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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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    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.2011-06-19
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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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    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.2011-06-19
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    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 $\mathbb{Z}^{2g}$.2011-06-19
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    Yes I agree that this should be a better way of defining the genus. In fact after I asked the question I found this [handout](http://www.google.co.cr/url?sa=t&source=web&cd=1&ved=0CBwQFjAA&url=http%3A%2F%2Ftartarus.org%2Fgareth%2Fmaths%2Fnotes%2Fii%2FCompAlgCurves_h.pdf&rct=j&q=The%20genus%20of%20an%20algebraic%20curve%20is%20invariant%20under%20isomorphisms&ei=izz-TbWqI-G_0AHj8dmTAw&usg=AFQjCNF22S2WVqOfum9BVfI0WEvfmTEHQg&cad=rja) about complex algebraic curves and the definition you propose for the genus and the point you make can be found there at the bottom of page 7 in the last paragraph.2011-06-19
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    $\phi^*(K_D) = K_C$ 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).2011-06-19
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    (In fact, for smooth curves over an algebraically closed field, $\phi^*(K_D)= K_C$ is a perfect characterization of \'etaleness, unless I am being dense here.)2011-06-19
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    @Akhil: you're right, I misspoke.2011-06-19
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    Why wouldn't $\mathrm{genus}(C_1)$ be well-defined? It will depend on $C_1$ which is all there is to know to know... $C_1$ :) It would not make sense to write, say, $\mathrm{genus}([C_1])$ if we were to write $[C_1]$ for the isomorphism class of $C_1$, though.2011-06-19
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    @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).2011-06-19
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    I probably don't understand what you mean by a curve. The curves I know (say, a concrete scheme of the appropriate dimension and with the correct properties over a field) have each sufficient information to compute their genus. I cannot see why you think that if I give you a curve concretely as the zero set of a family of polynomials one cannot compute its genus.2011-06-19
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    @Mariano: that's not what I mean. I mean that if one is using the notation $\text{genus}(C_1)$ for a curve $C_1$ then one already believes that "genus" is a property of a curve rather than a property of a presentation of a curve (which differs based on different presentations), so it doesn't make sense to then go and ask whether genus is invariant under isomorphism.2011-06-19
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    @Qiaochu: ah. you see a "curve" as something different from a "presentation of a curve". I don't.2011-06-19
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    @Mariano: maybe I'm not being clear about what I mean by "property of a curve." Here is a simple example. The quantity "number of generators" is a property of a presentation of a group. It is not a property of a group, since the same group may be presented with more than one cardinality of set of generators. More relevant to this discussion: "degree" is a property of an embedding of a curve, not a property of a curve. But the definition of genus Adrian is working with is not like these properties: it is already by definition a property of a curve.2011-06-19
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    @Mariano: maybe I'm also not being clear about what I'm responding to. The OP uses the notation $\text{genus}(C_1)$ but then asks for a proof that genus is invariant under isomorphism, which means the OP did not believe (when asking this question) that genus is a property of a curve. That means that the notation $\text{genus}(C_1)$ was not (from the OP's perspective) even well-defined for the OP when the OP was asking this question, and I was pointing out this inconsistency.2011-06-19
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    @Qiaochu: I hinestly do not follow you :) There are lots of properties of curves which are not invariant under isomorphism (for example, that one of the points in the underlying topological space is the set of real numbers---to give a stupid example)2011-06-20
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    @Mariano: that isn't a property of a curve, but of the set that the curve maps to in a particular concretization of the category of varieties. I really did not think I was using the word "property" in a non-standard sense here... for me a property of an object in a category $C$ is necessarily invariant under isomorphism by _definition_, or else it's a property of something more complicated than an object of $C$ (such as its image under some functor $C \to D$).2011-06-20
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    @Mariano: I guess you were really being serious about not distinguishing between a curve and a presentation of a curve. Huh. Okay.2011-06-20
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    @Qiaochu: I find your point of view quite exotic, in fact :)2011-06-20
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    @Mariano: I find _your_ point of view quite exotic! When someone asks you what a curve is, do you say "a variety of dimension $1$" or do you say "a collection of equations that define a variety of dimension $1$"?2011-06-20
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    For me a curve is something concrete, say "an integral scheme of dimension 1, proper over a field, with regular local rings" (as in Hartshorne's chapter on curves) (but in other context it could be an algebraic set of dimension $1$ in affine space) Such a thing has points, concrete points (one of which can be the set of real numbers!) A property of a curve is anything (a number, a boolean, ...) that can be produced out of the data that determines the curve (the underlying topological space, the structure sheaf, the map to $\operatorname{Spec}k$)...2011-06-20
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    ... Some of those are invariant under 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 $X$.