I read about so called "wild" automorphisms of the field of complex numbers (i.e. not the identity nor the complex conjugation). I suppose they must be rather weird and I wonder whether someone could explain in the simplest possible way (please) how I could imagine such wild automorphisms.

E.g. I suppose they are completely discontinuous. E.g. are the real rational numbers fixed or any other set of complex numbers? Can such an automorphism be pictured in a model?

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    $\begingroup$ There is no formula for such automorphisms. Their existence relies on Zorn's lemma, so it's not going to be constructive. If you know some field theory, then you can use Zorn's lemma to extend any isomorphism between two subfields to an automorphism of $\mathbf C$, usually in many ways, but even if the isomorphism between the subfields is concrete the extension to an automorphism of $\mathbf C$ won't be explicit. $\endgroup$
    – KCd
    Jun 5, 2013 at 14:37

2 Answers 2


Given any field automorphism of $\mathbb C$, the rational numbers are fixed. In fact, any number that is explicitly definable in $\mathbb C$ (in the first order language of fields) is fixed. (Actually, this means that we can only ensure that the rationals are fixed, I expand on this below.)

Any construction of a wild automorphism uses the axiom of choice. See here for a related open problem. In fact, there is a model of set theory first considered by Solovay (in this model the axiom of choice fails, but the model satisfies the axiom of "dependent choice", which suffices for classical analysis) where all sets of reals are Lebesgue measurable and have the property of Baire, and there the only automorphisms are the identity and complex conjugation.

Wild automorphisms are indeed far from continuous. Since choice is used explicitly in their construction, I am not sure there is an easy way to "imagine" them, though the example described below is in principle not too elaborate, given these caveats.

The first explicit construction in print seems to be in a paper by Kestelman,

H. Kestelman. Automorphisms of the field of complex numbers, Proceedings of the London Mathematical Society (2), 53, (1951), 1-12.

His paper however traces the first proof as being "implicitly" given by Steinitz, using a transcendence basis, call it $T$, of $\mathbb C$ (called $Z$ in the paper) over $\mathbb Q$ (called $R$ in the paper), so $\mathbb C$ is algebraic over $\mathbb Q(T)$. (Note that this is where choice is used, in verifying the existence of $T$ via, for example, Zorn's lemma.)

The point is that any such basis contains two points $x_0, x_1$ with $x_0\notin\{x_1,\bar x_1\}$. One can then consider any permutation $\pi$ of $T$ mapping $x_0$ to $x_1$, and there is a unique extension of $\pi$ to a field automorphism of $\mathbb Q(T)$, which then can be lifted to an automorphism of $\mathbb C$. Pages 4, 5 in the linked paper gives a few more details. The outline itself was pointed out by Rado.

After this is done, the paper discusses how very weak regularity conditions on an automorphism (continuity at a point, for example), trivialize it.

Let me close with some remarks. In particular, I want to expand on the remark on fixed points in the first paragraph.

The argument above indicates we can produce an automorphism by starting with a permutation of $T$, which gives rise to an automorphism of $\mathbb Q(T)$, and then lift this to an automorphism of $\mathbb C$. Note that different permutations of $T$ give rise to different automorphisms, that $|T|=\mathfrak c=2^{\aleph_0}$, and that there are $2^\mathfrak c$ permutations of $T$. This means that there are at least $2^\mathfrak c$ wild automorphisms. On the other hand, there are only $\mathfrak c^\mathfrak c=2^{\mathfrak c}$ functions from $\mathbb C$ to itself, regardless of whether they are field automorphisms or not. This means that there are precisely $2^{\mathfrak c}$ (wild) field automorphisms of $\mathbb C$.

The next thing to note is that there is some leeway here. We do not need to start with $T$. We could just as well take any subfield $\mathbb F$ of $\mathbb C$, take a transcendence basis over $\mathbb F$, and repeat the argument above. In fact, we see this way that given any automorphism of $\mathbb F$, there is a field automorphism of $\mathbb C$ that extends it. This is explained in more detail in the paper linked by kahen in a comment below,

Paul B. Yale. Automorphisms of the complex numbers, Math. Mag. 39 (1966), 135-141. (Lester R. Ford Award, 1967.)

From basic field theory we know that for any irrational algebraic $\alpha$ we can take $\mathbb F$ to be the smallest subfield of $\mathbb C$ containing all roots of the minimal polynomial of $\alpha$ over $\mathbb Q$, and that there are automorphisms of $\mathbb F$ that move $\alpha$. Since any such automorphism can be extended to one of $\mathbb C$, this shows that no irrational algebraic number is fixed by all automorphisms of $\mathbb C$.

