This is true only when the characteristic of $F$ is 0.
Suppose the characteristic is indeed 0. Then the *additive* subgroup generated by $1_F$, which is:
$\{\dots,-1_F+(-1_F)+(-1_F),\ -1_F+(-1_F),\ -1_F,\ 0_F,\ 1_F,\ 1_F+1_F,\ 1_F+1_F+1_F,\dots\}$
is an infinite cyclic group isomorphic to $\Bbb Z$, which the explicit isomorphism being:
$n \cdot 1_F \mapsto n$ (here, $n \cdot 1_F$ is equal to:
$1_F + 1_F + \cdots + 1_F$ ($n$ summands), if $n > 0$
$0_F$, if $n = 0$
$-1_F + (-1_F) + \cdots + (-1_F)$ ($-n$ summands) if $n < 0$).
Since we thus have an integral domain contained in $F$ (isomorphic to $\Bbb Z$), we can form its field of quotients, which is the *smallest* field containing this integral domain (in the sense that if $K$ is a field containing our isomorph of $\Bbb Z$, there is an injective ring-homomorphism from the field of quotients into $K$).
The image of this ring-homomorphism is then isomorphic to $Q(\Bbb Z) = \Bbb Q$, and since $F$ is such a field containing our isomorph of the integers, it thus has a subfield which is an isomorph of the rationals.
All of this is a rather long-winded way of saying, if $F$ has characteristic 0, then the smallest field we can build up starting with $1_F$ "acts just like the rational numbers":
By closure (of addition) we see that we must have all sums of $1_F$'s, and since the additive group of a field is an abelian group, we must also have all additive inverses of such sums.
Since a field must have all multiplicative inverses of $n\cdot 1_F$ (for any $n\neq 0 \in \Bbb Z$), with closure of multiplication we see we must also have all elements of the form:
$(m\cdot 1_F)(n\cdot 1_F)^{-1}$, for $m \in \Bbb Z, n\neq 0 \in \Bbb Z$
Defining addition and multiplication of these kinds of elements in the usual way:
$(m\cdot 1_F)(n\cdot 1_F)^{-1} + (m'\cdot 1_F)(n'\cdot 1_F)^{-1} = ((mn' + m'n)\cdot 1_F)((nn'\cdot 1_F)^{-1}$
$(m\cdot 1_F)(n\cdot 1_F)^{-1} \cdot (m'\cdot 1_F)(n'\cdot 1_F)^{-1} = ((mm')\cdot 1_F)((nn'\cdot 1_F)^{-1}$
and noting that $(m\cdot 1_F)(n\cdot 1_F)^{-1} = (m'\cdot 1_F)(n'\cdot 1_F)^{-1}$, whenever $mn' = m'n$
and that also we can identify $n\cdot 1_F$ with $(n\cdot 1_F)(1\cdot 1_F)^{-1}$,
we can verify (although it is tedious) such elements indeed form a subfield of $F$ isomorphic to $\Bbb Q$, with the isomorphism being:
$(m\cdot 1_F)(n\cdot 1_F)^{-1} \mapsto \dfrac{m}{n}$.
With a field of characteristic $p$, it is a slightly different story-in this case, the additive subgroup generated by $1_F$ is cyclic of prime order (hence the name "prime subfield"), and thus isomorphic to $\Bbb Z_p$, the field axioms then force us to conclude the multiplication of $F$ restricted to $\langle 1_F\rangle$ acts "just like multiplication modulo $p$".
