# Rigidity of the category of fields

Let's call a category rigid if every self-equivalence is isomorphic to the identity. For example, $\mathsf{Set}$, $\mathsf{Grp}$, $\mathsf{Ab}$, $\mathsf{CRing}$ (MO/106838), $\mathsf{Top}$ (SE/450193) are rigid, but $\mathsf{Mon}$, $\mathsf{Ring}$ and $\mathsf{Pos}$ are not (the automorphism class group has order $2$ in these examples). See (MO/56887) for some ideas how to approach the rigidity of $\mathsf{Sch}$. In each case the idea is that a category is rigid if every object can be defined in a categorical way, which is a quite interesting property. The paper "The automorphism class group of the category of rings" by Clark, Bergman (link) explains this in detail.

Question. Is the category of fields $\mathsf{Fld}$ rigid?

Here are some notions from field theory with categorical definitions (thus preserved by self-equivalences):

• The category of field extensions of $K$ is the comma category $K / \mathsf{Fld}$.

• Call $L/K$ locally finite if $\hom_K(L,L')$ is finite for all $L'/K$.

• $L/K$ is algebraic iff it is a directed colimit of locally finite extensions.
• $L/K$ is transcendent iff it is not algebraic.
• A field extension of $K$ is isomorphic to $K(x)$ iff it is transcendent and maps to every other trancendent extension (Lüroth's Theorem).
• For a field $K$, we have $\mathrm{PGL}_2(K) \cong \mathrm{Aut}_K(K(x))$ given by mapping $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \cdot K^*$ to $x \mapsto \dfrac{ax+b}{cx+d}$.

I've learned from MO/106838 that $\mathrm{PGL}_2(K)$ and $\mathrm{PGL}_2(L)$ are only isomorphic when $K \cong L$. Is there a direct proof for this? Does someone know at least a more modern reference? Is there perhaps even a reconstruction procedure of a field $K$ from $\mathrm{PGL}_2(K)$?

Thus, if $F : \mathsf{Fld} \to \mathsf{Fld}$ is an equivalence, then $F(K(x)) \cong F(K)(x)$ for every field $K$, inducing a group isomorphism $\mathrm{PGL}_2(K) \cong \mathrm{PGL}_2(F(K))$, which shows the existence of some isomorphism $F(K) \cong K$.

I hope that this is correct so far. But this is almost what we want. Namely, we have to find isomorphisms $F(K) \cong K$ which are natural in $K$. The methods from $\mathsf{CRing}$ (classification of coring structures on $\mathbb{Z}[x]$) don't apply here.

Some further categorical definitions which might be helpful (or interesting in their own right):

• $K$ is a prime field iff every morphism $K' \to K$ is an isomorphism
• $K \cong \mathbb{F}_p$ iff $\mathrm{PGL_2}(K)$ is finite and of order $p(p^2-1)$. Hence, $K \cong \mathbb{Q}$ iff $K$ is a prime field and not isomorphic to some $\mathbb{F}_p$. Hence, $\mathrm{char}(K)=p$ iff there is some $\mathbb{F}_p \to K$ and $\mathrm{char}(K)=0$ iff there is some $\mathbb{Q} \to K$.
• $K$ is algebraically closed if every algebraic extension $K \to L$ is an isomorphism. $L/K$ is an algebraic closure iff it is algebraic and $L$ is algebraically closed.
• $L/K$ is normal if for some (every) algebraic closure $\overline{K}/K$ the group $\mathrm{Aut}_K(L)$ acts transitively on $\hom_K(L,\overline{K})$.
• $L/K$ is purely inseparable iff $|\hom_K(L,\overline{K})|=1$ iff $K \to L$ is an epimorphism.
• $L/K$ is separable iff $L/K$ is algebraic and for all factorizations $K \to L' \to L$ with $L' \to L$ purely inseparable, $L' \to L$ is an isomorphism.
• $L/K$ is finite iff $L/K$ is algebraic and $\hom_K(L,-)$ preserves directed colimits (i.e. $L/K$ is a finitely presentable object of $K/\mathsf{Fld}$).
• An algebraic extension is simple iff it has only finitely many intermediate extensions.
• The degree $[L:K]$ of a finite extension $L/K$ is characterized in two steps. By the degree formula, it suffices to treat two cases: 1. $L/K$ is separable. Then $[L:K] := |\hom_K(L,\overline{K})|$. 2. $L/K$ is purely inseparable. Let $p=\mathrm{char}(K)>0$. Then one can prove $[L:K]=p^n$, where $n$ is the longest length of a chain of intermediate extensions.
• If we have a language $L$ with no relational symbols, and $T$ is some theory in the language $L$, can we axiomatize $T$ without using $\exists$ provided the category of all models of $T$ is rigid? – Camilo Arosemena-Serrato Jan 10 '14 at 20:07
• I think we can show that the characteristic zero subcategory is rigid, using the fact that every field of characteristic zero can be embedded in a rigid field. – Hurkyl Jan 11 '14 at 0:55
• @Hurkyl: I didn't know this result, thanks. But I doubt that it will help. The embedding isn't canonical, so naturality is again a problem. Meanwhile, I have ideas how to attack rigidity of $\mathsf{Fld}_p$ (for $p>0$), using a categorical characterization of the Frobenius. The case $p=0$ will be much harder. – Martin Brandenburg Jan 11 '14 at 9:01
• Hrm. I thought I saw a way to use the rigidity of one field to force $F$ to preserve field embeddings into it (I was thinking in terms of the skeleton), but I'm not seeing it anymore this morning. I suppose I was mistaken. – Hurkyl Jan 11 '14 at 12:11

