39

For any open subset $U\subseteq\mathrm{Spec}(A)$ let $S_U=A\setminus\bigcup_{\mathfrak p\in U}\mathfrak p$ and $\mathscr O'(U)=A[S_U^{-1}]$. It is obviously a presheaf.
Claim: For open subsets of the form $U=\mathrm{Spec}(A_f)$ with $f\in A$ we have $\mathscr O'(U)=A_f$. (This shows that the associated sheaf of $\mathscr O'$ is indeed $\mathscr O_{\mathrm{...

31

Let's say $k=\mathbb{C}$ (although something like this should work over any algebraically closed field). Let $V_1=\mathbb{P}^1-\{0,1,\infty,\pi\}$ and $V_2 = \mathbb{P}^1-\{0,1,\infty,e\}$. One can see that $V_1$ and $V_2$ are not isomorphic as varieties/schemes over $\mathbb{C}$: such an isomorphism would extend to a map $\mathbb{P}^1\to\mathbb{P}^1$ taking ...

25

Call $X $ your scheme over the field $k$, $P_1$ and $P_2$ the two special closed points, $A_1$and $A_2$ their respective open complements and $A_{12}=A_1\cap A_2$, so that $A_i\simeq \mathbb A^1_k$ and $A_{12}\simeq\mathbb G_m$, all affine schemes.
Here are some (not independent) proofs that $X$ is not affine.
Proof 1
The point $(P_1,P_2)\in X \times X $...

answered Feb 4 '12 at 9:43

Georges Elencwajg

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25

First, here are some things about the four generalizations you mention:
Monoids don't fall into Diers' framework: By his Proposition 1.4.1 the terminal object in his framework is strict, i.e. any morphism $1 \to A$ is an isomorphism, which is definitely not the case for monoids.
I also wouldn't expect Diers' examples to be instances of Toen/Vaquie's ...

23

This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":
Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $\alpha \in k$ let $R_\alpha = R[y_\alpha]/((x-\alpha)y_\alpha,y_\alpha^2)$. Then $R_\alpha$ is an affine line with an embedded prime $\...

answered Dec 9 '18 at 5:05

R. van Dobben de Bruyn

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21

Any scheme which is separated of finite type, has at least a triangulation, hence is, in particular, a CW-complex. In fact, by a theorem of Lojasiewicz, this is true for any semi-algebraic set (one can even get this for subanalytic sets, by a result of Hironaka, in Triangulation
of algebraic sets, Proc. Amer. Math. Soc. Inst. Algebra Geom. Arcata(1974)); ...

21

$\def\dr{d_{\rightarrow}}\def\du{d_{\uparrow}}$ This is true. I don't know a reference, but here is a proof. I will show, more strongly, that, at every stage in the Hodge-de Rham spectral sequence, we have a perfect pairing $E^{pq}_r \times E^{(n-p)(n-q)}_r \to k$. In particular, $E^{pq}_{\infty}$ and $E^{(n-p)(n-q)}_{\infty}$ are dual and, since $H^k_{DR}(X)...

answered Nov 2 '15 at 4:03

David E Speyer

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20

A brief answer.
First of all, the results are miraculous. Deligne's Hodge II and Hodge III give just a few example applications of the kind of results you can prove using mixed Hodge theory; these are great theorems which just fall out of the general theory. People quickly figured out more applications, like the Hodge-Deligne polynomial, which is itself a ...

18

It's inherently difficult to give a negative answer to a question like this, but here's a technical fact that pushes in that direction:
Let ZFC$_n$ be the subtheory of ZFC gotten by restricting Separation and Replacement to $\Sigma_n$ formulas. By the reflection principle,$^1$ for each $n$ the theory ZFC proves that there is an ordinal $\alpha_n$ such that $...

17

I shall assume that $X,Y$ are integral, locally noetherian schemes and that $f$ is dominant. Then the degree of $f$ is the degree of the corresponding extension of fields, namely
$$deg(f)=[Rat(X):Rat(Y)]$$.
We have for the fibers $X_y \; (y\in f(X))$ of $f$ the interesting result:
$$dim_{\kappa (y)} \mathcal O(X_y)\geq deg(f)$$
with equality for all ...

answered Oct 9 '11 at 13:02

Georges Elencwajg

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17

### Defining abstract varieties and their morphisms over a finitely generated subfield of the base field

This is treated (in much greater generality) in EGA IV$_3$, Théorème 8.8.2. The existence of $X_0$ and $Y_0$ follows from part (ii), whereas the existence of $f_0$ is part (i).
References.
[EGA IV$_3$] A. Grothendieck, Éléments de géométrie algébrique. IV: Étude locale des schémas et des morphismes de schémas. (Troisième partie). Publ. Math., Inst. Hautes ...

answered Oct 5 '18 at 15:50

R. van Dobben de Bruyn

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17

Here is a counterexample. Fix a field $k$, and let $Y$ be built from two copies of the affine nodal curve $y^2=x^3+x^2$, glued together on the complement of the singular point. In other words $Y$ is a nodal curve with doubled singular point. Then $Y$ is not affine because it is not separated. However, there is a homeomorphism $\mathbb{A}^1\to Y$, which is ...

16

If $C$ is a sufficiently general rational curve in $\mathbb P^3$ of degree $d$, then the vector space of degree $k$ homogeneous polynomials vanishing on $C$ has dimension precisely equal to
$$\max\left\{0,\binom{k+3}{3} - dk - 1\right\}\;.$$
More geometrically, there exists a degree $k$ surface containing $C$ if and only if $\binom{k+3}{3} - dk - 1 > 0$.
...

15

I think the easiest condition is the fact that the natural morphism
$$
(X, \mathcal O_X) \to \operatorname{Spec}(\mathcal O_X(X))
$$
is an isomorphism.

