Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ (resp. $Y$) be the affine $k$-scheme defined by the ideal $I$ (resp. $J$) in the polynomial ring $k[x_1,...x_n]$ (resp. $k[y_1,...,y_m]$). Let $Z$ be the affine scheme defined by the ideal $L$ in $k[z_1,...z_s]$, and let $f^\*:k[z]/L\rightarrow k[x]/I$ (resp. $g^\*:k[z]/L\rightarrow k[y]/J$) be $k$-homomorphisms, where $x=(x_1,...,x_n)$ and so forth, corresponding to scheme morphisms $f:X\rightarrow Z$ (resp. $Y\rightarrow Z$).

Then it should be possible to express the fiber product $X\times_{f,Z,g}Y$ via an ideal $W$ in the polinomial ring $k[x,y,z]$ [edit: actually, $W$ should be an ideal in $k[x,y]$] (where $x$ stands for the string of variables $x_1,...,x_n$, and so on).

Question: how to express $W\subseteq k[x,y,z]$ explicitely in terms of $I$, $J$, $L$, $f^\*$ and $g^\*$?

Edit: You can express things explicitely in terms of some polynomials $F_i$, $G_i$ and $H_i$ such that $I=(F_1,...,F_N)$, $J=(G_1,...,G_M)$ and $L=(H_1,...,H_S)$, and in terms of the components $(f_1,...,f_s)$ (resp. $(g_1,...,g_s)$) of $f$ (resp. $g$).

share|improve this question
    
I'm assuming that you're familiar with the fact that the fiber product of the affine schemes Spec(A) and Spec(B) over Spec(C) is given by $Spec(A \underset{C}{\otimes} B)$? So your question is about a nice description of this tensor product for various finitely generated k-algebras? Also, I'm not sure why you would expect your fiber product to be a closed subscheme of k[x,y,z]. I don't see why it's anything more than an algebra over k[z]/L. –  Mike Skirvin Mar 21 '10 at 17:45
    
@Mike: Yes, you can assume familiarity with the tensor product thing. The answer is "yes" also to your second question, in some sense. As far as I understand, the euristic interpretation of fiber product is: "construct a space $P$ over $Z$ such that its fiber $P_z$ over the point $z\inZ$ is just the absolute (over $Spec(k)=$ the residue field at $z$) product of the fibers: $X_z\timesY_z$". Does it sound correct? –  Qfwfq Mar 21 '10 at 20:17
    
Ah, can assume $k$ is a field if you want: just to simplify matters, but perhaps it's not relevant. –  Qfwfq Mar 21 '10 at 20:18
    
@unknown: not sure why you seem to think a space isn't needed between a LaTeX control sequence and a following letter! –  Reid Barton Mar 22 '10 at 6:13
add comment

1 Answer 1

up vote 5 down vote accepted

Dear unknown, let me first congratulate you on the clearness of your question and the quality of your notation, which I'm now going to use.

The fibre product $X\times_Z Y$ is the subscheme of $\mathbb A_k^n \times \mathbb A_k^m$ described by an ideal $\mathfrak A \subset k[x,y]$. That ideal is $\mathfrak A=I^e + J^e + D$, where

$I^e$ is the extension of $I\subset k[x]$ to $I^e\subset k[x,y]$,

$J^e$ is the extension of $J\subset k[y]$ to $J^e\subset k[x,y]$,

$D$ is the ideal generated by the $s$ differences $f_i(x)-g_i(y),\quad (i=1,\ldots,s)$

I find it clearer not to use generators for $I$ and $J$ and, strangely, $L$ is not used at all: this is because the fibre product is the same whether considered over $Z$ or over $\mathbb A_k^s$ !

share|improve this answer
    
Interesting answer: I will read more carefully later. Also, I suspect I made a mistake: the fiber product, as a scheme over $k$, should really sit inside the product (X\timesY), NOT inside $X\timesY\timesZ$ as it's written in my question. Should I edit? –  Qfwfq Mar 21 '10 at 20:30
    
Dear unknown, LaTeX doesn't render your comment very well for me: I get a message "Unknown control sequence" after your two words "NOT inside". Anyway, you may edit if you like, of course, but I think your question is fairly clear. Anyone interested in technical details about the exact scheme over which to take the fibre product might look them up in: EGA I,chapter 1, 3.3 Propriétés formelles du produit; changement de préschéma de base . –  Georges Elencwajg Mar 21 '10 at 21:57
    
Ok, I just wrote "NOT inside XxYxZ". Because, as you pointed out, it's a subscheme of XxY. –  Qfwfq Mar 22 '10 at 6:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.