Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
Try Bezout's theorem. –  Regenbogen Apr 1 '10 at 23:28
The book "Lectures on elliptic curves" by Cassels should give you some ideas for students interested in number theory. –  Regenbogen Apr 1 '10 at 23:28

11 Answers 11

up vote 41 down vote accepted

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.

  1. 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.

  2. 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.

  3. 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 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.

share|improve this answer
I don't think Emil Artin ever wrote a book called ALGEBRA, and I think Michael Artin's book ALGEBRA is too comparatively recent to be called a classic (but maybe that just makes me look old). Since you refer to geometric algebra ideas, could you mean Emil Artin's Geometric Algebra? No, I don't think topics in there would encourage someone to want to study algebraic geometry. Maybe you mean Michael Artin's book after all. –  KConrad Apr 2 '10 at 6:35
I don't think the ancient Greeks solved quadratic equations; perhaps you mean the Babylonians? –  Franz Lemmermeyer Apr 2 '10 at 12:35
KConrad's answer is great! –  Anonymous Apr 2 '10 at 15:31
@Franz: Solving quadratic equations appears in Diophantus' Arithmetica. –  Dror Speiser Apr 4 '10 at 12:03
@KConrad: These days, I think it's common to call Michael Artin's Algebra a classic. There also is some introductory material to algebraic geometry, so I assume he is talking about that. –  David Corwin Jul 12 '10 at 20:55

I saw Brian Conrad give an excellent one hour talk to undergraduates where he proved that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$ and $c(t) \in \mathbb{C}[t]$ such that $$a(t)^3 + b(t)^3 = c(t)^3.$$ He gave an elementary proof, then outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve.

share|improve this answer
Picard's Theorem says there are no nonconstant analytic functions that satisfy that equation. Corollary: No nonconstant polynomials can either. –  Matt Apr 2 '10 at 2:03
There are plenty of nonconstant entire functions solving $$a(t)^3 + b(t)^3=c(t)^3.$$ For your favourite non-constant entire function $a(t)$, set $b(t)=a(t)$ and $c(t)=2^{1/3}a(t)$. –  Robin Chapman Apr 2 '10 at 7:29
Perhaps the missing assumption is that $a,b,c$ are coprime. There is nice, quite elementary result, known as Mason's theorem, which implies that the equation $a^n+b^n=c^n$ (or even a more general one) doesn't have any solutions in coprime polynomials over arbitrary algebraic closed fields of characteristic zero (hence over arbitrary field of characteristic zero). You can see the details in the 11th paragraph of the Polynomial Excursions by Edward Barbeau. –  ifk Apr 2 '10 at 9:08
Thanks, ifk. I'll fix. –  David Speyer Apr 2 '10 at 11:10
@hilbertthm90: In "13 Lectures on Fermat Last Theorem" by Paolo Ribenboim (pp. 265-266) there is a proof (using Picard Theorem) of following result: If $f,g$ are entire functions, $p\neq 0$ is a polynomial, $\deg p \le n-3$ (he remarks that the result is still valid for $n-2$ here) satisfying equation $(f(z))^n+(g(z))^n=p(z)$ then $f,g,p$ are constant. An immediate consequence is that if $f(z),g(z),h(z)$ are entire functions, $h$ is nonvanishing and $(f(z))^n+(g(z))^n=(h(z))^n$ for $n\ge 3$ then $f=ah,g=bh$ for some $a,b\in \mathbb{C}$ such that $a^n+b^n=1$ –  ifk Apr 2 '10 at 21:52

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.

share|improve this answer
If you actually choose this topic, you may want the details of the computations I have skipped (although you can find them yourself). The only reference I know on this proof is the book "Geometria algebrica" by Marco Manetti, which you can find here: mat.uniroma1.it/people/manetti/dispense/dispense.html Unfortunately it is in italian, but the few parts you need should be understandable. You find the proof on page 221. –  Andrea Ferretti Apr 3 '10 at 20:47
Sono interessanti e ben scritti questi appunti di Manetti. Se conosce questo Professore, forse pottrebbe suggerirgli di tradurli in inglese e di pubblicarli. Comprerei volentieri un tal libro! –  Georges Elencwajg Apr 4 '10 at 12:54
Glielo suggerirò volentieri la prossima volta che lo incontro! –  Andrea Ferretti Apr 4 '10 at 13:35
the existence of a line argument is in Shafarevich, Basic algebraic geometry chapter one, and the discriminant quintic argument is detailed in Beauville's Complex algebraic surfaces, both in English. But I welcome the translation of the other notes as well. –  roy smith Nov 27 '10 at 4:49

This isn't a result so much as a perspective, but it is one of the main reasons I first got interested in algebraic geometry.

