. Set-theoretically, the Spec$R$= {pt} (if $R$ is a field!). The category of affine schemes over Spec$R$is (anti-)equivalent to the category of $R$-algebras. (See Hartshorne Chapter II.2) My notation was a bit bad. The $R$ in the second paragraph is an arbitrary ring. What you are doing is studying elliptic curves over $R$. The only reasonable way of doing this is using the language of schemes, unless $R$ is an algebraically closed field. Then you can do it as Hartshorne Ch I2011-10-04