Abstract Generality, Simplicity, Forgetting, and Discovery

Abstract

The paper contrasts two ways of generalizing and gives examples: probably most people think of examples like generalizing Cartesian coordinate geometry to differential manifolds. One kind of structure is replaced by another more complicated but more flexible kind. Call this articulating generalization as it articulates some general assumptions behind an earlier concept. On the other hand, by unifying generalization, I mean simply dropping some assumptions from an earlier concept or theorem. Hilbert, Noether, and Grothendieck were all known for highly non-trivial unifying generalizations.

Appendix: Naive Algebraic Geometry over the Integers

Let $y^2-x^3+x=0$ define a “curve” of integer points:

$$C_{\mathbb {Z}}\ =\ \{\langle x,y \rangle \in \mathbb {Z}\times \mathbb {Z}\ |\ y^2-x^3+x=0\}.$$

Efficient use of prime factorization, given below, shows $C_{\mathbb {Z}}$ has exactly three points:

$$\begin{aligned} C_{\mathbb {Z}}\ =\ \{\langle -1,0 \rangle , \langle 0,0 \rangle , \langle 1,0 \rangle \}. \end{aligned}$$

All these points have $y=0$ so the polynomials y and 0 give the same regular function on $C_{\mathbb {Z}}$ by criterion VALUES. Yet their difference y is obviously not divisible by the defining polynomial $y^2-x^3+x$ so they are not the same by criterion FACTORS. To study $C_{\mathbb {Z}}$ by tools of algebraic geometry, we must restore the role of FACTORS.

There are two ways to do this: 1) find new kinds of points for the “curve” $C_{\mathbb {Z}}$ so that the polynomials y and 0 do not agree at all these points or 2) give up the idea that a function is determined by its values at points. Scheme theory actually does both.

To keep the argument grounded in some mathematics, here is a proof that $C_{\mathbb {Z}}$ has just the three integer points.

Theorem 6.1

The curve $C_{\mathbb {Z}}$ has just three integer points: $\langle -1,0 \rangle$, $\langle 0,0 \rangle$, $\langle 1,0\rangle$.

Proof

Setting $y=0$ gives three trivial solutions to $y^2=0=x^3-x$, namely:

$$\begin{aligned} x=0,\ y=0\quad \text {or}\quad x=1,\ y=0\quad \text {or}\quad x=-1,\ y=0. \end{aligned}$$

This follows from factoring $x^3-x$ as $x\cdot (x-1)\cdot (x+1)$.

There are no solutions with $y\ne 0$. To see this, suppose $x^3-x$ is a non-zero square. Then $x\ne 0$ and $x\ne \pm 1$. And notice a prime factor of x cannot also be a prime factor of $x^2-1$. So, since their product $x^3-x$ is a square, both x and $x^2-1$ must be squares. (This is where the proof uses unique prime factorization of non-zero integers.) But then the successive integers $x^2-1$ and $x^2$ would both be squares. And this contradicts $x\ne \pm 1$ since the only successive integer squares are 0, 1.