In basic algebraic number theory you learn that some extensions of the integers, like $\mathbb{Z}[i]$, have unique factorization, but others, like $\mathbb{Z}[\sqrt{-3}]$, don't. You are then told to take a number field $K$ and consider the integral closure $\mathcal{O}_K$ of $\mathbb{Z}$ within it. This ring has the important property that it is a Dedekind domain, which means its ideals have unique factorization even though its elements don't. What is one to make of this definition? It's technically useful, but what is its conceptual content?

To see the geometry behind these definitions, replace $\mathbb{Z}$ with $\mathbb{C}[t]$. Finite extensions of $\mathbb{C}[t]$, such as those of the form $\mathbb{C}[t, x]/f(t, x)$, can be identified with Riemann surfaces $f(t, x) = 0$ by looking at the maximal spectrum if and only if the hypotheses of the implicit function theorem are satisfied; in particular, the partial derivatives of $f$ cannot simultaneously vanish anywhere (no singularities). Algebro-geometrically, all of the local rings need to be DVRs.

It turns out that this condition is equivalent to $\mathbb{C}[t, x]/f(t, x)$ being a Dedekind domain! This is because the property of being a Dedekind domain is local: it holds for a Noetherian ring if and only if it holds for each local ring. This isn't true for the property of being either a PID or being a UFD, so one can think of Dedekind domains as objects which look locally like PIDs in the same way that manifolds are objects which look locally like $\mathbb{R}^n$.

And in the end one is left with an important geometric intuition: rings of integers that aren't integrally closed correspond to "arithmetic varieties" with "singularities," and when we take integral closures we are resolving those singularities. More generally, I think the algebro-geometric perspective on number fields sheds a lot of light on the subject; Neukirch brings up this analogy towards the end of the first chapter, if I recall correctly.

share|improve this answer
I should also mention, if you're unfamiliar with the number/function field analogy, that the analogy usually has an intermediate step where you replace Z with F_q[t]. This field is even more similar to Z than C[t]; for example, a version of the prime number theorem holds! –  Qiaochu Yuan Apr 2 '10 at 6:07
I agree this is a nice point of view, but ifk said the students have not all had commutative algebra. So how can this be a topic that would be accessible to all the students? –  KConrad Apr 2 '10 at 6:14

Emerton mentioned introducing elliptic curves, but you can also introduce the group law on the smooth locus of a nodal or cuspidal cubic curve. The usual equations are $y^2 = x^3 + x^2$ and $y^2=x^3$. They are like the group laws for elliptic curves (i.e., you draw lines and the intersecting points sum to zero), but structurally simpler (being a torus and an additive group, respectively). The equations for torsion points are a bit easier to solve than the corresponding formulas for any particular elliptic curve. You can even try counting points over various finite fields for the nodal curve.

You might also want to look at some of David Speyer's posts on the Secret Blogging Seminar on algebro-geometric proofs and interpretations of classical theorems. Here are a few:

  1. Menelaus's Theorem
  2. Poncelet's Porism
  3. Quadratic reciprocity in the function field setting
share|improve this answer

A topic that you might enjoy learning yourself, and which you can certainly present in one or two 90 minutes lectures, is the group law on an elliptic curve. This is treated in an elementary fashion in Miles Reid's book "Undergraduate algebraic geometry" (and probably in many other places).

The advantages are that it is concrete and specific, but also not at all obvious. Furthermore, verifying associativity is quite tricky from an elementary viewpoint. (Reid gives a pretty complete discussion, if I remember correctly. In your talk, you would probably not be able to, or want to, cover associativity completely, though, since the details are quite elaborate. On the other hand, you can write down an explicit elliptic curve, find three explicit points, and explictly add them via the group operation in the two different ways necessary for verifying the associativity; this is always pretty entertaining to watch --- but make sure that you do the computations first, since you surely won't be able to work them out at the board; they will be too messy!)

