# Genus of a Curve - Riemann-Roch

I am currently working on the theory of elliptic curves and trying to understand what the genus a curve is.

Therefore I'm trying to understand the theory behind the Riemann-Roch Theorem. Hence the concept of Devisors.
But already the definition of a devisor makes me struggel.

"Definition 4.100: Let $C/K$ be a curve. The divisor group $Div_C$ of $C$ is the free abelian group over the places of $K(C)/K$. An element $D\in Div_C$ is called a divisor. It is given by $$D = \sum_{\mathfrak{p}_i\in\sum_{K(C)/K}}n_i\mathfrak{p}_i,$$ where $n_i\in\mathbb{Z}$ and $n_i=0$ for almost all $i$." ("Handbook of Elliptic and Hyperelliptic Cryptography" by Cohen and Frey)

$\mathfrak{p}$ is a place = the equivalence class of a valuation $v$ of $K(C)$
$\sum_{K(C)/K}$ is the set of places of $K(C)/K$
$K(C) = Quot(K[x_1,x_2]/f(x_1,x_2))$.

"Definition 4.102 Let $C/K$ be a curve and $f\in K(C)^*$. The divisor $div(f)$ of $f$ is given by $$div:K(C)\to Div_C$$ $$f\mapsto div(f) = \sum_{\mathfrak{p}_i \in \sum_{K(C)/K}}v_{\mathfrak{p}_i}(f)\mathfrak{p}_i$$" ("Handbook of Elliptic and Hyperelliptic Cryptography" by Cohen and Frey)

What does this really mean? I am looking at the definitions but I get no feeling for this theory.
I would be really happy if someone could explain a bit the concept of the divisors, the genus and finally the resulting Riemann Roch Theorem.
Or maybe someone knows a good source to work with, to understand this theory.

"Theorem 11.15 (Riemann Roch) Given and algebraic curve C, there exists an integer $g$ (called the genus of $C$) such that $$l(D)-l(\mathscr{K}-D) = deg(D)-g+1$$ for all divisors $D$." (Washington)
with $l(D) = dim\mathscr{L}(D)$, where $\mathscr{L}(D) = \{fct. f|div(f)+D\geq 0\}\cup\{0\}$

I'm working mainly with the books "Handbook of Elliptic and Hyperelliptic Cryptography", "The Arithmetic of Elliptic Curves" by Silverman and "Elliptic Curves Number Theory and Cryptography" by Washington.

• These definitions will seem abstract without some basic background in algebraic geometry, e.g. Shafarevich's Basic Algebraic Geometry I. A divisor on a curve is a finite formal sum of points, and the $\mathrm{div}$ map sends a global rational function on the curve to the formal sum of its zeros and poles (counted with appropriate sign and multiplicity). Riemann-Roch takes some more work, and you should have some idea of what a linear system is before trying to understand it. Nov 14, 2016 at 20:29
• I think you should look also at the Abel-Jacobi map (before the Riemann-Roch theorem) Nov 14, 2016 at 20:42

As has already been pointed out, it might be worthwhile learning a bit of algebraic geometry first but I'll try to answer this without going too deep. For most of this I will follow Silverman, as I think elliptic curves make divisors a lot more approachable than abitrary varieties or schemes.

Recall some basics of topology, every compact orientable 2-manifold looks like (is homeomorphic to) an $n$-torus, i.e. either a sphere or a doughnut shape with $n$ holes. This $n$ happens to be equal to the genus of this surface. How do we connect this to elliptic curves though?

Over $\mathbb{C}$, the solution set of an elliptic curve actually looks like a torus topologically (this is called the uniformisation theorem), as we quotient $\mathbb{C}$ by some lattice. Since this is simply a torus with 1 hole, the genus of an elliptic curve is $1$. We can also do this for hyperelliptic curves (although the genus will be different).

On to divisors: Every place of the funtion field of an elliptic curve corresponds to a point on this curve, so we can just think of a divisor as a formal sum of points. The tiny catch is that this sum needs to be finite but that is perfectly fine.

