If you want to teach something intriguing, you should do something that introduces a new geometric idea while also involving algebra in an essential way. I recommend that you give an introduction to the projective plane, showing the other students that it is a natural extension of ordinary space which makes some geometric properties more uniform (such as intersection properties of curves), gives a fruitful new way to think about old topics (like asymptotes), and lets you do things that are impossible to conceive without it (reducing rational points mod $p$). There should be substantial interplay between algebra and geometry, but make sure to draw pictures to emphasize the geometric aspects.
In algebra, we can conceive of the quadratic formula in a uniform manner, but the ancient Greeks [Edit: Babylonians, not Greeks] couldn't do this because they didn't have the idea of negative numbers. So they had several quadratic formulas on account of not being able to write something as simple as $ax^2 + bx + c = 0$ at one stroke (for any signs on $a, b$, and $c$, with $a$ nonzero). Our extended skill at algebra lets us work with one case where the ancients had to take multiple cases. We can also say with complex numbers that any quadratic equation has two roots, allowing for a double root to count as one root with multiplicity two.
The thrust of what comes next is to extend the plane so that geometric properties become nicer in a similar way the algebra is becoming nicer when we use more general number systems.
Consider the intersection properties of lines in the plane. There is a dichotomy: usually two lines in the plane meet in one point, but some pairs of lines (the parallel ones) meet in no points. Let's see what this looks like under stereographic projection. Lines in the plane become circles through the north pole, but not including the north pole itself. It's natural to close up the image and take that whole circle as a substitute for the original line. So we can see that lines in the plane naturally close up into circles through the north pole. Under stereographic projection, the old dichotomy between parallel and non-parallel lines takes on a new appearance: a pair of non-parallel lines corresponds under stereographic projection to a pair of circles intersecting in two different points, one of which is the north pole, while a pair of parallel lines corresponds under stereographic projection to a pair of circles which are tangent at the north pole. It is natural to think of two tangent circles as having their point of tangency be an intersection point of multiplicity two, much like a quadratic polynomial can have a double root. So after stereographic projection we can "see" two points of intersection for any pair of lines. This geometric construction is something like the algebraic use of more general number systems to find roots to all quadratic equations. The moral to take from this example is that in a larger space, curves that used to not intersect may now intersect (or rather, their natural closures in the new space intersect) with a uniform count of the number of intersection points. If the students agree that enlarging number systems to create solutions to polynomial equations is good, they should agree that enlarging space to make intersection properties more uniform is good too. Another important feature is that the sphere, like the plane, is a homogeneous object: we can transform (rotate) the space to carry one point to any other point. On the sphere as a space in its own right, there is truly nothing special about the north pole.
An even better geometric extension of the plane is the projective plane, although at first it will feel unfamiliar and strange because you can't see it all at once.
You should introduce it in a uniform manner as points described with homogeneous coordinates $[x,y,z]$ where $x$, $y$, and $z$ are not all 0 and, say,
$$
[3,6,2] = [1,2,2/3] = [1/2,1,1/3] = [3/2,3,1] \text{ and } [0,5,0] = [0,1,0].
$$
Although it is impossible to see the whole projective plane at once, we can get glimpses of large parts of it using three different charts: $U_0$ is the points where $x \not= 0$, $U_1$ is the points where $y \not= 0$ and $U_2$ is the points where $z \not= 0$. These three charts together cover the projective plane. Any nonzero coordinate can be scaled to 1 and that fixes the other two homogeneous coordinates of the point, e.g., $[x,y,1] = [x',y',1]$ if and only if $x = x'$ and $y = y'$. This means we can identify each of $U_0$, $U_1$, and $U_2$ with the usual plane (e.g., identify $U_2$ with ${\mathbf R}^2$ by identifying $[x,y,1]$ with $(x,y)$). This means the projective plane locally looks like the plane, much like the sphere does, except we can't see all of it at the same time as we can with the sphere.
