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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.

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    @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 $C$.2011-06-19

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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!).

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    ... Some of those are invariant $u$nder isomorphism, some are not.2011-06-20
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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 $X$.