4

Are there any nontrivial field endomorphisms of $\mathbb{H}$ that are not the identity map when restricted to $\mathbb{R}$? I realize that the only nontrivial endomorphism on $\mathbb{R}$ is the identity map, so if the desired function $f$ exists, it must be that $f(\mathbb{R}) \ne \mathbb{R}$. Are there any such mappings?

Tyler
  • 2,223
  • There are "wild" automorphisms of $\mathbb{C}$ . These may help you find one. https://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers – badjohn Jul 12 '18 at 07:55
  • Thinking a little more. It should be easy to take one of these wild automorphisms of $\mathbb{C}$ and extend it to $\mathbb{H}$. – badjohn Jul 12 '18 at 10:25
  • 1
    Unless there is a weakness in the argument I used here the answer is No! Wild automorphisms of $\Bbb{C}$ don't really help here because $\Bbb{R}$ is the center of $\Bbb{H}$. Once we can deduce that an endomorphism maps $\Bbb{R}$ to itself, the rest is standard: identity is the only endomorphism of $\Bbb{R}$. – Jyrki Lahtonen Jul 12 '18 at 10:26
  • Thinking a bit more... Possibly this question should be closed as a dupe of the one I linked to. But, this feels like a bad time to use my dupehammer. Blowing one's own trumpet that way should be done very sparingly on this site. – Jyrki Lahtonen Jul 12 '18 at 10:30
  • I'll defer to that. – badjohn Jul 12 '18 at 14:23
  • 1
    @JyrkiLahtonen I see the connection with that other question; your answer was very helpful. I left a comment on your answer there about something that seemed like a typo, but otherwise I think I understand now. I'm still fairly new to the site; what is the appropriate way to resolve my question here? – Tyler Jul 12 '18 at 16:32
  • I'm not sure what's best, and because I'm personally involved I should exercise a bit of care. I will ask other users with a lot of experience. For now, I will post a stub answer. You may post one also (if you want feedback on your understanding of the steps). – Jyrki Lahtonen Jul 12 '18 at 17:35
  • @JyrkiLahtonen I'm trying to understand why $\mathbb{C}$ doesn't have the same problem - i.e. why the wild automorphisms exist there but not in $\mathbb{H}$. Obviously, the center of $\mathbb{C}$ is $\mathbb{C}$ itself, not $\mathbb{R}$, but re-reading your answer, I'm not sure what relevance the center has. $\mathbb{C}$ is certainly finite-dimensional over $\mathbb{R}$, so why do we not run into the same problem here? – Tyler Jul 12 '18 at 18:50
  • 1
    The difference is that in $\Bbb{C}$ everything commutes with everything. My proof for the fact $f(\Bbb{R})\subseteq\Bbb{R}$ relied on $\Bbb{R}$ being the intersection of two copies of $\Bbb{C}$ inside $\Bbb{H}$.(each such copy is its own centralizer). We don't have that when $f$ is an endomorphism of $\Bbb{C}$. Consequently $f$ need not map reals to reals, and that's where the wild homomorphisms come from. Heck, an endomorphism of $\Bbb{C}$ need not be surjective. – Jyrki Lahtonen Jul 12 '18 at 20:45

1 Answers1

4

There are no such automorphisms of $\Bbb{H}$. The argument from an earlier answer of mine implies that any ring homomorphism $f:\Bbb{H}\to\Bbb{H}$

  • is surjective (this is non-trivial),
  • is injective (a division ring has no non-trivial ideals, so this is trivial), and
  • hence maps the center $\Bbb{R}$ to itself.

You also already knew that the identity is the only endomorphism of $\Bbb{R}$, so the question is settled.

amWhy
  • 209,954
Jyrki Lahtonen
  • 133,153