0
$\begingroup$

Let $V, W$ be algebraic varieties and $F: V \rightarrow W$ be an affine morphism: my definition of affine is that there exists an open cover $\{ U_i \}_i$ of $W$ with each $U_i$ (isomorphic to) an affine variety, such that $F^{-1}(U_i)$ is (isomorphic to) an affine variety for any $i$.

I'm trying to prove that for any $U \subseteq W$ open and affine, $F^{-1}(U)$ is affine. The first step in the proof I'm following is to assume that $W$ itself is affine. I don't understand why we can do this.

I know something about how the preimages of affine subvarieties look through morphisms of affine varieties, but have no clue about such preimages through morphisms of algebraic varieties. So what guarantees that $F^{-1}(U)$ is an algebraic variety and that $F|_{F^{-1}(U)} : F^{-1}(U) \rightarrow U$ is still an affine morphism of algebraic varieties? Or is this even needed?

$\endgroup$
4
  • $\begingroup$ What exactly are your definitions here? For instance, is an affine variety the spectrum of a finitely-generated algebra over a field for you? $\endgroup$ – KReiser May 26 at 18:06
  • $\begingroup$ @KReiser An affine variety is the zero set of some polynomials over $K^n$ with the Zariski topology and the structure of space with functions (local quotients of polynomial functions). I guess strictly topologically (ignoring the functions) this is the same as the spectrum of a f.g. algebra without nilpotents with the Zariski topology, but I don't know how to turn this into a space with functions, I think that would go into scheme theory, right? An algebraic variety is a space with functions which has a cover with open affine varieties and is Hausdorff (in the spaces-w/-functions sense). $\endgroup$ – rosecabbagedragon May 27 at 8:20
  • 1
    $\begingroup$ Yeesh. I'm pretty firmly in the "please use schemes" camp, but I understand that there is some cost to switching and folks do so at their own pace. As far as your problem, it seems to me that one reasonable way to prove this is to show the claim for $W$ affine and then bootstrap this to the general case by applying that case to each $U_i$. If you include a link to the material you're working out of, this could be confirmed. $\endgroup$ – KReiser May 27 at 10:11
  • 1
    $\begingroup$ @rosecabbagedragon - In chapter I.3 in Hartshorne he defines the structure sheaf $\mathcal{O}_X$ for any quasi projective variety $X$ over an algebraically closed field. As I mention in my post: There is a criterion for a scheme to be affine and you may try to prove a similar result for a quasi projective variety in the sense of Chapter I.3 using the structure sheaf from this chapter. $\endgroup$ – hm2020 May 27 at 12:15
0
$\begingroup$

Question: "So what guarantees that F−1(U) is an algebraic variety and that F|F−1(U):F−1(U)→U is still an affine morphism of algebraic varieties? Or is this even needed?"

Answer: Let "algebraic variety" be the one defined in Hartshorne, Chapter I: This means $k$ is an algebraically closed field and $V,W$ are quasi projective varieties of finite type over $k$ (in the sense of HH.I.3). If $F: V \rightarrow W$ is a morphism of algebraic varieties and if $U\subseteq W$ is an open subset, it follows $F^{-1}(U) \subseteq V$ is an open subset, and any open subset of $V$ has canonically the structure of an algebraic variety. Hence $F^{-1}(U)$ is canonically an algebraic variety. I believe you will need a result in HH.II.2 - Exercise 2.17 which is a criterion for a "scheme" to be affine to conclude.

You must formulate and prove a similar result for algebraic varieties in the sense of Chapter I. Let $(X, \mathcal{O}_X)$ be an algebraic variety in the sense of HH.I

You must prove that $X$ is an affine algebraic variety iff there are elements

$$f_1,..,f_l \in H^0(X, \mathcal{O}_X):=A$$

such that the open subsets $X_{f_i}$ are affine algebraic varieties and the ideal $I:=(f_1,..,f_l)\subseteq A$ generate the unit ideal,

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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