Let $C$ be a projective plane curve, defined by a polynomial in $Z[x,y]$, over a field $K$. Does the geometric genus of $C$ depend on the choice of $K$?

I think the answer to this question is obviously 'yes', since the genus can change when you reduce modulo a prime (i.e, change the base field from $Q$ to $F_p$), i.e. "bad reduction".

I guess my real question is: can the genus change when I change the base field from from $Q$ to $\bar{Q}$ (algebraic closure of $Q$)?

My first answer is still 'yes', because the genus-degree formula tells us that we can write the genus as $\frac{(d-1)(d-2)}{2} - \sum \delta_P$, where $\delta_P$ is the delta invariant at a singular point $P$, so I'm thinking that changing my base field from $Q$ to $\bar{Q}$ might introduce some new singularities.

On the other hand, it might be possible that all of the singularities over $\bar{Q}$ are already present over $Q$.

Anybody know for sure?


1 Answer 1


The arithmetic genus of a curve $X/k$ is defined by $p_a(X) = 1 - \chi(\mathcal{O}_X)$. By base change and cohomology this is invariant under extension of scalars, so the arithmetic genus does not change.

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    $\begingroup$ Thanks. I wasn't aware of that formula, but I think the arithmetic genus is (d-1)(d-2)/2. It certainly makes sense that it would be invariant. But what about the geometric genus? $\endgroup$ Apr 27, 2018 at 5:16

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