6

The field $\mathbb{C}$ of complex numbers has an involution, and the same is true of the field of algebraic numbers (the algebraic closure of $\mathbb{Q}$ as a subfield of $\mathbb{C}$) and of the field of complex ruler-and-compass constructible complex numbers; in all three cases the involution is the complex conjugation. Any quadratic extension of a field $F$ has an involution that fixes every element of $F$.

Question. $~$If $F$ is a field of characteristic different from $2$ whose every element is a square, does it follow that it has an involution? In particular, does every algebraically closed field of characteristic not $2$ have an involution?

I suspect that the answers are NO and NO$\ldots$ but do not really know, of course, otherwise I would not be asking.

Are there any theorems that say if a field is such and such, then it has an involution, or, perhaps, that under certain conditions there are no involutions? That is, are there theorems that supply fields with involutions (which are not among the examples given above), or fields without an involution? I would appreciate any hints or references. I hit on this question during my current work; it created quite a crater which I want to somehow fill in.

chizhek
  • 3,059
  • 4
    Am I misunderstanding? Doesn't every field have involutions? For example, $x\mapsto -x$? And of course $x\mapsto x$? – MPW Jul 28 '15 at 14:27
  • 7
    An involution is an automorphism of order $2$. The map $x\mapsto-x$ is an automorphism of order $2$ of the underlying additive group of a field (of characteristic not $2$, remember), but is not an automorphism of the field. The identity mapping is an automorphism of order $1$, the only one. – chizhek Jul 28 '15 at 15:51

1 Answers1

4

No. A counter-example is given by the algebraic closure $\overline{\mathbb{F}}_p$ of a finite field $\mathbb{F}_p$ of characteristic $p>2$.

Indeed, since any automorphism $\sigma$ of $\overline{\mathbb{F}}_p$ fixes all elements of the prime field $\mathbb{F}_p$, then this $\sigma$ is an element of the absolute Galois group of $\mathbb{F}_p$, which is isomorphic to the inverse limit of all $\mathbb{Z}/n\mathbb{Z}$ and contains no non-trivial element of finite order. In particular, $\sigma$ cannot be an involution.

I am not aware of conditions guaranteeing the existence of an involution for a given field.

Community
  • 1
Pierre-Guy Plamondon
  • 11,138
  • 2
    So this is it for fields of characteristic $p>2$. How about fields of characteristic $0$? The algebraic closure $\overline{\mathbb{Q}}$ does have an involution. – chizhek Jul 29 '15 at 10:18
  • @chizhek Good question. I haven't found a counter-example in characteristic zero. – Pierre-Guy Plamondon Jul 29 '15 at 10:50
  • An idea. Let $F$ be a field of characteristic $0$ every element of which is a square. There is an $i\in F$ such that $i^2=-1$. Let $E$ be a maximal subfield of $F$ that does not contain $i$. If we can prove that $F=E(i)\ \ldots$ – chizhek Jul 29 '15 at 15:14
  • @chizhek Good point. According to this MO question, you can always find such a subfield $E$ if $F$ is algebraically closed. There still remains the case when $F$ is not algebraically closed... – Pierre-Guy Plamondon Jul 29 '15 at 15:58
  • The MO question asks for a maximal proper subfield of a field (in this particular question, an algebraically closed one), which may exist or not. But the field $E$ in my previous comment always exists: the union of a nonempty chain of subfields of the field $F$ that exclude $i$ is a subfield excluding $i$. The question here is, is the subfield $E$ "almost at the top", meaning $E(i)=F$. – chizhek Jul 29 '15 at 18:05
  • Ah, now I see: I said "Let $E$ be a maximal subfield of $F$ that does not contain $i$." This was sloppily formulated. I should say "Let $E$ be a subfield of $F$ maximal in the family of all subfields of $F$ that do not contain $i$." Or perhaps, "$\ldots$ maximal with respect to the property of excluding $i$." – chizhek Jul 29 '15 at 18:18
  • @chizhek Yes, that is how I understood your comment. But unless I understand the MO answers wrongly, they state that any algebraically closed field of characteristic zero admits a maximal proper subfield, and that this maximal subfield never contains $i$; moreover, adding $i$ yields the whole algebraically closed field. – Pierre-Guy Plamondon Jul 29 '15 at 19:30
  • Yes, it is so. But the field $F$ in my question (now of characteristic $0$) has the weaker property that every element is a square. Considering maximal $i$-excluding subfields guarantees their existence. I also tried another approach: maximal formally real subfields, which likewise exist. In either case I was able to get a few steps ahead, but then run into a wall. – chizhek Jul 30 '15 at 08:48
  • The complex numbers $\mathbb{C}$ contain uncountably many involutions, exactly one of which is continuous (with the usual metric topology on $\mathbb{C}$) and it has $\mathbb{R}$ as its fixed field. This is one definition of the real numbers, in addition to it being the unique topologically complete, linearly ordered field. (This sort-of spells the continuous involution idea out https://mathoverflow.net/questions/110428/nondegenerate-involutions-on-the-complex-numbers-always-just-conjugation) – Jeffrey Rolland Jul 16 '21 at 03:06
  • What is the state-of-the-art in characteristic 2? Sorry if this is an ignorant question. I guess every $[\mathbb{F}{2^r}: \mathbb{F}{2^{r-1}}] = 2$; what about the existence of involutions of $\overline{\mathbb{F}}_2$? – Jeffrey Rolland Jul 16 '21 at 03:07
  • I have one more dumb question: $\mathbb{C}$ is not $\overline{\mathbb{Q}}$ but its metric completion with respect to the usual modulus metric/valuation, $[\overline{\mathbb{Q}}]$; are there involutions on the metric/valuation completions of the various $\overline{\mathbb{F}}_p$'s with respect to their usual valuations, $[\overline{\mathbb{F}}_p]$, for both $p=2$ and $p>2$? Many thanks in advance. – Jeffrey Rolland Jul 16 '21 at 17:27
  • (If one could prove any involution on $[\overline{\mathbb{F}}_p]$ would have to send all and only algebraic elements to algebraic elements -- maybe that's obvious? -- then any involution on $[\overline{\mathbb{F}}_p]$ would restrict to an involution on $\overline{\mathbb{F}}_p$; if every involution on $\overline{\mathbb{F}}_p$ extends to an involution on $[\overline{\mathbb{F}}_p]$ -- again, maybe this is obvious -- then it does so uniquely, so "No involutions on $\overline{\mathbb{F}}_p$" implies "No involutions on $[\overline{\mathbb{F}}_p]$".) – Jeffrey Rolland Jul 16 '21 at 17:50