42

If you add cuspidal curves, then $\overline{\mathcal{M}}_{1,1}$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families
$$y_1^2 = x_1^3 + t^6 \ \mbox{and}\ y_2^2 = x_2^3 + 1$$
(so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of ...

answered Sep 25 '19 at 19:13

David E Speyer

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31

Riemann combines what is called Riemann-Roch and Riemann-Hurwitz nowadays.
He considers the dimension of the space of holomorphic maps of degree $d$ from the Riemann surface of genus $g$ to the sphere. He computes this dimension in two ways. By Riemann-Roch this dimension is
$2d-g+1$, for a fixed Riemann surface. (Indeed, Riemann-Roch says that
the dimension ...

answered Dec 1 '19 at 16:44

Alexandre Eremenko

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27

This is a special case of the main theorem in the paper by
I. N. Baker, J. A. Deddens, and J. L. Ullman,
A theorem on entire functions with applications to Toeplitz operators,
Duke Math. J.
Volume 41, Number 4 (1974), 739-745.
They proved a similar statement for arbitrary entire functions.

answered Jan 24 '16 at 13:36

Alexandre Eremenko

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26

The definition using exponential of such an ad hoc looking series is admittedly not too illuminating. You mention that the series looks vaguely logarithmic, and that's true because of denominator $m$. But then we can ask, why include $m$ in the denominator?
A "better" definition of a zeta function of a curve (more generally a variety) over $\mathbb ...

25

No. If a curve embeds into $\mathbb{P}^3$, its tangent space at every point has dimension $\leq 3$. This is a strong restriction on the possible singularities. For instance,
a curve locally isomorphic to the union of the lines $x=y=z=0$, $x=y=t=0$, $x=z=t=0$, $y=z=t=0$ in $\mathbb{A}^4$ cannot be embedded in $\mathbb{P}^3$.

24

The "cyclic cover trick" can refer to more than one thing. One example is as follows. Let $L$ be an invertible sheaf on a smooth projective scheme such that some power $L^{\otimes d}$ has a global section $s$ whose zero scheme $D$ is a smooth Cartier divisor (all of these smoothness conditions are not strictly necessary). Let $\nu:Y\to X$ be the ...

23

The answer to question 1 is yes, and is known as Harris's Uniform Position Lemma. It was proved in Harris's 1980 paper Galois groups of enumerative problems.
You can find a nice exposition in Chapter 9 of Solving Polynomial Equations, by Bronstein, Dickenstein and Emiris.
You should be warned that the analogous statement is not true in characteristic $p$, ...

answered Nov 15 '13 at 0:57

David E Speyer

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23

1) The genus of a complete intersection of multidegree $(d_1,\ldots ,d_{n-1})$ in $\mathbb{P}^n$ is $g=1+\frac{1}{2} d_1\cdot \ldots \cdot d_{n-1}(\sum d_i-n-1)$ (just compute the degree of the canonical bundle). This gives very particular values for $g$: $0, 1, 3, 4, 5, 6, 9, 10, 13, 15, 16,\dotsc $ . Any curve whose genus is not in this list cannot be ...

22

Yes, every proper 1-dimensional complex-analytic space $X$ admits a closed immersion (in the sense of locally ringed spaces over $\mathbf{C}$) into an analytic projective space and more specifically is the analytification of a 1-dimensional projective $\mathbf{C}$-scheme (uniquely determined up to unique isomorphism by the GAGA theorems). It also holds in ...

21

The question seems fine to me. Off the top of my head:
1) The Jacobian is a group, and in fact an abelian variety, whereas the curve usually isn't. This gives you a lot of structure to play with that you didn't have initially. For example, to show that a general curve doesn't map onto a curve of smaller positive genus, you can use the fact that the ...

