Is $\mathbb Z/p\mathbb Z$ a subfield of every finite field? I translate this from a German book: "For every finite field 
$K$ there exists a prime number $p$ such that $\mathbb Z/p\mathbb Z$ is a subfield of $K$"
But how is this possible? For example the field $K = \{0,1\}$ contains integers but $\mathbb Z/p\mathbb Z$ contains equivalence classes. To be a subfield it would also have to be a subset of $K$.
 A: Every field contains a subfield isomorphic to either $\mathbb{Z}/p\mathbb{Z}$ (for some prime $p$) or $\mathbb{Q}$. 
This follows from basic laws of additive exponents. 
Let $\mathbb{K}$ be a field with multiplicative identity $1_\mathbb{K}$. Then consider the map $f:\mathbb{Z}\to\mathbb{K}$ defined by $n \mapsto n1_\mathbb{K}=\underbrace{1_\mathbb{K}+1_\mathbb{K}+\cdots+1_\mathbb{K}}_{n-\mbox{times}}$.
Basic laws of (additive) exponents tell us that $f(n+m)=(n+m)1_\mathbb{K} = n1_\mathbb{K}+m1_\mathbb{K}=f(n)+f(m)$ and $f(nm)=(nm)1_\mathbb{K} = (n1_\mathbb{K})(m1_\mathbb{K})=f(n)f(m)$ so $f$ is a ring homomorphism. Thus by the first isomorphism theorem $\mathbb{Z}/\mathrm{Ker}(f) \cong \mathrm{Im}(f)$.
The kernel of $f$ is either $\{0\}$ (this means $\mathbb{K}$ has characteristic $0$) and so $\mathbb{Z} \cong \mathrm{Im}(f)$. Thus $\mathbb{K}$ has a subring isomorphic to the integers and so (since it's a field) must have (multiplicative) inverses for these elements and so has a sub*field* isomorphic to $\mathbb{Q}$.
Otherwise the kenerl of $f$ is a nonzero ideal of $\mathbb{Z}$ and so has the form $p\mathbb{Z}$ where $p$ must be prime (otherwise $\mathbb{K}$ would a contain a subring which has zero divisors). Thus $\mathbb{K}$ has a subring isomorphic to $\mathbb{Z}/p\mathbb{Z}$ (for some prime $p$). 
If we begin with the assumption that $\mathbb{K}$ is finite, this rules out characteristic zero (there's not enough room to fit the infinitely large copies of $\mathbb{Z}$ or $\mathbb{Q}$). So finite fields must contain a subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$ for some prime $p$.
By the way, these subfields (the subfield generated by $1_\mathbb{K}$) are called prime subfields.
For your particular example, $K=\{0,1\} \cong \mathbb{Z}/2\mathbb{Z}$ (this field is of characteristic 2 and is equal to its prime subfield).
A: $\mathbb Z / 2\mathbb Z$ contains $K=\{0,1\}$. It means $\mathbb Z / 2\mathbb Z$ and $K=\{0,1\}$ are isomorphism.
