I recently started writing up some material on finite presentation for the CRing project. There seems to be a folk “finitely presented” approach in mathematics: to prove something over a big, scary uncountable field like $\mathbb{C}$, one argues that the problem descends to some much smaller subobject, for instance a finitely generated subring of the complex numbers. It might be possible to prove using elementary methods the analog for such smaller subobjects, from which one can deduce the result for the big object.

One way to make these ideas precise is the characteristic p principle of Abraham Robinson, which I blogged about in the past when describing the model-theoretic approach to the Ax-Grothendieck theorem. Today, I want to describe a slightly different (choice-free!) argument in this vein that I learned from an article of Serre.

Theorem 1 Let ${F: \mathbb{C}^n \rightarrow \mathbb{C}^n}$ be a polynomial map with ${F \circ F = 1_{\mathbb{C}^n}}$. Then ${F}$ has a fixed point.

We can phrase this alternatively as follows. Let ${\sigma: \mathbb{C}[x_1, \dots, x_n] \rightarrow \mathbb{C}[x_1, \dots, x_n]}$ be a ${\mathbb{C}}$-involution. Then the map on the ${\mathrm{Spec}}$‘s has a fixed point (which is a closed point).

In fact, this result can be proved using directly Robinson’s principle (exercise!). The present argument, though, has more of an algebro-geometric feel to it, and it now appears in the CRing project — you can find it in the chapter currently marked “various.

Let us now prove this result, following Serre. It is clear that the presentation of ${\sigma}$ involves only a finite amount of data, so we can construct a finitely generated ${\mathbb{Z}}$-algebra ${R \subset \mathbb{C}}$ and an involution $\displaystyle \overline{\sigma}: R[x_1, \dots, x_n] \rightarrow R[x_1, \dots, x_n]$

such that ${\sigma}$ is obtained from ${\overline{\sigma}}$ by base-changing ${R \rightarrow \mathbb{C}}$. We can assume that ${\frac{1}{2} \in R}$ as well.

To see this explicitly, we simply need only add to ${R}$ the coefficients of the polynomials ${\sigma(x_1), \dots, \sigma(x_n)}$, and ${\frac{1}{2}}$, and consider the ${\mathbb{Z}}$-algebra they generate.

Suppose now the system of equations ${\sigma(x_1, \dots, x_n) - (x_1, \dots, x_n)}$ has no solution in ${\mathbb{C}^n}$. This is equivalent to stating that a finite system of polynomials (namely, the ${\sigma(x_i) - x_i}$) generate the unit ideal in ${\mathbb{C}[x_1, \dots, x_n]}$, so that there are polynomials ${P_i \in \mathbb{C}[x_1, \dots, x_n]}$ such that ${\sum P_i \left( \sigma(x_i) - x_i \right) = 1}$.

Let us now enlarge ${R}$ so that the coefficients of the ${P_i}$ lie in ${R}$. Since the coefficients of the ${\sigma(x_i)}$ are already in ${R}$, we find that the polynomials ${\sigma(x_i) - x_i}$ will generate the unit ideal in ${R[x_1, \dots, x_n]}$. If ${R'}$ is a homomorphic image of ${R}$, then this will be true in ${R'[x_1, \dots, x_n]}$.

Choose a maximal ideal ${\mathfrak{m} \subset R}$. Then ${R/\mathfrak{m}}$ is a finite field, and ${\sigma}$ becomes an involution $\displaystyle (R/\mathfrak{m})[x_1, \dots, x_n] \rightarrow (R/\mathfrak{m})[x_1, \dots, x_n].$

If we let ${\overline{k}}$ be the algebraic closure of ${R/\mathfrak{m}}$, then we have an involution $\displaystyle \widetilde{\sigma}: k[x_1, \dots, x_n] \rightarrow k[x_1, \dots, x_n].$

But the induced map by ${\widetilde{\sigma}}$ on ${k^n}$ has no fixed points. This follows because the ${\widetilde{\sigma(x_i)} - x_i}$ generate the unit ideal in ${k[x_1, \dots, x_n]}$ (because we can consider the images of the ${P_i}$ in ${k[x_1, \dots, x_n]}$). Moreover, ${\mathrm{char} k \neq 2}$ as ${\frac{1}{2} \in R}$, so ${2}$ is invertible in ${k}$ as well.

So from the initial fixed-point-free involution ${F}$ (or ${\sigma}$), we have induced a polynomial map ${k^n \rightarrow k^n}$ with no fixed points. We need only now prove:

Lemma 2 If ${k}$ is the algebraic closure of ${\mathbb{F}_p}$ for ${p \neq 2}$, then any involution ${F: k^n \rightarrow k^n}$ which is a polynomial map has a fixed point.

Proof: This is very simple. There is a finite field ${\mathbb{F}_q}$ in which the coefficients of ${F}$ all lie; thus ${F}$ induces a map $\displaystyle \mathbb{F}_q^n \rightarrow \mathbb{F}_q^n$

which is necessarily an involution. But an involution on a finite set of odd cardinality necessarily has a fixed point (or all orbits would be even).