# Tag Info

4

Sorry for resurrecting such an old post, but Section §1.6 of Hartshorne annoys me in the sense that the concept of "abstract nonsingular curve" is not used very much (if at all) in algebraic geometry. We provide an alternative version of that section (specifically, Lemma 1.6.5 through Theorem 1.6.9) that avoids use of said object. We start with a ...

4

As it is pointed out in Matsumura, CRT, page 31, for a noetherian local domain $A$ the equality $\operatorname{ht}\mathfrak p+\dim A/\mathfrak p=\dim A$ holds for any prime ideal $\mathfrak p$ iff $A$ is catenary. So, we are looking for a noetherian local domain which is not catenary. There are no trivial examples, but you can find one at ...

4

They agree. In fact, you only need to assume quasicoherence; the key point is that the étale cohomology of a quasicoherent sheaf on an affine scheme vanishes in degrees $> 0$, just as for the Zariski topology. For more details, see tag 03DW in the Stacks project.

3

If one of the quadrics has maximal rank $4$ (or better one of the quadrics in the pencil, which is in fact true), then it is projectively equivalent to the Segre variety. In this case, the other quadric cuts out a curve of bidegree $(2, 2)$ on $\mathbb{P}^1 \times \mathbb{P}^1$. As the twisted cubic has bidegree $(1, 2)$, it follows that we get not only the ...

3

Here is a (translated) quote from G. Castelnuovo, "Sur les intègrales de différentielles totales appartenant à une surface irrégulaière", Comptes rendus hebdomadaires des séances de l'Acadeémie des sciences, Paris. 140, 23 Jan 1905. pp. 220-222: out of respect for Picard's profound research on surfaces admitting a group of birational automorphisms, ...

2

Are you asking for integer pairs? If so, $(x+3)^2+y^2=13=2^2+3^2$, so the integer pairs on this circle can only be $(-3 \pm 2, 0 \pm 3)$, and $(-3 \pm 3, 0\pm 2)$.

2

There are prime ideals $\mathfrak p$ in noetherian local rings $A$ such that $\operatorname{ht}\mathfrak p+\dim A/\mathfrak p<\dim A$. For instance, let $A=K[X,Y,Z]_{(X,Y,Z)}/(XY,XZ)$, and $\mathfrak p=(y,z)$. Then $\operatorname{ht}\mathfrak p=0$ (since $\mathfrak p$ is minimal), and $\dim A/\mathfrak p=1$ (since $A/\mathfrak p\simeq K[X]_{(X)}$). On ...

2

For a quick and dirty explanantion that offers no insight (as requested), take a $3\times 3$ permutation matrix and draw a star in every unoccupied spot that is not above (in the same column) or to the right (in the same row) of any $1$. Put $0$s everywhere else. That's the Schubert cell corresponding to the permutation matrix. For example, for the ...

2

Your first isomorphism seems believable: just plug in any open set $U$ on both sides. You need to check things like $$(f^*\mathscr G)\big|_{f^{-1}U} = \left(f\big|_{f^{-1}U}\right)^* \left(\mathscr G\big|_U\right),$$ etc, because you want to use the adjunction $\left(f\big|_{f^{-1}U}\right)^* \dashv \left(f\big|_{f^{-1}U}\right)_*$. But pulling it through ...

2

Yes, that sounds about right. Yes. The superscript means taking the invariant subspace under the action of $\tau$. In general, if $V$ is a vector space and $G$ is a group acting on it, then $V^G = \{ v \in V \mid gv = v \forall g \in G \}$.

2

I think the following works, maybe there is an easier answer... First note that $\det E_{|L}=\bigotimes_{i=1}^{\dim V-1}\mathcal{O}_L(a_i(L))=\mathcal{O}_L(\sum a_i(L))$. Note also that $\det E_{|L}\otimes \mathcal{O}_L(2)=\det (V\otimes\mathcal{O}_{\mathbb{P}^n})_{|L}=0$. So $\det E_{|L}=\mathcal{O}_L(-2)$ and $\sum a_i(L)=-2$. Then, note that, because ...

