2 thoughts on “A two-liner

  1. \Bbb{C} is such a ring. Take the non-trivial automorphism of \Bbb{Q}[\sqrt{2}]/\Bbb{Q} and extend it to an automorphism \Bbb{C}/\Bbb{Q}. The restriction of the latter automorphism to \Bbb{R} gives a non-trivial injection.

    Notice that the image of \Bbb{R} by this automorphism cannot be \Bbb{R}, since there are no non-trivial automorphisms of \Bbb{R}.

    Liked by 1 person

Leave a comment