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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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    $\begingroup$ You have not asked a question, really. Do you want to know what what means? $\endgroup$ Dec 12, 2011 at 15:12
  • $\begingroup$ Please ask the question in the question body, not I'm the comments! :) $\endgroup$ Dec 13, 2011 at 0:22
  • $\begingroup$ Dear Nadori, Mariano is absolutely right. However while adding the new question, you have completely deleted the old one, so that shaye's and my answer now look strangely irrelevant. Please roll back, reestablish the old question,and follow it by the new one. That said, I have addressed your new question in an Edit to my answer. $\endgroup$ Dec 13, 2011 at 9:45
  • $\begingroup$ Dear Georges, you're right. I'm sorry. I have changed it again. $\endgroup$
    – Nadori
    Dec 13, 2011 at 11:01

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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  • $\begingroup$ Nice explanations ! $\endgroup$
    – user18119
    Dec 12, 2011 at 22:35
  • $\begingroup$ Thanks, @QiL: I really appreciate that coming from you, an expert in arithmetic geometry who knows these questions infinitely better than I ever will ! $\endgroup$ Dec 12, 2011 at 22:55
  • $\begingroup$ Dear Georges, you are too modeste ! $\endgroup$
    – user18119
    Dec 12, 2011 at 23:02
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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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    $\begingroup$ 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 I $\endgroup$
    – shaye
    Oct 4, 2011 at 17:09

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