> * I'm sorry I didn't reply to your email at the time. I'm not sure that I received it. But I just checked, and I did get at least two messages on the topic that I didn't reply to, one from someone named Briggs and one from someone named Barlotti. If either of these is you, I apologize. I do try to answer blog-related email.

Neither is me, but it's no problem! You have often responded to my blog-related e-mails (my pessimistic Reddit comment that I didn't expect a reply was because it was probably my first time writing to you, and most bloggers without comments don't respond), and I have no proof that I sent this, so I might well have thought I did but left it sitting in draft, or something.

> * That article was #4 in a series. I did add a correction to article 3 the same day that 4 was published, and very soon afterward (same day, perhaps) I published a detailed followup. A few months later I added cross-links between all the articles. (See https://blog.plover.com/math/i-5.html )

I do see you mentioning in part 3 that there are many more additive-group automorphisms of ℝ, even continuous group automorphisms, than you had originally expected, and also mentioning (there and explicitly in part 5) that continuity is needed to say that there aren't even more. But I think that I probably did not make my point very clear, or else I am still misunderstanding.

I think (but can't say for sure—I refer to “the linked article”, because I missed that it was just a link to the math tag, so that I was just seeing whatever article was current at the time) that I was referring to https://blog.plover.com/math/i-4.html, which still seems to say:

> … the only automorphisms of the complex numbers are the identity function and the function a + bi → a - bi.

Here the context makes clear that "automorphism" means (at least) "field automorphism". It is this statement that I was disputing. You mention a streamlined argument that 1 is sent to 1 and −1 to −1, hence i to ±i; but that doesn't allow us to conclude—it only says that f(a + bi) equals f(a) ± f(b)i, and we are left to wonder what f is doing on ℝ. (The same issue crops up in the same way in part 3, which again seems more or less to conclude with the observation that f(i) = ±i.) Without continuity, we need not have that f restricts to the ‘standard’ embedding of ℝ in ℂ, and so have more than two field automorphisms of ℂ; but the construction of discontinuous automorphisms here is even scarier, using not just a Hamel basis of ℝ over ℚ but a transcendence basis (https://en.wikipedia.org/wiki/Transcendental_extension#Trans...) for ℂ over the algebraic closure of ℚ. A cute consequence that, to me, is proof enough that transcendence bases are a worthwhile concept: the field ℂ(X) of rational functions in one variable over ℂ admits a field embedding into ℂ; or, more precisely and to some people cuter, the algebraic closure of ℂ(X) is isomorphic to ℂ.

It is certainly true that, all possible fanciness aside, there are only two ℝ-algebra endomorphisms of ℂ, where the definition of an algebra map requires it to take multiplicative identity to multiplicative identity. In particular, we find that every ℝ-algebra endomorphism of ℂ is an automorphism, which sounds fancy when you say it with that many syllables.