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"Does the (arithmetic) genus of a variety depend on the base field?"

So the question arises from a definition of the elliptic curve, the Hartshorne's book just says that the elliptic curve is a curve of genus 1. (316p) Indeed, this definition is probably for the elliptic curve over C. Then I found the wikipedia says so for general fields. (https://en.wikipedia.org/wiki/Elliptic_curve)

Formally, an elliptic curve is a smooth, projective, algebraic curve of genus one

For this definition to make sense, the (arithmetic) genus should be invariant over base fields since we call $y^2z=x^3+axz^2+bz^3$ as an 'elliptic curve', no matter what the base field is. Then I got to wonder if the genus is invariant over base field in general.

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    $\begingroup$ What definition of genus are you using? This is not hard to see for any of them, but the exact proof varies depending on the definition. $\endgroup$
    – Will Sawin
    Commented Sep 7, 2020 at 12:50
  • $\begingroup$ @WillSawin I'm talking about the arithmetic genus (en.wikipedia.org/wiki/Arithmetic_genus). I found the definition at the problem 5.3 (230p) of the Algebraic Geometry(Hartshorne). $\endgroup$
    – droptable
    Commented Sep 7, 2020 at 12:56
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    $\begingroup$ Use Proposition III.9.3 in Hartshorne: "cohomology commutes with flat base extension". $\endgroup$
    – abx
    Commented Sep 7, 2020 at 13:37
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    $\begingroup$ On the other hand, if you use "genus" for the relative genus, so the number it appears in the Riemann-Roch formula $\ell(D)=deg(D)+1-g$ for divisors $D$ defined over the base field, with degree suffiently large and $\ell(D)$ the dimension as $K$-vector spaces, of the space of functions on the curve, defined over $K$ and with polar divisor bounded by $D$, then this genus can change under finite inseparable extensions. $\endgroup$
    – Nulhomologous
    Commented Sep 7, 2020 at 14:28
  • $\begingroup$ I have a feeling that the arithmetic genus does depend on the base field. For instance if you look over $\overline{\mathbb{F}_3}(t)$ the curve given by $y^2=x^3+t$ should have genus one. However, after inseparable base change such as $\overline{\mathbb{F}_3}(s)$ with $t=s^3$ it becomes genus zero: $y^2=x^3+s^3=(x+s)^3$. This agrees with @Nulhomologous comment. $\endgroup$
    – amateur
    Commented Sep 7, 2020 at 16:46

1 Answer 1

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I will try to answer the question of the title (so, if genus changes under base extension), but just for curves.

Answer 1: If the curve is smooth, projective and geometrically irreducible over a field, the genus does not change under a base extension.

For a reference, you can see Lemma 53.8.2. plus Lemma 33.26.2., which shows that such curves verify the hypothesis of the lemma 53.8.2.

Since this is what you asked, I could stop here. But I want to explain why there are a lot of results about genus changing under base extension. And the explanation is:

Answer 2: If the curve is regular and projective over a nonperfect field, then the genus can change under an inseparable extension. Such curves that change the genus are called "non-conservative".

Notice that over a nonperfect ground field, one needs to distinguish between regular (all the local rings are regular) and smooth (Jacobian condition or, equivalently, the base change to the algebraic closure is regular).

Why does one care about regular but non-smooth curves? Because of the famous equivalence between "extension $K/k$ of transcendence degree 1" and "regular projective curves over $k$", or, if you like, because normal projective curves are regular, but not necessarily smooth (see the stacks project: 53.2 Curves and function fields).

The curve given by @amateur in the comments, so the projective regular curve determined by the equation $y^2=x^3+t$ over $\mathbb{F}_3(t)$, is an example: it is regular, but not smooth. It has arithmetic genus 1, but when base changed to $\mathbb{F}_3(t^{\frac 13})$ (and desingularized) it has arithmetic genus 0. So, using the definition "an elliptic curve is a smooth, projective, algebraic curve of genus one with a fixed point", such a curve is not an elliptic curve, although it has a nice looking Weierstrass equation.

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