# Tag Info

## Hot answers tagged algebraic-curves

40

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 ...

28

Here are some observations. I include the case $g=1$ (even if $X$ has no rational point). Denote by $\hat{\mathcal X}$ the (proper) minimal regular model of $X$ over the $O_K$ and let $\mathcal X$ be the smooth locus of $\hat{\mathcal X}$. (1) If the Néron model exists, it is equal to the smooth locus $\mathcal X$ of the minimal regular model. (2) If the ...

28

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 ...

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.

23

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 $p$, ...

23

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 $\mathbb{P}^3$.

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

There does not exist a map of a smooth complete genus 2 curve to $M_3$. Such a map would give rise to a surface $S$ (of general type) which violates the Bogomolov-Miyaoka-Yau inequality $c_1(S)^2 \leq 3c_2(S)$. This inequality is equivalent to $3\sigma (S) \leq e(S)$ where $\sigma$ and $e$ are the signature and topological euler characteristic of the ...

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

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 ...

21

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 involves an Euler ...

20

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 $F_i$...

19

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 ...

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 ...

18

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 ...

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 ...

16

There are already good answers by quid and by Dustin Clausen here. I thought, though, that I'd take the time to write out something more leisurely and expository. To get from Weil Reciprocity to Quadratic Reciprocity, one must make some things more general and some things less general, and there is a choice of which order to do these things in. I will first ...

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 ...

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 $$K_X^... 15 I have always felt this was due to Riemann himself: especially in: Theory of Abelian Functions, 1857. Of course the association of a Riemann surface to an algebraic curve is generally attributed to him, and there he also proves conversely there is a plane curve associated to a (compact) Riemann surface. He proves there too the Riemann part of the Riemann ... 15 The reason is that there are two ways of thinking about "points". Let A be a ring. Then, define: A scheme-theoretic/topological point of Spec A is a prime ideal of A. A geometric/functorial point of Spec A is an equivalence class of morphisms Spec K\rightarrow Spec A, where K is a field, and two morphisms:$$p_1 : \text{Spec }K_1\rightarrow \...

15

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

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 ...

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 $H^0(C_4,... 13 Let$E,E'$be elliptic curves over the residue field. Then$E \times E'$has good reduction but is not a Jacobian. However, any principally polarized abelian surface that reduces to$E \times E'$but is not a product of elliptic curves is a Jacobian of some genus-$2$curve$C\$ that cannot have good reduction. Explicitly (when the residue characteristic is ...

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