For example, if $E: y^2=x^3+1$ over $\mathbb{C}$, then $Q=(2,3)$ and $R=(0,1)$ are points on $E$. We then have divisors

$D_1 = 2[R], D_2=-2[R]+[Q], D_3=[R]+[Q]$, and these form an abelian group under the obvious addition so $D_1+D_2=D_3$. (Forget about the group law on elliptic curves if you know about it). We also have the concept of the degree of a divisor, which is just the sum of the $n_i$. In the above the degrees of $D_1,D_2$ and $D_3$ are $2,-1$ and $1$ respectively.

Now consider a nonzero function $f$ in the function field of the curve. We say $v_P(f)=n$ if $f$ has a zero of order $n$ at $P$ (note $n$ can be negative if it has a pole instead. Checking for all poles and zeroes of this function, we can construct a divisor out of $f$, which is called $div(f)$.

For example, let $f=x-2$. Then $f=0$ if $x=2$ which occurs for the points $Q=(2,3)$ and $Q'=(2,-3)$ and we can easily see that these are simple zeroes. This means $v_Q(f)=v_{Q'}(f)=1$.

We can also see that $f$ does not have a pole at any point in the affine plane, however, elliptic curves really live in projective space and it turns out that $f$ has a pole of order $2$ at $\infty$. So $v_{\infty}(f)=-2$.

Having found all the zeroes and poles, we now see that the divisor of $f$ is $div(f)=[Q]+[Q']-2[\infty]$. Note that the degree of this is zero and this is always the case for divisors of functions.

For the Riemann Roch space $\mathcal{L}(D)$, we want to find all functions whose divisor is at least the negative of another divisor. For example, if we take $D=[Q]+[Q']$, then we are saying that we are only allowing functions which have at most simple poles at $Q$ and $Q'$ and nowhere else. It is an easy check that this forms a complex vector space.

Adjusting our idea above, $\dfrac{1}{x-2}$ lives in this space (as well as any scale multiple by the vector space structure), but no others do. Hence, we have $1$ basis element so $l(D) \geq 1$. We can also check that $\dfrac{x}{x-2}$ lives in here and that is it so $l(D)=2$. It is useful to note that since all nonconstant functions have poles, then if the degree of $D$ is negative then $l(D)=0$.

Now the Riemann Roch theorem can either be viewed as the definition for the genus, or if you already know this, it can be used to find $l(D)$, up to some correction term.

Now $\mathcal{K}$ is what is known as the canonical divisor and comes from a differential form which I won't attempt to explain here, but has degree $2g-2$.

So in the above, $\mathcal{K}-D$ has degree $-2$ which is negative hence $l(\mathcal{K}-D)=0$ and we can see that the Riemann Roch formula holds.

• You meant "over $\mathbb{C}$" not "over $\mathbb{Q}$" ? And "since all nonconstant functions have poles" : Liouville's theorem for compact Riemann surfaces Nov 14, 2016 at 22:04
• @user1952009 I'm not sure where you think I should(n't) have written $\mathbb{C}$? And if a nonzero function has no poles (ie is holomorphic) then by Liouville's, it is constant as you say, so I'm sorry but I'm not sure what you're getting at. Nov 14, 2016 at 22:12
• If the OP isn't sure of what the genus is, then a reference for holomorphic $\implies$ constant is required. And you wrote an elliptic curve over $\mathbb{Q}$, which is weird (we are talking of Riemann surfaces, and the isomorphism between complex elliptic curve and complex torus) Nov 14, 2016 at 22:17
• Ah apologies; thank you for the reference. I'm much more arithmetically minded and it makes more sense to me to consider (closed) points than to treat it geometrically, but I am happy to concede this point if OP prefers a more geometric intuition Nov 14, 2016 at 22:18
• Also, from the Riemann surface perspective, the $\infty$ point of the elliptic curve $E$ is the point that make it compact (so we have to take it in account prior to applying Riemann-Roch). Nov 14, 2016 at 22:22