If a is invertible with inverse b, then ab = 1. Applying f gives f(ab) = f(1) = 1 and so f(a) * f(b) = 1, i.e., f(a) is invertible and its inverse is f(b) = f(a^-1).

Clear from the fact that f is a bijection between R and R'.

We prove that f^-1 respects multiplication and leave other details to the reader.

Suppose a', b' R'. As f is surjective, there are a, b R such that f(a) = a' and f(b) = b'. The fact that f is a morphism implies

f(ab) = f(a) * f(b).

Applying f^-1 to both sides gives

ab = f^-1(f(a) * f(b)).

Substituting f^-1(a') for a and f^-1(b') for b in the left-hand side, and a' for f(a) and b' for f(b) in the right-hand side, we find

f^-1(a')f^-1(b') = f^-1(a' * b'), as required.

This is direct from the conditions given before the theorem.

The first statement (that Ker f is an additive subgroup of R) follows directly from the conditions given before the theorem.

Let r R and a Ker f. Then

f(r * a) = f(r) * f(a) = f(r) * 0 = 0,

whence r * a Ker f.

If is injective and a belongs to Ker f, then f(a) = f(0) = 0, and injectivity implies a = 0.

Conversely, if Ker f = {0}, and f(a) = f(b), then f(a - b) = 0 so that a - b = 0 and a = b.