Equivalent Definitions of Prime Subfield I found two definitions for a prime subfield $K$ of a field $F$. 
1.  Wolfram $-$ $K$ is the subfield of $F$ generated by the multiplicative identity $1$ of $F$. 
2.  ProofWiki $-$ $K$ is the intersection of all the subfields, say $\{ K_c \}$, of $F$.
I am aware that in the first definition, $\langle 1 \rangle = \{ n \cdot 1 : n \in \mathbb Z \}$.  Since $1$ is in each $K_c$, $\langle 1 \rangle \subset K$.  But, how are they equal?  I tried two approaches and they ran into the same problem.
One, prove directly that $K \subset \langle 1 \rangle$.  But how do I know each $x \in K$ is of the form $1 + 1 ... +1$?  Another approach is to show that $\langle 1 \rangle$ is a subfield of $F$, ie an element in $\{ K_c \}$.  But how do I show that the inverse, $x^{-1}$, of each $x \in \langle 1 \rangle$ is also of the form $1 + 1 ... +1$?

Let's write $K$ for the field "generated by $1$" and $L$ for the intersection of all subfields of $F$.
It should be clear that $L\subseteq K$. This is because $K$ is a subfield of $F$, and is therefore one of the fields in the intersection that created $L$.
What does it mean to be "generated by $1$", and why is $K\subseteq L$?
Saying that $K$ is generated by $1$ means that $K$ is the smallest subfield containing $1$. Since $K$ is closed under addition, it must contain every element of the form $1 + 1 + \cdots + 1$. Since $K$ is closed under taking negatives, it must contain every element of the form $-(1 + 1 + \cdots + 1)$. There are now two possibilities. 


*

*There is some number $n$ such that
$$
\underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0.
$$
In this case, it turns out that $n$ has to be prime and $L=\mathbb F_p$, the finite field with $p$ elements (I'll supply details if you're interested). We don't have to add multiplicative inverses in because they're already present.

*There is no number $n$ such that
$$
\underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0.
$$
In this case the subring generated by $1$ is $\mathbb Z$. We think of the number $n$ as being
$$
\underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0.
$$
You should then check that this definition makes sense with multiplication. (This is a consequence of the distributive law.)
But since a subfield must contain multiplicative inverses, we're forced to add these in and we get that $L=\mathbb Q$.
(Note: we say that $F$ has characteristic $p$ in the first case and $0$ in the second.)
Finally, since $L$ is a field it contains $1$ and so it has to also contain the field generated by $1$, as that field is the smallest subfield containing $1$.
