"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
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> 
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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.



Does the genus of a variety depend on the base field?
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.](https://stacks.math.columbia.edu/tag/0BY9) plus [Lemma 33.26.2.](https://stacks.math.columbia.edu/tag/04L2), 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](https://stacks.math.columbia.edu/tag/0BXX)).


The curve given by @amateur in the [comments](https://mathoverflow.net/questions/371091/does-the-genus-of-a-variety-depends-on-the-base-field#comment938276_371091), 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.


