# The genus of an algebraic curve is invariant under isomorphisms

I would like to know how to prove (or even better to see a full proof) of the following "fact".

Let $C_1$ and $C_2$ be two smooth curves and let $\phi : C_1 \rightarrow C_2$ be an isomorphism. Then $$\text{genus}(C_1) = \text{genus}(C_2)$$

I am not completely sure this is true since I haven't seen this result explicitly stated, but I Imagine it has to be true.

The motivation for this comes from an exercise from Silverman's book The Arithmetic of Elliptic Curves. I was doing the following exercise and I found that I needed the above mentioned fact in order for my argument for $(i) \implies (ii)$ to work.

2.5 Let $C$ be a smooth curve. Prove that the following are equivalent (over $\bar{K}$):

(i) $C$ is isomorphic to $\mathbb{P}^1$.

(ii) $C$ has genus $0$.

(iii) There exist distinct points $P, Q \in C$ satisfying $(P) \sim (Q)$

I've thought about it but unfortunately I don't really see how to easily relate the dimensions of the Riemann Roch spaces associated to each curve.

I would really appreciate some help with this.

Thank you.

• What is your definition of the genus? By the way, for the exercise, if $(P) \sim (Q)$, then there is a rational function with one zero at $P$ and one pole at $Q$ and no other zeros and poles; so we get a map from the curve to $\mathbb{P}^1$ of degree one. – Akhil Mathew Jun 19 '11 at 16:38
• @Akhil Thank you. The definition of genus I'm using is the one Silverman gives in the section about the Riemann Roch Theorem in chapter II of his book. Namely, that the genus $g$ is the natural number such that for every divisor $D \in \text{Div}(C)$ $$\ell(D) - \ell(K_C - D) = \deg{D} - g + 1$$ where $K_C$ is a canonical divisor on the curve $C$. – Adrián Barquero Jun 19 '11 at 16:45

The notation $\text{genus}(C_1)$ doesn't even make sense unless you know that the genus is invariant under isomorphism; if it isn't, the genus must depend on information other than $C_1$ which you haven't provided.
In any case, the definition of genus you are given implies that it is unique, and since the various dimensions $\ell(D)$ are defined independently of any choices they are automatically invariant under isomorphism, so the definition you have been given already comes with a guarantee that $g$ is invariant under isomorphism. But if you want a "proof" anyway, then setting $D = 0$ gives $\ell(K_C) = g$, so it suffices to show that the canonical divisor is invariant under isomorphism (that's why it's called the canonical divisor!).
• Yes you're absolutely right that it would not make sense to talk about the genus of $C$ if it weren't invariant. In fact the result you quote that $\ell(K_C) = g$ is the corollary following the statement of the Riemann-Roch theorem in the book. Probably my confusion comes from the fact that there's no proof of this or of other important results in the book because the first two chapters are supposed to be just a review of some algebraic geometry basics that the reader should know. Then I'll get to proving that result about the canonical divisors. Thank you very much. – Adrián Barquero Jun 19 '11 at 18:05
• In fact I think $\ell(K_C) = g$ is a much more transparent definition of the genus and it's the one you should keep in mind for intuition. The point here is that $\mathcal{L}(K_C) = \Omega^1_{k(C)/k}$, so $g$ is really the complex dimension of the space of $1$-forms. $1$-forms are dual to $1$-cycles and the integral homology of a compact oriented surface of topological genus $g$ is $\mathbb{Z}^{2g}$. – Qiaochu Yuan Jun 19 '11 at 18:11
• $\phi^*(K_D) = K_C$ isn't true for a general separable map because there may be ramification. (For instance, consider a rational function on an elliptic curve, giving a map to $\mathbb{P}^1$: the pull-back of the canonical divisor on $\mathbb{P}^1$ (which has degree $-2$) won't be the one on $\mathbb{P}^1$ (which is everywhere holomorphic). – Akhil Mathew Jun 19 '11 at 19:53
• @Mariano: I can understand $C_1$ being used to denote, for example, a particular equation of a curve, but in the OP it only denotes a curve. So if you think the notation $\text{genus}(C_1)$ is well-defined, you already believe that the genus is a property of a curve rather than a presentation of a curve (hence invariant under isomorphism). – Qiaochu Yuan Jun 19 '11 at 20:38
The genus of a curve, viewed as the dimension of the space of holomorphic form $H^0(C,\Omega_C^1)$ is even invariant under birational morphisms. Indeed, if $f:X \to Y$ is a birational morphism between smooth complex projective varieties, then $f^*$ induces an isomorphism $H^0(Y,K_Y^{\otimes m}) \to H^0(X,K_X^{\otimes m})$ for each integer $m \geqslant 0$, where $K_X = \Lambda^{\dim X} T_X^*$ is the canonical bundle on $X$.