Let L be the subfield generated by 0 and 1. For every positive integer m the element
belongs to L, and therefore also the element
Put
A = {m 
N |
m > 0, m · 1 = 0}.
We
distinguish two cases:
A is the empty set.
A is not empty.
In this case the map Z -> L
that sends m to m · 1 is an injective morphism. It is easy to see
that this map extends to an injective morphism
Q -> L,
m/n -> (m · 1)/(n · 1).
This map identifies Q with L.
A contains a smallest positive element p. Since 0 and 1
are distinct, p > 1. If p were not prime, then there
exist positive integers b, c < p such that
bc = p. It follows that
(b ·
1)(c · 1) = (bc) · 1 = p · 1
= 0
so that at least one of b · 1, c
· 1 equals 0, contradicting the minimality of p. But
then the obvious map Z/pZ -> L is
injective and maps isomorphically onto L.