21

Yes, this argument can be made rigorous. One needs three steps.
Step 1. Show that there is at least one smooth plane curve of degree $d$ with the expected genus. Essentially, the proof is given by your heuristic topological argument (deform the union of $d$ lines in general position).
Step 2. Show that if one slightly perturbs the coefficients of a ...

answered Jun 5 '14 at 22:04

Francesco Polizzi

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21

The correct setting for this construction turns out to be projective varieties, so let me suppose we have a smooth variety $X$ inside $\mathbf{P}^N$, for some $N \ge 1$, defined by the vanishing of some homogenous polynomials $F_1, \dots, F_r$ in variables $x_0, \dots, x_N$, with the $F_i$ having coefficients in $\mathbf{Q}$. Actually, let me assume the $F_i$...

answered Jan 22 '17 at 10:18

David Loeffler

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21

It suffices to consider the case when $\Omega$ is a circumcircle, so let it be.
At first, the points $A_b, A_c, B_c, B_a, C_a, C_b$ lie on a conic if and only if
$$
\frac{AB_a\cdot AB_c}{AC_a\cdot AC_b}\cdot \frac{BC_a\cdot BC_b}{BA_b\cdot BA_c}\cdot
\frac{CA_b\cdot CA_c}{CB_a\cdot CB_c}=1\quad\quad\quad\quad(\heartsuit)
$$
(by Pascal theorem, they lie on a ...

19

You can start here "Fermat's last theorem" and anabelian geometry??
In particular, I mention there that: At some point Deligne thought he had a proof that the section conjecture implied Mordell, but the proof doesn't work. This is all explained in an appendix by Deligne to a paper of Stix: http://arxiv.org/abs/0910.5009
Finally, for an actual ...

answered Aug 29 '14 at 14:26

Felipe Voloch

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18

If you're taking the definition of rational to be: birational to $\mathbb{P}^1$ over the field $k$, then the stated property is not even true. There are conics which have no rational points, and so are not rational, but are rational over a quadratic extension. For example, the affine conic $x^2 + y^2 + 1 = 0$ over the field $\mathbb{Q}$.
Added: since you ...

17

The connection came from the paper by Dedekind and Weber "Theorie der algebraischen Functionen einer Veranderlichen", Crelle's Journal, 1882. In this paper the authors recover the theory by Riemann (including the famous Riemann-Roch theorem) by the abstract procedure of assigning a "curve" to an degree one transcendental field extension $\Omega/\mathbb{C}$ ...

17

That is certainly not true. Consider the case that $C$ is an elliptic curve. Then $\text{Aut}(C\times C)$ contains $\text{GL}(2,\mathbb{Z})$ as a subgroup.

17

The projective form of your curve is $3y^{2} z = 4x^{3} - z^{3}$. This has three obvious points: $(1 : 1 : 1)$, $(1 : -1 : 1)$, and $(0 : 1 : 0)$.
Your curve is isomorphic over $\mathbb{Q}$ to the Fermat cubic, $x^{3} + y^{3} = z^{3}$. This latter curve has only three rational points on it: $(1 : -1 : 0)$, $(1 : 0 : 1)$ and $(0 : 1 : 1)$, and so your curve ...

17

I found the lemma on page 63 of Study's Einleitung in die Theorie der Invarianten linearer Transformationen auf Grund der Vektorenrechnung (1923).
The source cited for the proof is page 202 of Study's Methoden zur Theorie der ternären Formen (1889), where it appears as an endnote and refers to a "less elementary" proof by
O. Hölder, Zum ...

answered Oct 21 '20 at 15:01

Carlo Beenakker

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16

For all pairs. Suppose $n\geq m$. Take for $C_1$ a curve with an inflection point of order $m$, say $F=0$ with $F(X,Y,Z)=ZY^{m-1}+X^m+Z^m$. Then take $C_2$ defined by $G(X,Y,Z)F(X,Y,Z)+Z^n=0$, where $G$ is general of degree $n-m$. Then $C_1\cap C_2$ is reduced to the point $p:=(0,1,0) $. To make sure that $C_2$ is smooth, we apply Bertini's theorem to the ...

