Suppose that x, y K satisfy f(x) = f(y). Then f(x - y) = 0.

If x y, then x - y is nonzero, hence invertible. But then 1 = f((x - y)^-1(x - y)) = f((x - y)^-1) f(x - y) = f((x - y)^-1) 0 = 0, a contradiction.

Hence x = y.

This is a direct consequence of Part 1.

Put L = {x K | f(x) = x}.

Clearly, f(0) = 0 and f(1) = 1, so 0,1 L.

Suppose that x, y L. Then, as f is a morphism,

1. f(-x) = f(0 - x) = f(0) - f(x) = 0 - x = -x,
2. 1 = f(1) = f(xx^-1) = f(x) f(x^-1) = xf(x)^-1, so f(x)^-1 = x^-1,
3. f(xy) = f(x)f(y) = x y,
4. f(x + y) = f(x) + f(y) = x + y,

whence -x, x^-1, xy, x + y L.

This suffices to establish that L is a subfield of K.