I have seen a Theorem:
every field contains exactly one prime subfield $K_0$.
$K_0$ has to contain 1 and all its multiples: $n \cdot 1 = 1 + \ldots + 1$
Is this because it has to contain 0 and 1 by definition, and then it must also holds that $1 + 1$ belongs to K, and $(1 + 1) + 1$ too because of two-arguments $+$ operation of its abelian group?
so we can prove it by induction?