1

Great question! Here is a counterexample: Let $(A,\mathfrak m)$ be a DVR, and let $A \subseteq B$ be a finite extension of domains such that $B$ has exactly two primes above $\mathfrak m$. For example, $A = \mathbb Z_{(5)}$, and $B = \mathbb Z_{(5)}[i]$, with the primes $(1+2i), (1-2i)$ lying above $(5)$. Then $Y = \operatorname{Spec} A$ is the space ...

1

There is probably a more direct way to prove this, since it is apparently an exercise in Silverman, but one way to see it is: if $C$ has genus $1$, then its Jacobian $E$ is an elliptic curve, and $C$ and $E$ become isomorphic over any field $L$ over which $C$ obtains a rational point.

1

Let $R$ be the quotient of a polynomial ring $k[x_1,x_2,\dots]$ in infinitely many variables over a field by the ideal generated by all products $x_ix_j$ for $i\neq j$. Note that if $P\subset R$ is a prime ideal, then there can be at most one $i$ such that $x_i\not\in P$. It follows that if $P_i$ is the ideal generated by all the $x_j$ for $j\neq i$, ...

1

First, your even more basic question. Indeed $\pi_*\mathbb{Z}_{\mathbb{A}^1}$ and $i_*\mathbb{Z}_x$ are not coherent sheaves. In fact, they are not even sheaves of $\mathcal{O}_X$ modules. And as Remy points out, $H^1_{ét}(X,\pi_*\mathbb{Z}_{\mathbb{A}^1})$ and $R^1\pi_*\mathbb{Z}_{\mathbb{A}^1}$ are not the same kind of objects, the first being a group, the ...

1

Lemma. Let $f \colon X \to Y$ be a finite morphism, and let $\mathcal F$ be any sheaf on $X_{\operatorname{ét}}$. Then $R^if_* \mathcal F = 0$ for all $i > 0$. Proof. It suffices to show that $(R^if_* \mathcal F)_{\bar y} = 0$ for all geometric points $\bar y \to Y$. But the stalk is computed by $H^i_{\operatorname{ét}}(X', \mathcal \pi^* F)$, where $X' ... 1 By your definition of regular functions ("A regular function on$U$is a rational function that is well-defined at all points of$U$"), they indeed do not form a sheaf, as$O_X(\emptyset)$is the entire field of rational functions on$X$, rather than$0$. An easy (if inelegant) way to fix this is to say that your definition is only the definition of a ... 1 Usually Schubert cells are constructed using the structure theory of semisimple Lie groups (in this case$SL(3,\mathbb C)$), but for this case, one can give an explicit description as follows. Start by fixing one flag, say the standard one$\mathbb C\subset\mathbb C^2\subset\mathbb C^3$. This will be the unique Schubert-cell of dimension$0$. The further ... 1 For the even case, you are showing that the image is isomorphic to$\mathbb{A}^1$. It is not necessary (and not true) that$\varphi_n$is the isomorphism. In fact, your answer almost contains the map from$\mathbb{A}^1$to$\varphi_n(\mathbb{A}^1)$and the map in the other direction. (You will need to check that they are mutual inverses of course). In the ... 1 Careful. You have shown that a the sections of a presheaf$\mathscr G$over a base for the topology on some space$X$are isomorphic to those of a sheaf$\mathscr F$($\mathscr G$,$\mathscr F$are assumed to be presheaves on$X$). This is not enough to conclude abstractly that$\mathscr G$is a sheaf isomorphic to$\mathscr F$; such a thing is clearly not ... 1 There is a problem in the conjecture. Over$\mathbb C$it is really easy to prove. I believe you mean$\mathbb{A}^n(\mathbb{C})$. Here is a way of proving it that I like and that can work for proving many things over rings whence you know it over$\mathbb{C}$First notice that for countability reasons,$\mathbb{C}\$ has an infinite transcendance basis. This ...

Only top voted, non community-wiki answers of a minimum length are eligible