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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.
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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 Jan 10 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 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 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 at 12:11

1 Answer 1

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.

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