Field of Characteristic Zero has Unique Prime Subfield/Proof 2

Theorem
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 $\left({\Q, +, \times}\right)$.

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 $\left({F, +, \circ}\right)$ be a field such that $\operatorname{Char} \left({F}\right) = 0$.

Let $P$ be a prime subfield of $F$.

From Intersection of Subfields, this has been show 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
 * $\forall n \in \Z, n \ne 0: \left({n \cdot 1_F}\right)^{-1} \in P$

So $P$ contains all elements of $F$ of the form:
 * $\left({m \cdot 1_F}\right) \circ \left({n \cdot 1_F}\right)^{-1}$

where $m, n \in \Z, n \ne 0$.

which using division notation can be expressed more clearly 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:

So $P\,'$ is closed under $+$.

Next:

So $P\,'$ is closed under $\circ$.

Next:

So $P\,'$ is closed under taking inverses of $+$.

Next, assuming that $m \ne 0$:

So $P\,' \setminus \left\{{0_F}\right\}$ 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:


 * $\forall m, n \in \Z: n \ne 0: \phi \left({\dfrac m n}\right) = \dfrac {m \cdot 1_F} {n \cdot 1_F}$

First we show that $\phi$ is well-defined.

Suppose $\dfrac {m_1} {n_1} = \dfrac {m_2} {n_2}$.

Then:
 * $m_1 n_2 = m_2 n_1$

by multiplying both sides by $n_1 n_2$.

We need to show that $\phi \left({\dfrac{m_1} {n_1}}\right) = \phi \left({\dfrac{m_2} {n_2}}\right)$.

That is, $\phi \left({\dfrac{m_1} {n_1}}\right) = \phi \left({\dfrac{m_2} {n_2}}\right)$.

So $\phi$ is well-defined.

Next, we need to show that $\phi$ is a ring isomorphism.

So:

and:

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 $F$, and $P = \operatorname{Im} \left({\phi}\right)$ as defined above.

We know that $1_F \in K$.

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 $P_1 = P_2$.