Similarly, if $\alpha$ and $\beta$ are transcendental and algebraically independent, then there is a transcendence basis $T$ with $\alpha,\beta\in T$, and there is an automorphism of $\mathbb Q(T)$ that maps $\alpha$ to $\beta$. Again, this extends to an automorphism of $\mathbb C$, so no transcendental number is fixed by all automorphisms of $\mathbb C$.

It follows that only the rational numbers are fixed by all automorphisms. On the other hand, again from basic field theory, we have that if $\alpha$ is algebraic, then any automorphism must map $\alpha$ to one of its conjugates, that is, to a root of the minimal polynomial of $\alpha$ over $\mathbb Q$. This means that there are only finitely many possible values the image of $\alpha$ may take.

Finally, a technical remark that I had on a comment, but probably deserves better visibility: To construct Solovay's model referred to above requires an inaccessible cardinal. On the other hand, as shown by Shelah, no additional consistency strength is required to show that there are models of set theory without choice, where all sets of reals have the property of Baire. Now, if a field automorphism of $\mathbb C$ is Lebesgue measurable, then it is trivial (the identity, or complex conjugation). Same if it is Baire measurable. In any model of set theory where all sets of reals have the property of Baire, all functions $f:\mathbb C\to \mathbb C$ are Baire measurable. It follows that in these models, the only automorphisms of $\mathbb C$ are trivial.

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    $\begingroup$ A more readable exposition is probably this one by Paul B. Yale, 1966(?) $\endgroup$
    – kahen
    Jun 5, 2013 at 14:47
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    $\begingroup$ If all sets are Baire then there are no wild automorphisms, and that doesn't require the consistency of an inaccessible cardinal. $\endgroup$
    – Asaf Karagila
    Jun 5, 2013 at 18:27
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    $\begingroup$ @Gerard: I have edited the answer to address this issue. $\endgroup$ Jun 9, 2013 at 2:19
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    $\begingroup$ @Gerard: Is it obvious that $\pi$ and $e$ can be exchanged? They are not known to be algebraically independent. $\endgroup$ Feb 13, 2018 at 16:18
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    $\begingroup$ “[...] which then can be lifted to an automorphism of $\mathbb{C}$.” Here it is crucial that $\mathbb{C}$ is algebraically closed. Indeed, the same conclusion does not even hold for automorphisms of $\mathbb{R}$: no matter what you do with transcendence bases and the like, the fact remains that every ring automorphism of $\mathbb{R}$ is the identity! (The non-negative elements of $\mathbb{R}$ are precisely the ones having a real square root, so any ring automorphism of $\mathbb{R}$ is automatically monotone.) $\endgroup$ Feb 22, 2018 at 20:21

The complex numbers can be characterized algebraically as the unique algebraically closed field of transcendence degree continuum over $\mathbb{Q}$. I think the key to understanding automorphisms lies in thinking of $\mathbb{C}$ in this way and forgetting any other structure like topology or embedding of $\mathbb{R}$.

There are also lots of "good" subfields of $\mathbb{C}$, like, say, algebraic extensions of $\mathbb{Q}$, and any automorphism of a subfield can be extended to an automorphism of $\mathbb{C}$ (this requires AC, of course). Moreover, in fact there are lots of endomorphisms that are not automorphisms, and lots of subfields of $\mathbb{C}$ that are isomorphic to $\mathbb{C}$.

All of them are quite "bad" analytically - in particular, not Lebesgue measurable (because even any measurable group homomorphism $\mathbb{C} \to \mathbb{C}$ is automatically continuous).

  • $\begingroup$ How can you have a proper field endomorphism? Wouldn't that imply that the field is isomorphic to a proper subfield of itself? $\endgroup$ Jan 26, 2014 at 6:17
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    $\begingroup$ @AviSteiner: It does imply that, and for fields of infinite transcendence degree it shouldn't be surprising. For instance, $\mathbb{Q}(x_1,x_2,\dots) \simeq \mathbb{Q}(x_2,x_3,\dots)$ is an example of isomorphism of a field with its subfield. Here the same happens, except that the transcendence degree is continuum. $\endgroup$ Jan 26, 2014 at 7:47
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    $\begingroup$ @AviSteiner: Actually, even for finite transcendence degree there's nothing wrong with a field being isomorphic to its proper subfield. For instance, $\mathbb{Q}(x) \simeq \mathbb{Q}(x^2)$ by an isomorphism that maps, guess what, $x$ to $x^2$. $\endgroup$ Jan 26, 2014 at 7:59

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