(In case you want to show students that the projective plane is a really natural model of something they have known in another context, think about nonzero ideals in ${\mathbf R}[x]$. Any ideal has a generator, but the polynomial generator is only defined up to a nonzero scaling factor. Usually we normalize the generator to be monic, but if we don't want to insist on a particular choice of generator then the right model for the generator is a point in projective space. In particular, for any nonzero ideal $(f(x))$ where $\deg f(x) \leq 2$, write $f(x) = ax^2 + bx + c$; the coefficients $a, b, c$ are only defined up to an overall scaling factor, so the point $[a,b,c]$ is one way to think about that ideal.)
Next introduce curves in the projective plane as solutions to homogeneous polynomial equations in $x$, $y$, and $z$ and explain what the algebraic process of homogenization and dehomogenization of polynomials is, e.g., it makes $y = 2x + 1$ into $y = 2x + z$ or $x^2 - y^2 = x+ 1$ into $x^2 - y^2 = xz + z^2$. In particular a line in the projective plane is the solution set to any equation $ax + by + cz = 0$ where the coefficients are not all 0.
Now let's look at what a point on a specific curve in the projective plane looks like in each of the three standard charts, carry out the same kind of calculus computation in each chart, and compare the results with each other. We will use the curve $C : x^2 + y^2 = z^2$ in the projective plane (not to be confused with a surface in 3-space given by the same equation) and the points $P = [3,4,5]$ and $Q = [1,0,1]$ which lie on $C$. How do $C$, $P$, and $Q$ appear in each of the charts $U_0$, $U_1$, and $U_2$?
a) In $U_0$, which is identified with the plane by $[x,y,z] \mapsto (y/x,z/x)$, $C$ becomes the hyperbola $z^2 - y^2 = 1$, $P$ becomes $(4/3,5/3)$, and $Q$ becomes $(0,1)$. Here we identify $U_0$ with the usual $yz$-plane. By calculus, the tangent line to $z^2 - y^2 = 1$ at the point $(4/3,5/3)$ is $z = (4/5)y + 3/5$ and the tangent line at $(0,1)$ is $z = 1$.
Note that we actually miss two points from $C$ when we look at the intersection of it with $U_0$: $[0,1,\pm 1]$.
b) In $U_1$, $C$ becomes the hyperbola $z^2 - x^2 = 1$ in the $xz$-plane, $P$ becomes the point $(3/4,5/4)$ with tangent line $z = (3/5)x + 4/5$, and $Q$ doesn't actually live in this chart (kind of like the north pole under stereographic projection not going to anything the in the plane). Here two points from $C$ are missing: $[1,0,\pm 1]$.
c) In $U_2$, $C$ becomes the circle $x^2 + y^2 = 1$, $P$ becomes $(3/5,4/5)$ with tangent line $y = (-3/4)x + 5/4$, and $Q$ becomes $(1,0)$ with tangent line $x = 1$. Every point from $C$ lies in $U_2$, so no points are missing here. We see the "complete" curve in this chart.
It is essential to draw three pictures here (of the $yz$-plane, $xz$-plane, and $xy$-plane) and mark $P$ and $Q$ in each (except you don't see $Q$ in the $xz$-plane).
Now comes the beautiful comparison step: in all three charts the homogenization of the tangent line at $P$ is exactly the same equation: $3x + 4y = 5z$. The tangent line at $Q$ in $U_0$ and $U_2$ homogenizes in both cases back to $x = z$. This suggests there should be an intrinsic concept of tangent line in the projective plane to the curve $C$ at the points $P$ and $Q$, and you can compute the tangent line by looking at any chart containing the relevant point of interest, doing calculus there, and then homogenizing back. The homogenization of your ordinary linear equation to a homogenuous linear equation will always be the same, and its solutions in the projective plane define the tangent line to the projective curve at that point.
As further evidence of the consistency of this new space and the geometry in it, look at the intersections of the two tangent lines at $P$ and $Q$: in $U_0$ -- the $yz$-plane -- the tangent lines meet in $(1/2,1)$ while in $U_2$ -- the $xy$-plane -- the tangent lines meet in $(1,1/2)$. These points both homogenize back to the same point $[2,1,2]$, which is the unique (!) point in the projective plane satisfying $3x + 4y = 5z$ and $x = z$.