There is also a very strong connection with the theory of elliptic functions, but you probably wouldn't want to try to fit this into the same lecture. But if you want to learn it yourself, you will want to read about Weierstrass's elliptic functions.

Actually, one thing that comes out of the theory of elliptic functions is that the complex-valued points on the elliptic curve, as a topological group, are isomorphic to $S^1\times S^1$ (a product of two circle groups). So the algebraic description of the group law on the elliptic curve gives a very complicated, but very interesting, way of describing $S^1\times S^1$!

Why do I say "very interesting"? Well, the fact that it lives in the world of algebra gives it a dimension of richness that you can't obtain just by talking about $S^1\times S^1$. For example, there are beautiful connections to the theory of Diophantine equations.

As one example, if you had enough time, you could mention that if the elliptic curve is defined over $\mathbb Q$, then the set of $\mathbb Q$-valued points on the curve is closed under the group operation, and is (by a famous theorem of Mordell) a finitely generated abelian group. (Note: trying to develop tools to determine the precise structure of this group remains one of the central problems in modern number theory. To learn about this, try googling "Birch--Swinnerton-Dyer conjecture".) (Also note: number theorists refer to the set of $\mathbb Q$-valued points of an elliptic curve over $\mathbb Q$ as the Mordell--Weil group of the elliptic curve. If you google Mordell--Weil group, you will find a statement of Mordell's theorem mentioned above, and a lot more besides.)

share|improve this answer

I am not an algebraic geometer, but I have often covered the topic of finding rational points on the circle and certain cubic curves with undergraduates. This is elementary (find appropriate chords and tangents) but it opens the door to elliptic curves.

share|improve this answer

Bezout's theorem can be very entertaining. I saw a elementary proof (essentially using linear algebra and commutative algebra) of this as an undergrad.(Unfortunately I don't recall the proof or the reference, but here is a reference) And the statement was quite astonishing to me at that point ( like 2 circles intersect in 4 points ...).Also it introduces the idea of intersection multiplicity.

share|improve this answer
Actually ALL circles share a comlpex double point at infinity - write down the equations to see for yourself. This is why 3 points in the plane determine a circle: They already share two points, so if they share 3 more they must be the same circle! –  Steven Gubkin Apr 2 '10 at 12:25

If you're willing to not be completely careful with some of the details of projective geometry and skip over the worst of the computations, it is certainly possible to solve the simplest nontrivial Schubert problem (count lines meeting 4 given lines in (complex projective) 3-space) in about an hour.

You don't need any intersection theory beyond what a high school student could in theory figure out; the relevant equations are one degree 2 equation in 6 variables for the Grassmannian, and 4 linear equations for the Schubert varieties. You can derive those equations from the vanishing of determinants.

Things I like about this presentation: it answers a question which is elementary (at least if you ignore the "complex projective" part) but whose answer is not obvious, and it introduces the very important idea of a moduli space, in this case the Grassmannian, in a concrete way.

share|improve this answer
there is also the elementary argument by specialization. i.e. assume the 4 lines meet in pairs. then there are clearly two lines meeting all 4. the line joining the two common points of the two pairs, and the line joining the two points where one pair meets the plane of the other pair. Then one discusses the art of specializing a problem having a lower semicontinuous answer, to a case where the answer remains the same, but is easier to solve. –  roy smith Nov 27 '10 at 4:57

Although the theory is much cleaner working over algebraically closed fields, I think Sturm's Theorem (counting real roots of a polynomial) is a very nice result in real algebraic geometry which is at a level appropriate for undergraduates.

share|improve this answer
How would a lecture on Sturm's theorem convince students to take a course in algebraic geometry? That is what ifk wants to achieve as a result of his lecture(s). –  KConrad Apr 2 '10 at 6:16

There is an algebraic proof that a revolution torus intersect a bitangent plane into the reunion of two circles (Villarceau circles). I only have a reference in French: http://denisfeldmann.fr/PDF/cercles.pdf

The torus is easily seen as a degree $4$ algebraic surface. Its intersection with a plane must therefore be a curve of degree $4$. The point is then to prove that this curve is the union of two conics (using its multiple points), then that the conics are circles (using the circular points).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.