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I have two questions:

  1. Can a irreducible rational curve have infinitely self intersections?

  2. If $f$ has rational coefficients and the solutions for $0=f(x,y)\in k[x,y]$ are parametrized by rational functions with rational coefficients of some parameter $t$, then the image of this parametrization over the rationals miss only finitely many rational points.

It is not clear to me why only finitely many points were missed. My first guess is some what related to rational points are dense. Can someone elaborate a bit geometric intuition on 2 and how to prove it(I think hint will suffice)?

nowhere dense
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user45765
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  • Are these your own questions or do they come from a textbook ? – Rene Schipperus Nov 04 '16 at 02:27
  • 1 is not from textbook. 2 is from shafarevich claim that missing finite points which I think he will cover in later chapters. However, I do not have geometrical intuition on 2. So I need some input. – user45765 Nov 04 '16 at 02:39
  • Where in shafarevich is this claim ? – Rene Schipperus Nov 04 '16 at 02:48
  • I think you want to read ch I 1.3 prop. – Rene Schipperus Nov 04 '16 at 02:57
  • @ReneSchipperus pg 7(latest edition) Basic Algebraic Geometry Vol.1 Sec 1.2 Rational Curves, toward the end of second paragraph. – user45765 Nov 04 '16 at 02:58
  • @ReneSchipperus Yes. That is what I want. – user45765 Nov 04 '16 at 02:58
  • At the end of that section he gives a counter example to the general case as you have in the question. – Rene Schipperus Nov 04 '16 at 03:02
  • I give a bounty to this question because I had the same doubt today and I couldn't answer it. I have an idea of how to approach it: Let $C=V(f)$ and $(\phi,\psi): \mathbb{A}^1\rightarrow C$ the parametrization of $C$ defined over $\mathbb{Q}$. It is easy to see that the parametrization is surjective (except maybe for finitely many points). Hence, if $(a,b)\in C$ is a rational point then there is an element $t_0\in k$ such that $$\phi(t_0)=a \text{ and } \psi(t_0)=b.$$ So we have to prove that this two conditions imply $t_0$ is rational for all but finitely many $(a,b)\in C(\mathbb{Q})$. – nowhere dense Apr 12 '20 at 14:11
  • If $\phi=\phi_1/\phi_2$ and $\psi=\psi_1/\psi_2$ where $\phi_i$ and $\psi_i$ are polynomials then we have that $t_0$ satisfy both polynomial equations $$\phi_1(t_0)-a\phi_2(t_0)=0 \text{ and }\psi_1(t_0)-b\psi_2(t_0)=0 \tag{$\star$}.$$ Hence, if $t_0$ is not rational, then the minimal polynomial of $t_0$ would be a nontrivial factor of both polynomials in $(\star)$. Therefore, the resultant $R(a,b)$ of both polynomials in $(\star)$ must be 0. – nowhere dense Apr 12 '20 at 14:22
  • So if a point $(a,b)$ is not in the image of the parametrization it should be in $C\cap V(R)$ where $R$ is the resultant. As $C$ is irreducible it is therefore enough to prove that $$f(x,y)\not\mid R(x,y)$$ And this looks quite difficult as the resultant is really ugly... – nowhere dense Apr 12 '20 at 14:22
  • @nowheredense I am going to use a stronger result. Since the map is birational, you can see there are open sets with isomorphism as ringed space. WLOG, you can shrink the target open set a bit which can be treated as $D(f)$ non vanishing points of $f$ for the corresponding variety. Denote the variety coordinate ring as $A(X)$. Then $A(X)/(f)$ describes $V(f)=D(f)^c$. Note that $A(X)/f$ is 0 dimensional and noetherian which forces $A(X)/f$ artinian. Artinian gives you $A(X)/f$ is a finite product of local rings. That yields finite points missing. – user45765 Apr 12 '20 at 14:44
  • @user45765 If I undertand correctly, I think you are trying to prove that a birrational map between $\mathbb{A}^1$ and the curve induce an isomorphism between an open set of $\mathbb{A}^1$ and an open set of the curve, and the complement of the open set in the curve is finite. This is true but the problem is that this isomorphism is a priori in the bigger field $k$ so it doesn't give you a correspondence between rational points necessarily. – nowhere dense Apr 12 '20 at 17:06
  • @nowheredense You can always enlarge the field and then take the rational points afterwards. – user45765 Apr 12 '20 at 18:05
  • @user45765 Yes, but if the isomorphism has coefficients in the large field then it will not preserve the rational points. I mean, something like what I asked here may happen – nowhere dense Apr 12 '20 at 18:09
  • @nowheredense I think there is some issue here. The morphism you take over algebraic closure is exactly the same morphism with same coefficient over $Q$. For the problem in your link, I think it certainly could happen as you do not know what is coefficient of morphism over $C$ to start with. However, here you know what to start with by just naively taking the same morphism. – user45765 Apr 12 '20 at 20:02