Remember that $Q$ went missing in the chart $U_1$? Well, its tangent line did not go missing: the projective line $x = z$ in the projective plane meets the chart $U_1$ in the ordinary line $x = z$ of the $xz$-plane, which is an asymptote to the piece of $C$ we can see in $U_1$. This is really amazing: asymptotes to (algebraic) curves in the usual plane are "really" the tangent lines to missing points on the complete version of that curve in the projective plane. To see this from another point of view, move around $C$ clockwise in the chart $U_2$ (where it's a circle) and figure out the corresponding motion along the piece of $C$ in the chart $U_0$ (where it's a hyperbola): as you pass through the point $Q = (1,0)$ in $U_2$, what happens in the chart $U_0$ is that you jump off one branch of the hyperbola onto the other branch by skipping through an asymptote, sort of. (There is a second point on $C$ in $U_2$ that you don't see in $U_0$ -- the point $R = [-1,0,1]$ is $(-1,0)$ in $U_2$ -- and paying attention to that point may help here.)
The conic sections -- parabolas, hyperbolas, and ellipses -- which look quite different in ${\mathbf R}^2$, simplify in the projective plane because they all look like the same kind of curve (once you close them up): $y = x^2$ becomes $yz = x^2$, $xy = 1$ becomes $xy = z^2$, and $x^2 + y^2 = 1$ becomes $x^2 + y^2 = z^2$, which is the same as $x^2 = (z-y)(x+y) = z'y'$, where $z' = z-y$ and $y' = z+y$. I like to think about this as a fancy analogue of the Greek [Edit: Babylonian] use of many forms of the quadratic formula because they didn't have the right algebraic technique to realize there is one quadratic formula. Using the projective plane we see there is really one conic section.
You might want to show by examples the nicer intersection properties of lines in the projective plane: any two lines in the projective plane meet in exactly one point. This is just a glimpse of the fact that curves in the projective plane have nicer intersection properties than in the ordinary plane, but to get the correct theorem in that direction for curves other than lines, you need to (a) work over the complex numbers and (b) introduce an appropriate concept of intersection multiplicity for intersection points of curves, somewhat like the idea of tangent circles intersecting in a point of multiplicity two which I mentioned earlier. The relevant theorem here is Bezout's theorem, but to state it correctly is complicated precisely because it is technical to give a good definition of what the intersection multiplicity is for two curves meeting at a common point.
For the student who wants to be a number theorist, compare reduction mod $p$ in the usual plane and the projective plane. In the study of Diophantine equations (e.g., to show $y^2 = x^3 - 5$ has no integral solutions), it is very useful to reduce mod $p$, and there is a natural way to reduce a point in ${\mathbf Z}^2$ modulo $p$ However, there's no reasonable way to reduce all points in ${\mathbf Q}^2$ modulo $p$: when the rational numbers have denominator divisible by $p$, you can't make sense of them mod $p$: we can reduce $(-7/4,51/8)$ mod 5, for example, but not mod 2. In the projective plane, however, we can reduce rational points mod $p$ by the idea of choosing a set of primitive integral coordinates, where the homogeneous coordinates are relatively prime. For example, $[-7/4,51/8,1] = [-14,51,4]$ in ${\mathbf P}^2({\mathbf Q})$, and this can be reduced mod $p$ for any $p$ at all. For example, in ${\mathbf P}^2({\mathbf F}_2)$ it becomes $[0,1,0]$.
(There is another primitive set of homogeneous coordinates for the point, namely $[14,-51,-4]$, but that reduces mod $p$ to the same thing as before, so this reduction mod $p$ process is well-defined.) This suggests that the projective plane has better mapping properties than the usual plane, in some sense.
Best Answer
RRT There is a big difference in difficulty between the compact Riemann surface case and the projective curve case, for reasons already mentioned. Namely a projective curve comes equipped with a large supply of meromorphic functions, but the proof that they exist for a compact Riemann surface is a major step.
I suggest you decide what your goal is. E.g., do you want to make clear why the theorem is true, without giving all steps of verification, or do you want to show how some of the trivial steps are derived easily using modern sheaf techniques, or perhaps give complete derivations of some significant parts of the statement?
if you use sheaf theory, it is trivial to show that chi(D)-chi(O) = deg(D) for any divisor D, where chi is the holomorphic Euler characteristic: chi(D) = h^0(D)-h^1(D). To get full RRT from this one needs to compute chi(O) = 1-g, where g is the topological genus, and then to prove duality, that h^1(D) = h^0(K-D), where K is the canonical divisor.