15

No, it's not true in general (EGA 2, (1.2.3)).
The following example is taken from EGA 2, (1.3.3). Over a field $K$, let $S$ be the affine plane with a doubled origin. Then $S$ is the union of two affine open subsets $Y_1$ and $Y_2$, each of them is isomorphic to the affine plane, glued along the complementary subset of their origin. In particular, $Y_1$ ...

15

TL;DR The higher étale homotopy groups are the homotopy groups of the profinite completion of the shape of the étale topos. As such they are profinite groups. If you choose to see profinite groups as topological groups, group schemes or pro-systems is largely a matter of choice.
How is the étale homotopy group defined? There are many ways, but possibly the ...

14

This is pretty much explained in the comments, but let me put it into an answer. One wants algebraic de Rham cohomology to be isomorphic to the usual de Rham cohomology (using $C^\infty$ forms) when $k=\mathbb{C}$, and have similar properties when $k$ is an arbitrary field of characteristic zero. From this point of view, as Grothendieck observed in the 1960'...

14

One very important point, implicit in the comments by @zzy and @abx, is that valuative criteria allow to check a property (e.g. properness) of (say) an $S$-scheme $X$ directly in terms of the functor of points of $X$. This is of course especially convenient if $X$ is defined by this functor, classical examples being Picard, Quot or various moduli functors. ...

13

Let $U \cong \mathbb{A}^n$ be the image of $j$ and let $\Delta = \mathbb{A}^n \setminus U$. Since $U$ is affine, $\Delta$ is purely codimension $1$, so $\Delta$ is the zero locus of some $f \in k[x_1, \ldots, x_n]$. But then $f$ is a unit on $U$, contradicting that $U \cong \mathbb{A}^n$.

answered May 25 '13 at 13:01

David E Speyer

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13

In the spirit of You Could Have Invented Spectral Sequences by T.Chow, I claim you could have invented $\operatorname{Ext}^{1}(\mathbb Q(0),M)$ as group of "rational points" of a motive. Here is how.
The motivation comes from conjectures on special values of $L$-function. Before the general conjectures were formulated by S.Bloch and K.Kato, there were two ...

13

What is true is that there is an antiequivalence between the category of schemes affine over $S$ (that is $S$-schemes for which the preimage of an open affine of $S$ is an open affine) and quasi-coherent $\mathcal{O}_S$-algebras.
The anti-equivalence is realized by the pushforward of the structure sheaf and the relative spectrum (see Exercise 5.17 in ...

13

Not exactly a book, but there are course notes on abelian varieties from a course that Brian Conrad taught a few years ago. It is definitely from the perspective of scheme theory/functor of points.

13

The laudatio for the Wolf prize explains it like this:
Central to modern algebraic geometry is the theory of moduli, i.e.,
variation of algebraic or analytic structure. This theory was
traditionally mysterious and problematic. In critical special cases,
i.e., curves, it made sense, i.e., the set of curves of genus greater
than one had a natural ...

answered Oct 12 '18 at 16:29

Carlo Beenakker

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13

The answer is no. For instance, take the quadratic cone
$$
Y = \mathrm{Spec}(\Bbbk[x,y,z]/(xz-y^2))
$$
and let $X$ be its blowup at the vertex. Then $X$ is regular, but
$$
H^0(X,\mathcal{O}_X) = H^0(Y,\mathcal{O}_Y) = \Bbbk[x,y,z]/(xz-y^2)
$$
is not.

12

Note that if $T\to S$ is also quasicompact, then $S$ must be locally noetherian: this boils down to the well-known fact that if $A\to B$ is a faithfully flat ring homomorphism and $B$ is noetherian, then so is $A$.
This proves that Jason's example above is indeed a counterexample. More generally, any non-noetherian local scheme $S$ is a counterexample: if $...

answered Dec 16 '11 at 8:06

Laurent Moret-Bailly

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12

Suppose the structure morphism $g: X\to \operatorname{Spec}(A)$ is separated and of finite type, and $f: \operatorname{Spec}(B)\to \operatorname{Spec}(A)$ is faithfully flat; furthermore, assume $A, B, X$ are all Noetherian. Denote $X\times_A B$ as $X_B$ and let $f': X_B\to X$ be the base change of $f$ along $g$, and $g': X_B\to \operatorname{Spec}(B)$ the ...

12

Here's a theorem I find useful:
Theorem. Let $\phi \colon X \to Y$ be a smooth morphism of schemes of relative dimension $d$. Then there exists an open cover $X = \bigcup U_i$ of $X$ such that each $U_i \to Y$ factors as
$$U_i \stackrel \pi \to \mathbb A^d_Y \to Y,$$
with $\pi$ étale. Mnemonic: smooth morphisms have étale coordinates.
See ...

12

You can prove essential smallness of the category of finite type $S$-schemes (without further assumptions), where $S$ is any scheme, as follows:
If $S$ is affine and we only consider affine finite type $S$-schemes, these correspond to finite type $\Gamma(S)$-algebras. These are isomorphic to algebras of the form $\Gamma(S)[x_1,\dotsc,x_n]/I$ for some ...

12

I am just posting my comment as one answer. So long as you are only asking about schemes (rather than complex analytic spaces), you can avoid the hard analysis from the previous MathOverflow answer. The argument below sketches this. The main additional detail beyond Hilbert scheme techniques is a strong variant of Chow's Lemma. Let $k$ be an ...

11

I suppose that "additive" means that "additive over short exact sequences". If so, this is does not seem too hard, at least if $X$ is separated.
By noetherian induction, you may assume that for all proper integral subscheme $Y$ of $X$, the restriction of $g$ to $Y$ is given by a multiple of the generic rank at $Y$. But every coherent sheaf with support on $...

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