Genus of a Curve - Riemann-Roch

Question

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.

Answer 1

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.