Theorem: Field of Characteristic Zero has Unique Prime Subfield

Let $F$ be a field, whose zero is $0_F$ and whose unity is $1_F$, with characteristic zero.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a subfield of $F$
:$(2): \quad P$ is isomorphic to the field of rational numbers $\struct {\Q, +, \times}$.
That is, $P \cong \Q$ is a unique minimal subfield of $F$, and all other subfields of $F$ contain $P$.
This field $P$ is called the prime subfield of $F$.

Proof:
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
      | q = \forall k \in \Z
      | l = k \cdot 1_F
      | o = \in
      | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall m, n \in \Z: n \ne 0
      | l = \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}}