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What is the Hasse-Weil inequality (in particular, a lower bound) for singular projective curves over finite fields which are not geometrically irreducible?

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This has nothing to do with the Hasse-Weil inequality: I assume your curve is arithmetically irreducible. Let $C_1, \ldots,C_r$ be the geometric components of the curve. Each rational point is on some $C_i$ and therefore by Galois conjugacy on each $C_i$. In particular, the rational points are on the intersection of two geometrically irreducible curves, so there are only finitely many of them. Bezout's Theorem gives an upper bound for them.

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  • $\begingroup$ But I want a lower bound. $\endgroup$
    – Nanjun Yang
    Commented Sep 29 at 2:43
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This is to address the updated question asking specifically for a lower bound.

The best lower bound you can get in general is zero. Let $a$ be a quadratic nonresidue in $\mathbb F_p$, let $\alpha\in \mathbb F_{p^2}$ be a square root of $a$, and let $f(x) \in \mathbb F_p[x]$ be any monic irreducible polynomial of large even degree. Then the union of $\alpha y^2 = f(x)$ and $-\alpha y^2 = f(x)$ is defined over $\mathbb F_p$ and has no $\mathbb F_p$-rational points: Such a point would have to satisfy $y = 0$, but this can't happen with $x\in\mathbb F_p$ since $f$ is irreducible. (The points at infinity have field of definition $\mathbb F_p(\sqrt{\alpha}) \ne \mathbb F_p$.)

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    $\begingroup$ The phrase "points at infinity" is a bit ambiguous, since it depends on which projective embedding you are thinking of. If you're thinking of $\mathbb{P}^2$, then both curves intersect in the rational point $(0:1:0)$. If you're instead considering the normalization, then you're no longer in $\mathbb{P}^2$, and you might as well take any union of Galois conjugate curves in projective space and perform repeated blow-ups so as to create a disjoint union, which of course never contains a rational point. $\endgroup$
    – R.P.
    Commented Sep 29 at 18:43
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    $\begingroup$ Ah, right, I was implicitly thinking of the points at infinity on the smooth models of the individual curves. $\endgroup$
    – John Doyle
    Commented Sep 29 at 18:52
  • $\begingroup$ I wonder if you can get 0 as a lower bound by tweaking the example. But in any case a lower bound 1 essentially makes the same point. $\endgroup$
    – R.P.
    Commented Sep 29 at 19:13
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    $\begingroup$ Maybe replacing $y^2$ with $y^d$, where $d$ is the degree of $f$, would work. $\endgroup$
    – John Doyle
    Commented Sep 29 at 19:32

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