2 Answers2

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  1. I think it not possible for any plane algebraic curve $C$ to have infinitely many self-intersections. If $p$ is a self-intersection then locally around $p$ curve looks like $f=f_1f_2=0$, thus $df|p=0$. So, infinitely many points of self-intersection give us infinitely many points where $df=0$ or infinitely many zeroes of the system of equations $f=0$, $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$. But if two polynomials of two variables have infinitely many common zeroes they have a common irreducible component. Therefore, $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$ along $C$ a and any point of $C$ is singular. That is not possible because singular points form a closed subset of a variety.
  2. If you get such parametrization you curve is necessarily rational. This is because existence of rational map from $\mathbb{P}^1$ to the curve tell us that the curve is unirational. On the other hand Luroth's theorem tells us that any unirational curve is rational.
Alex
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For the second question the idea of @user45765 was correct.

Denote by $C$ the curve given by $f(x,y)=0$. Then, if there is a birrational map $\varphi:\mathbb{A}^1\rightarrow C$ (that is, a parametrization by rational functions) there are open sets $U\subseteq \mathbb{A}^1$ and $V\subseteq C$ such that $\varphi_{|U}:U\rightarrow V$ is an isomorphism. As $\varphi_{|U}$ is defined over $\mathbb{Q}$ it's inverse must be defined over $\mathbb{Q}$ as well (see here). Hence, $\varphi$ gives a bijection between the rational points in $U$ and $V$. As $C\setminus V$ is finite (for example, because of the description of the Zariski topology on a curve) we get the result above.

nowhere dense
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  • The reliance on my answer using descent theory is excessive - this can be resolved with the definition of a birational morphism and the topology on an irreducible curve. Two varieties $X,Y$ are birational iff there are open sets $U\subset X$ and $V\subset Y$ so that $U\cong V$. Since the open subsets in an irreducible curve are exactly the complements of finitely many closed points and rational points are closed, saying that the parameterization is a birational morphism implies that it misses only finitely many points. – KReiser Apr 12 '20 at 23:10
  • I think that would imply that $\varphi(\mathbb{C})$ misses only finitely many rational points. But I would like to show that $\varphi(\mathbb{Q})$ misses only finitely many rational points and that follows from $U\cong_{\mathbb{Q}} V$. – nowhere dense Apr 12 '20 at 23:16
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    I still think this is excessive. If two integral $\Bbb Q$-varieties $X$ and $Y$ are birational over $\Bbb Q$, then their function fields $\Bbb Q(X)$ and $\Bbb Q(Y)$ are isomorphic and we can define an isomorphism of open affine subsets defined over $\Bbb Q$ by looking at the map on function fields and being moderately clever. This gives you (with no need for fancy technology) an isomorphism of two open subsets defined over $\Bbb Q$, each of which has finite complement and the isomorphism preserves $\Bbb Q$-points. – KReiser Apr 12 '20 at 23:33
  • (Though I guess it depends on how much you think Luroth, which implies that the parameterization is a birational map, is "fancy" - my vote is that it's not very fancy because one can prove it at an undergraduate level using only concepts from basic field theory.) – KReiser Apr 12 '20 at 23:35
  • You are right, that approach convince me that it is not necessary to use the other result :) – nowhere dense Apr 13 '20 at 01:05