A curve has infinitely many points

Question

Let $f\in k[x,y]$, where $k$ is an algebraically closed field. I would like to prove the curve $f(x,y)=0$ has infinitely many points. What I know is $k$ is infinite, but I don't know how to use this to prove this curve has infinite points.

Shafarevich speaks about this in Basic Algebraic Geometry 1: Varieties in Projective Space on the page 4, so I suppose this should be trivial.

Thanks

Related: https://math.stackexchange.com/questions/286352/polynomial-with-infinitely-many-zeros — Watson, Oct 23 '18 at 09:39

score 8 · Accepted Answer · answered Apr 14 '15 at 04:55

8

We need the hypothesis that $f$ is (either zero or) nonconstant. Assume WLOG that $f(x, y)$ is nonconstant in $x$. Write $f(x, y) = x^n g(y) + \text{lower terms}$ where $x^n$ is the highest order term in $x$ and $g(y)$ is a nonzero polynomial in $y$. For all but finitely many values $y_0$ of $y$, $g(y_0)$ is nonzero, so for all but finitely many values $y_0$ of $y$, $f(x, y_0)$ is a nonconstant polynomial in $x$ and hence has a root.

answered Apr 14 '15 at 04:55

Qiaochu Yuan

468,795

1

What about the non-constant polynomial $f(x,y)=nx+1$ over $\mathbb{C}$? The corresponding curve has only one point. – uniquesolution Aug 18 '17 at 07:19
1

@uniquesolution: no, the points are $(-1/n, y)$ for any $y$. – Qiaochu Yuan Aug 21 '17 at 21:19

score 0 · Answer 2 · answered Apr 14 '15 at 04:25

0

Hint : for all $y \in k$, consider the polynomial $P_y(X) = f(X,y) \in k[X]$.

answered Apr 14 '15 at 04:25

Abelmondo

493

A curve has infinitely many points

2 Answers2