In 4 pages of the notes on my web page, I take an argument from Bill Fulton to do one of these for plane curves, namely that chi(O) = 1-g, with some hand waving over the computation of topological Euler characteristics by deformation.
see pages 38-42: http://alpha.math.uga.edu/%7Eroy/rrt.pdf
If you just want to show why the result is true with some arguments omitted, I feel nothing beats Riemann’s own exposition. Riemann himself proved the theorem in clear natural stages: 1st the theorem in the special case of the canonical divisor, i.e. he proved that there are exactly g independent holomorphic differential forms on a compact Riemann surface of genus g. Then he proved that for each point p, there is one meromorphic differential with a double pole at p and zero residue, equivalently he proved the RRT for divisors of form K+2p, i.e. that h^0(K+2p) = g+1, so that in addition to the g holomorphic forms there is one with a double pole at p, (and necessarily zero residue). (Riemann also allows himself the simplification of assuming all points of the divisor considered are distinct.) Then, from the existence of these basic types of differentials, one deduces the converse of the residue theorem.
Even if you barely read German, you can get the idea of what is most important just from deciphering the headings of the first two paragraphs in Riemann's treatment of the theorem in his great paper "On Abelian Integrals"; (since differentials the things you integrate, he speaks of their integrals): "Integrale erster Gattung" (integrals of first kind, i.e. of holomorphic differentials); "Integrale zweiter Gattung" (integrals of second kind, i.e. of differentials with double pole at one point)....
With this information, one can deduce the RRT in two steps, as clearly explained in Griffiths and Harris’s book, (see in particular pages 233, 244-5). Namely for each effective divisor D, to compute the number of independent meromorphic functions with pole divisor dominated by D, is equivalent to computing the number of meromorphic differential forms that would occur as their differentials. The assumed facts give the number of meromorphic differentials with the right singularities, and then one has only to compute how many of those differentials are exact, which is a period computation.
As in Riemann’s original paper they give this calculation in terms of the periods of integrals, and as in Roch’s follow-up these periods are computed in terms of residues using Green’s theorem. All this is done nicely in Griffiths Harris, where they depend on the Kodaira vanishing theorem for the requisite existence of differential forms of first and second kinds. (They also discuss this deep theorem earlier in their book.)
The point is that the deduction of RRT from the known existence of the right number of differential forms of first and second kinds is very clear and elementary complex calculus. Hence it only remains to provide an elementary proof of these prerequisite facts about forms. Now if one restricts attention to plane curves, at least the holomorphic forms can be immediately written down in coordinates, as Riemann himself points out. He also says one can write them down in the meromorphic case as well, but does not do so. I believe this can indeed be done in an elementary way, as indicated in the book of Brieskorn and Knorrer, but I have not done it except in special cases.
In fact one can finesse the existence of the forms of second kind by the trick of Brill and Noether; i.e. one can use duality to rely exclusively on the existence of holomorphic forms for the theorem. This argument is explained in the books of Arbarello, Cornalba, Grifiths and Harris (appendix A, chapter 1), and in the book on algebraic curves by Griffiths. I have also written out this argument in the notes for my course in pdf form, at the link below: http://alpha.math.uga.edu/%7Eroy/8320.pdf
If you want a complete argument for compact Riemann surfaces, a nice treatment using sheaves is in the book Lectures on Riemann surfaces by Robert Gunning, including some nice Hilbert space arguments to deduce the finiteness of cohomology groups. I like the old book, but a more current version of his notes are posted on his web site at: https://web.math.princeton.edu/~gunning/
Finally David Mumford recommended reading the argument by George Kempf, in Crelle’s Journal, and reproduced in his book Algebraic Varieties, using sheaf theory heavily for the algebraic case. I found this rather terse, but aspire to understanding it. Here is a link to it: http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002194015&physid=PHYS_0046
Please forgive the somewhat rushed answer, I hope some of the references are useful. I apologize if my correction of an error causes a bump, but I hope it does not.