Let $\struct {F, +, \circ}$ be a field such that $\Char F = 0$. Let $P$ be a prime subfield of $F$. From Field has Prime Subfield, this has been shown to exist. As $P$ is a subfield of $F$, we apply Zero and Unity of Subfield and see that the unity of $P$ is $1_F$. As $P$ is closed: :$\forall m \in \Z: m \cdot 1_F \in P$ of which $0 \cdot 1_F = 0_F$ the only one that is zero :$\forall n \in \Z, n \ne 0: \paren {n \cdot 1_F}^{-1} \in P$ So $P$ contains all elements of $F$ of the form: : $\paren {m \cdot 1_F} \circ \paren {n \cdot 1_F}\^{-1}$ where $m, n \in \Z, n \ne 0$. This can be expressed more clearly in division notation as: :$\dfrac {m \cdot 1_F} {n \cdot 1_F}$ Now let $P'$ consist of all the elements of $F$ of the form: :$\dfrac {m \cdot 1_F} {n \cdot 1_F}$ Let $\dfrac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \in P'$ and $\dfrac {m_2 \cdot 1_F} {n_2 \cdot 1_F} \in P'$. Then: {{begin-eqn}} {{eqn | l = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}       | r = \frac {\paren {\paren {m_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} } + \paren {\paren {m_2 \cdot 1_F} \circ \paren {n_1 \cdot 1_F} } } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }       | c = Addition of Division Products }} {{eqn | r = \frac {\paren {\paren {m_1 n_2} \cdot \paren {1_F \circ 1_F} } + \paren {\paren {m_2 n_1} \cdot \paren {1_F \circ 1_F} } } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }       | c = Product of Integral Multiples }} {{eqn | r = \frac {\paren {\paren {m_1 n_2} \cdot 1_F} + \paren {\paren {m_2 n_1} \cdot 1_F} } {\paren {n_1 n_2} \cdot 1_F}       | c = as $1_F$ is the unity of $F$ }} {{eqn | r = \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}       | c = Integral Multiple Distributes over Ring Addition }} {{eqn | o = \in       | r = P'       | c = by definition of $P'$ }} {{end-eqn}} So $P'$ is closed under $+$. Next: {{begin-eqn}} {{eqn | l = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \circ \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}       | r = \frac {\paren {m_1 \cdot 1_F} \circ \paren {m_2 \cdot 1_F} } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }       | c = Product of Division Products }} {{eqn | r = \frac {\paren {m_1 m_2} \cdot \paren {1_F \circ 1_F} } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }       | c = Product of Integral Multiples }} {{eqn | r = \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}       | c = as $1_F$ is the unity of $F$ }} {{eqn | o = \in       | r = P'       | c = by definition of $P'$ }} {{end-eqn}} So $P'$ is closed under $\circ$. Next: {{begin-eqn}} {{eqn | l = -\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F}       | r = \frac {-\paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}       | c = Negative of Division Product }} {{eqn | r = \frac {-1 \cdot \paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}       | c = {{Defof|Integral Multiple}} }} {{eqn | r = \frac {-m_1 \cdot 1_F} {n_1 \cdot 1_F}       | c = Integral Multiple of Integral Multiple }} {{eqn | o = \in       | r = P'       | c = Definition of $P'$, as $-m_1 \in \Z$ }} {{end-eqn}} So $P'$ is closed under taking inverses of $+$. Next, assuming that $m \ne 0$: {{begin-eqn}} {{eqn | l = \paren {\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} }^{-1}       | r = \frac {n_1 \cdot 1_F} {m_1 \cdot 1_F}       | c = Inverse of Division Product }} {{eqn | o = \in       | r = P'       | c = Definition of $P'$ }} {{end-eqn}} So $P' \setminus \set {0_F}$ is closed under taking inverses of $\circ$. Thus by Subfield Test, $P'$ is a subfield of $F$. It follows that $P = P'$, and so $P$ contains precisely the elements of the form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$. We can consistently define a mapping $\phi: \Q \to P$ by: :$\forall m, n \in \Z: n \ne 0: \map \phi {\dfrac m n} = \dfrac {m \cdot 1_F} {n \cdot 1_F}$ First we show that $\phi$ is well-defined. Suppose $\dfrac {m_1} {n_1} = \dfrac {m_2} {n_2}$. Then by multiplying both sides by $n_1 n_2$: :$m_1 n_2 = m_2 n_1$ We need to show that: :$\map \phi {\dfrac {m_1} {n_1} } = \map \phi {\dfrac {m_2} {n_2} }$ So: {{begin-eqn}} {{eqn | l = \map \phi {\frac {m_1} {n_1} } + \paren {-\map \phi {\frac {m_2} {n_2} } }       | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \paren {-\frac {m_2 \cdot 1_F} {n_2 \cdot 1_F} }       | c =  }} {{eqn | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {-m_2 \cdot 1_F} {n_2 \cdot 1_F}       | c = from above: negative of element of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ }} {{eqn | r = \frac {\paren {m_1 n_2 - m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}       | c = from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ }} {{eqn | r = \frac {0 \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}       | c = as $m_1 n_2 = m_2 n_1$ }} {{eqn | r = 0_F       | c =  }} {{end-eqn}} That is: :$\map \phi {\dfrac {m_1} {n_1} } = \map \phi {\dfrac {m_2} {n_2} }$ demonstrating that $\phi$ is well-defined. Next, we need to show that $\phi$ is a ring isomorphism. So: {{begin-eqn}} {{eqn | l = \map \phi {\frac {m_1} {n_1} } + \map \phi {\frac {m_2} {n_2} }       | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}       | c =  }} {{eqn | r = \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}       | c = from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ }} {{eqn | r = \map \phi {\frac {m_1 n_2 + m_2 n_1} {n_1 n_2} }       | c =  }} {{eqn | r = \map \phi {\frac {m_1} {n_1} + \frac {m_2} {n_2} }       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map \phi {\frac{m_1} {n_1} } \circ \map \phi {\frac {m_2} {n_2} }       | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \times \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}       | c =  }} {{eqn | r = \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}       | c = from above: product of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ }} {{eqn | r = \map \phi {\frac {m_1 m_2} {n_1 n_2} }       | c =  }} {{eqn | r = \map \phi {\frac {m_1} {n_1} \circ \frac {m_2} {n_2} }       | c =  }} {{end-eqn}} thus proving that $\phi$ is a ring homomorphism. From Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ is a ring monomorphism. It is also clear that $\phi$ is a surjection, as every element of $P$ is the image of some element of $\Q$. It follows that $\phi$ is a ring isomorphism. Now let $K$ be a subfield of $F$. Let $P = \Img \phi$ as defined above. We know that $1_F \in K$. {{begin-eqn}} {{eqn | l = 1_F       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = \forall k \in \Z: k \cdot 1_F       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = \forall m, n \in \Z: n \ne 0: \paren {m \cdot 1_F} \circ \paren {n \cdot 1_F}^{-1}       | o = \in       | r = K       | c =  }} {{eqn | ll= \leadsto       | l = P       | o = \subseteq       | r = K       | c =  }} {{end-eqn}} Thus $K$ contains a subfield $P$ such that $P$ is isomorphic to $\Q$. The uniqueness of $P$ follows from the fact that if $P_1$ and $P_2$ are both minimal subfields of $F$, then $P_1 \subseteq P_2$ and $P_2 \subseteq P_1$, thus $P_1 = P_2$. {{Qed}}