TITLE: Interesting results in algebraic geometry accessible to 3rd year undergraduates
QUESTION [55 upvotes]: On another thread I asked how I could encourage my final year undergraduate colleagues to take an algebraic geometry or complex analysis courses during their graduate studies.
Willie Wong proposed me following idea - to show them some interesting results in this fields with relatively simply proofs and some consequences in other fields.
Thus by 'interesting' result in algebraic geometry I here mean the result which may convince 3rd year undergraduate student to study algebraic geometry.
In fact I'm supposed to give some talk during the seminar dedicated to final year undergraduates, and I can propose my own topic, so I thought that it could work.
But my problem is that I'm just wanna-be student of algebraic geometry and I don't have enough insight and knowledge to find a topic which 'could work'.
Also I doubt whether it is possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field.
So in short my first question is as above:

Is it possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field?

To be more precise - the talk should take a one or two meetings, 90 minutes each one.
The audience will be, as I said 3rd year undergraduate students, all of them after two semester course in algebra, some of them after one or two semesters in commutative algebra. All taking the course in one complex variable, and after one semester introductory course in differential geometry.
Some of them may be interested in number theory.
The second, related question is as follows:

If You think that the answer to the previuos question is positive, please try to give an example of idea/theorem/result which would be accessible in such time for such audience, and which You find interesting enough to make them consider possibility of studying algebraic geometry. Try to think about the results which shows connections of AG with some other fields of mathematics.

REPLY [28 votes]: If you are willing to cheat over some details, you could present a "proof" of the statement that every smooth cubic surface in $\mathbb{CP}^3$ contains exactly 27 lines. The fact that lines on such a surface can be counted, and that the result turns out not depend on the cubic seemed to me like magic before I learned algebraic geometry.
I think this topic is good for a number of reasons. The statement is elementary and suprising. The proof is fairly elementary as well (granting some things), but is far from trivial and shows many interesting ideas typical of algebraic geometry, like the use of parameter spaces. Moreover, it is not so elementary that you will be able to give a complete proof in one lecture, so interested students will want to know more. Finally the most delicate concepts from an algebraic geometry point of view are those of smoothness and dimension, but you can do with the analog concept in differential geometry, thereby providing a link with something they already know.
I have written all the steps just to be sure, and I don't think there is anything too complicated for your audience. Of course, depending on the time, you will want to skip over some details and present them as black boxes.
The proof
First, you can define smoothness in terms of differential geometry, and make the claim that an algebraic variety is smooth in a point if and only if the Jacobian of the defining equations has maximal rank. Remark that one implication always holds in the differentiable case, but there may be smooth manifolds defined by equations which are singular. Polynomials are rigid enough to prevent this. You could actually prove this fact for hypersurfaces, which is the case you are interested in. So, students will be able to understand what a smooth subic surface is.
The first step of the proof is the fact that every cubic surface, smooth or not, contains lines. This is the least elementary part, and you may want to skip this until the end of the talk.
Once you have a line $l$ consider planes $H$ containing $l$. The intersection of $H$ with the cubic is a plane cubic. Factoring the polynomial, it must be the union of $l$ with a smooth conic, or three lines. Moreover you can show that in the second case the three lines are distinct and do not meet in one point, otherwise the cubic would not be smooth there. This is completely elementary: you just prove that if the three lines do not form a triangle, all partial derivatives of the defining equation for your cubic surface vanish at a point.
The third step is to give a bijection between the set of planes containing $l$ and $\mathbb{P}^1$. A simple (but a bit long) computation tells you when the residual conic in the plane is smooth. Namely the vanishing of the determinant of the residual conic is an equation of degree $5$ on this $\mathbb{P}^1$. One can show that this equation has distinct roots, again because the cubic is smooth.
We conclude that for a given line $l$ there are exactly $5$ planes through $l$ on which the cubic decomposes as the union of three lines. Said differently, every line meets exactly $10$ other lines.
Finally, this implies the total number of lines is $27$. Indeed take any of these planes containing $3$ lines, call them $r$, $s$ and $t$. Any other line on the surface meets the plane, hence it meets one between $r$, $s$ and $t$. There are no triple of lines meeting in a point, because such a point would be singular. Hence each of $r$, $s$ and $t$ meets $8$ other lines, giving a total of $3 \times 8 + 3 =27$ lines.
The black box
It remains the black box that you can find at least one line on the surface. Here you will have to be sketchy, but hopefully this will wet your students appetite to see more about algebraic geometry.
First, you define the Grassmannian as a set, and make the claim that it is an algebraic variety. This is particularly easy for the Grassmannian of lines in $\mathbb{P}^3 = \mathbb{P}(V)$, since it is defined by the unique quadratic equation $\alpha \wedge \alpha = 0$ inside $\mathbb{P}(\bigwedge^2 V)$. Similarly you show that the parameter space for cubic surfaces is again an algebraic variety itself, namely a projective space $\mathbb{P}^{19}$.
Then you define the incidence variety of couples $(l, X)$ of a line and a cubic such that $l \subset X$; this is a subvariety of the product. The argument now uses the theorem on the dimension of the fibres of an algebraic morphism. Of course you cannot prove it, but it is rather plausible. Make the remark that such a neat statement can possibly hold only because polynomials are rigid enough. For the notion of dimension, you can use the dimension as complex manifolds, or half the real dimension.
Show that the set of cubics containing a given line is itself a projective space $\mathbb{P}^{15}$. Since the Grassmannian has complex dimension $4$ (this is plausible, being a hypersurface in $\mathbb{P}^5$), the incidence variety has dimension $19$, like the parameter space for cubics. By the theorem on the dimension of the fibres, to show that the projection is surjective you only need to exhibit a cubic which has a fibre of dimension $0$, that is, a cubic with a finite number of lines.
This is easy to do explicitly. Note that it is easier to do with a singular cubic, so to obtain a result about smooth cubics it will be easier to work with a parameter space containing singular ones.