18 1. THE CONSTRUCTION OF REAL AND COMPLEX NUMBERS Exercise 1.7.1. Under this assumption, (i) show that f(1) = 1, and (ii) show that f is an injection. (iii) Must such an f be a surjection? Consider f satisfying only the additive property above in the case when F = Q. We have f(x) = f(x + 0) = f(x) + f(0), so f(0) = 0. Next we have, for n N, f(n) = nf(1) and f(m/n) = (m/n)f(1) for any positive rational number m/n. While we are not assuming the multiplicative property here, to avoid the trivial, we assume that there exists an x Q, so that f(x) = 0. From this, it follows that f(1) = 0. Also, for any positive rational number r, we have f(−r) = −f(r). Thus, f(r) = rf(1) for all r Q. Let us see what happens for R. We assume that f(1) = 0. We have f(r) = rf(1) for all r Q. However, as we point out after the definition of Hamel basis in Appendix B, things go completely awry unless we im- pose further properties. So, assume that f preserves multiplication, that is f(xy) = f(x)f(y) for all x, y R. Then, it follows from the above exercise that f(1) = 1, and f(a−1) = f(a)−1 if a = 0. The next thing to note here is that if a R and a = 0, then a2 0 and f(a2) = (f(a))2, so f(a2) 0. Since all positive real numbers have unique positive square roots, we can conclude that if c 0, then f(c) 0. Thus, if a b, then f(a) f(b) since b a 0. Now take any real number c. If c Q, then f(c) = c. If c Q and f(c) = c, then there are two possibilities. If c f(c), choose a rational number r so that c r f(c). Then f(c) f(r) = r, which is a contradiction. If f(c) c, we run into the same problem. So we conclude that f is the identity. Definition 1.7.2. Let F be a field. An automorphism of F is a bijection, f : F −→ F , such that (a) f(x + y) = f(x) + f(y) for all x, y F , (b) f(xy) = f(x)f(y) for all x, y F . Exercise 1.7.3. If F is a field, show that the automorphisms of F form a group under composition of functions. This group is called the automor- phism group of the field F and is denoted by Aut (F ). Theorem 1.7.4. The groups Aut (Q) and Aut (R) consist only of the identity map. Exercise 1.7.5. Find a field F such that Aut (F ) = {1}. Exercise 1.7.6. (i) Let F be a field and let f be an element of Aut (F ). Define Hf = {x F | f(x) = x}. Show that Hf is a subfield of F . (ii) Suppose that F is a field and that Q is a subfield of F . If f Aut (F ), show that Q is a subfield of Hf.
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