16

As the title to Mumford's famous paper "Toward an enumerative geometry..." suggests, knowing the cohomology / cycle theory of the moduli space of curves allows one to answer enumerative geometry questions for curves.
Here is an (very concrete) example that came up in real life for a student of mine. He had a family of genus 2 curves over $\mathbb{P}^2$ ...

15

It seems to me that this is not true and that a counterexample can be constructed as follows.
Take a double cover $\alpha \colon X \longrightarrow A$ of an abelian surface $A$, branched over a smooth divisor $B \in |2 L|$, with $L$ very ample. We have $$K_X=\alpha^* L, \quad \alpha_* \mathcal{\omega}_X = \mathcal{\omega}_A \oplus \omega_A (L),$$ hence $$K_X^...

answered May 12 '13 at 8:37

Francesco Polizzi

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15

I have always felt this was due to Riemann himself: especially in: Theory of Abelian Functions, 1857. Of course the association of a Riemann surface to an algebraic curve is generally attributed to him, and there he also proves conversely there is a plane curve associated to a (compact) Riemann surface.
He proves there too the Riemann part of the Riemann ...

15

The reason is that there are two ways of thinking about "points".
Let $A$ be a ring. Then, define:
A scheme-theoretic/topological point of Spec $A$ is a prime ideal of $A$.
A geometric/functorial point of Spec $A$ is an equivalence class of morphisms Spec $K\rightarrow$ Spec $A$, where $K$ is a field, and two morphisms:
$$p_1 : \text{Spec }K_1\rightarrow \...

15

The original paper of Riemann is his celebrated "Theorie der Abel'schen Functionen" in Crelle's Journal of 1854. This paper can be found online at https://www.maths.tcd.ie/pub/HistMath/People/Riemann/AbelFn/
There is an English translation of Riemann's Collected Papers (Kendrick Press, I believe).
Modern accounts can be found in several textbooks or ...

14

There are two different things which are called "Riemann surface" in the literature.
The modern notion (introduced by Hermann Weyl): complex 1-dimensional manifold.
In older literature this is sometimes called "Abstract Riemann surface".
"Riemann surface spread over the plane" (or over the sphere, or over some other surface). Surface de Riemann etalee in ...

answered Jul 5 '13 at 7:51

Alexandre Eremenko

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14

No. In fact more is true: the locus of all $n$-pointed curves of genus $g-1$ with a single elliptic tail $E$, such that $\mathrm{Aut}(E)=\mathbf Z/6$, has codimension two in $\overline M_{g,n}$ and consists of noncanonical singularities. This was famously determined by Harris and Mumford in their paper on the Kodaira dimension of the moduli space of curves (...

14

Yes. Start from a genus 2 curve $C_2$, and choose a point of order 3 in $JC_2$, giving rise to an étale $\mathbb{Z}/3$-covering $C_4\rightarrow C_2$.
Then $C_4$ cannot be hyperelliptic: a $g^1_2$ on $C_4$ would be stable under the covering automorphism $\sigma $, hence descend to $C_2$, which is impossible for degree reasons. $\sigma $ acts on $H^0(C_4,...

14

The answer is positive if and only if $\mathbb{A}^1\setminus S$ is an arithmetic curve, i.e., $\pi_1(\mathbb{A}^1\setminus S)\subset \mathrm{Aut}(\mathbb{H}) = PSL_2(\mathbb{R})$ is an arithmetic subgroup.
This however does not happen "very often". Let me be more precise.
Note that the Euler characteristic of $\mathbb{A}^1\setminus S$ equals $1-\# S$.
...

answered Jan 6 '18 at 10:30

Ariyan Javanpeykar

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13

Let $C$ be a non hyperelliptic curve of genus 3, let $f\colon C'\to C$ be an e'tale double cover and let $A$ be the Prym variety of $f$. Then $A$ is a principally polarized surface and the Abel-Prym map embeds $C'$ in $A$ as an element of $|2\Theta|$. Multiplication by $-1$ on $A$ restricts on $C'$ to the involution induced by $f$, hence the image of $C'$ ...

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