Just of list of further properties for the characteristic zero case:

• A field is real closed iff it is not algebraically closed itself but of finite degree in its algebraic closure.

• A field is formally real iff it has a morphism into a real closed field.

• A real closure of $K$ is an extension $K\to L$ that is algebraic and $L$ is real closed. Since any real closed field has a unique ordering and for any ordering on $K$ there is a unique real closure w.r.t to that ordering, the possible orderings on $K$ are in bijection with the real closures of $K$.

• An ordered field is euclidean if every positive element has a square root. If $K\to L$ is a real closure, then the euclidean closure of $K$ is the colimit of all intermediate fields $K\subseteq M\subseteq L$ whose degree is a power of two. (in other words: orderings in $K$ are also in bijection to euclidean closures)

• A field is called pythagorean if the sum of two squares is again a square. The pythagorean closure of $K$ is the intersection of all its euclidean closures (within a fixed algebraic closure say).

In characteristic $p>0$ Martin and I have worked out the Center of $Fld_d$: It is isomorphic to $\mathbb{N}$ and generated by the Frobenius. In particular the Frobenius is uniquely determined.

Proof: Let $F: id_{Fld_p} \to id_{Fld_p}$ be a natural transformation and let $f(X)\in\mathbb{F}_p(X)$ be the image of $X$ under $F$.

Then by naturality $F_K(T)=f(T)$ for all transcendent elements $T\in K$. By looking at $K=\overline{\mathbb{F}_p}(X)$,$T_1=X$ and $T_2=X+c$ for some $c\in\overline{\mathbb{F}_p}$ one obtains $f(X+c)=F_K(X+c)=F_K(X)+F_K(c) = f(X)+F_K(c)$. By considering the possible zeros of a denominator this implies that $f(X)\in\mathbb{F}_p[X]$. Now consider $K=\mathbb{F}_p(X,Y)$, $T_1=X$, $T_2=Y$, $T_3=X+Y$, $T_4=XY$ and obtain $f(X+Y)=f(X)+f(Y)$ as well as $f(XY)=f(X)f(Y)$. Since these are equations in $\mathbb{F}_p[X,Y]$, they imply $f(0)=0$ and $f(1)=1$. It follows from this thread that $f(X)=X^{p^k}$ for some $k\in\mathbb{N}$. Therefore $F(x)=x^{p^k}$ for all transcendent $x$ of all fields $K$. Now consider $K(y)$, an algebraic element $a\in K$ and the two transcendent elements $y$ and $y+a$. We have the equations $F(y)=y^{p^k}$ and $F(y+a)=(y+a)^{p^k}=y^{p^k}+a^{p^k}$. Therefore $F(a)=a^{p^k}$ for all algebraic elements of $K$ too. QED.

• Are the bijections in points 3 and 4 natural in any sense? – user153312 Feb 4 '15 at 14:19
• Yes. [and some more letters so I can post this comment...] – Johannes Hahn Feb 4 '15 at 14:42
• Could you please elaborate a little? I just want to write these facts down in detail so I can look into them when I learn field theory. – user153312 Feb 4 '15 at 14:47
• "When I learn field theory" sounds like some introduction to algebra course, i.e. the field theory you'll learn would mostly consist of some basic definitions that are necessary to learn about Galois theory. Ordered fields do not usually show up in such a course. When you start learning about ordered fields, then the two points in my post become very easy exercise. – Johannes Hahn Feb 4 '15 at 14:52
• Understood, thanks. Do you have a particular introductory textbook you'd recommend? – user153312 Feb 4 '15 at 15:24