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Wed, 13 Sep 2006
Automorphisms of the complex numbers
Robert C. Helling points out that there is a much simpler proof that this is the case. Suppose that f is an automorphism, and that x^{2} = y. Then f(x^{2}) = (f(x))^{2} = f(y), so that if x is a square root of y, then f(x) is a square root of f(y). As I pointed out, f(1) = 1. Since -1 is a square root of 1, f(-1) must be a square root of 1, and so it must be -1. (It can't be 1, since automorphisms may not map two different arguments to the same value.) Since i is a square root of -1, f(i) must also be a square root of -1. So f(i) must be either ±i, and the theorem is proved. This is a nice example of why I am not a mathematician. When I want to find the automorphisms of C, my first idea is to explicitly write down the general automorphism and then start bashing away on the algebra. This sort of mathematical pig-slaughtering gets the pig cut up all right, but mathematicians are not interested in slaughtering pigs. By which I mean that the approach gets the result I want, usually, but not new or mathematically interesting results. In computer programming, the pig-slaughtering approach often works really well. Most programs are oversubtle, and can be easily improved by doing the necessary tasks in the simplest and most straightforward possible way, rather than in whatever baroque way the original programmer dreamed up. [ Previous articles in this series: Part 1 Part 2 Part 3 Followup article: Part 5 ] [Other articles in category /math] permanent link |