Field of Characteristic Zero has Unique Prime Subfield/Proof 2

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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 $\struct {\Q, +, \times}$.


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 Intersection of Subfields is 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:

\(\displaystyle \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) \(=\) \(\displaystyle \frac {\paren {\paren {m_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} } + \paren {\paren {m_2 \cdot 1_F} \circ \paren {n_1 \cdot 1_F} } } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }\) $\quad$ Addition of Division Products $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {\paren {m_1 n_2} \cdot \paren {1_F \circ 1_F} } + \paren {\paren {m_2 n_1} \cdot \paren {1_F \circ 1_F} } } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }\) $\quad$ Product of Integral Multiples $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {\paren {m_1 n_2} \cdot 1_F} + \paren {\paren {m_2 n_1} \cdot 1_F} } {\paren {n_1 n_2} \cdot 1_F}\) $\quad$ as $1_F$ is the unity of $F$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) $\quad$ Integral Multiple Distributes over Ring Addition $\quad$
\(\displaystyle \) \(\in\) \(\displaystyle P'\) $\quad$ by definition of $P'$ $\quad$

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


Next:

\(\displaystyle \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \circ \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) \(=\) \(\displaystyle \frac {\paren {m_1 \cdot 1_F} \circ \paren {m_2 \cdot 1_F} } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }\) $\quad$ Product of Division Products $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {m_1 m_2} \cdot \paren {1_F \circ 1_F} } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }\) $\quad$ Product of Integral Multiples $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) $\quad$ as $1_F$ is the unity of $F$ $\quad$
\(\displaystyle \) \(\in\) \(\displaystyle P'\) $\quad$ by definition of $P'$ $\quad$

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


Next:

\(\displaystyle -\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F}\) \(=\) \(\displaystyle \frac {-\paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}\) $\quad$ Negative of Division Product $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1 \cdot \paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}\) $\quad$ Definition of Integral Multiple $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-m_1 \cdot 1_F} {n_1 \cdot 1_F}\) $\quad$ Integral Multiple of Integral Multiple $\quad$
\(\displaystyle \) \(\in\) \(\displaystyle P'\) $\quad$ Definition of $P'$, as $-m_1 \in \Z$ $\quad$

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


Next, assuming that $m \ne 0$:

\(\displaystyle \paren {\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} }^{-1}\) \(=\) \(\displaystyle \frac {n_1 \cdot 1_F} {m_1 \cdot 1_F}\) $\quad$ Inverse of Division Product $\quad$
\(\displaystyle \) \(\in\) \(\displaystyle P'\) $\quad$ Definition of $P'$ $\quad$

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:

\(\displaystyle \map \phi {\frac {m_1} {n_1} } + \paren {-\map \phi {\frac {m_2} {n_2} } }\) \(=\) \(\displaystyle \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \paren {-\frac {m_2 \cdot 1_F} {n_2 \cdot 1_F} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {-m_2 \cdot 1_F} {n_2 \cdot 1_F}\) $\quad$ from above: negative of element of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {m_1 n_2 - m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) $\quad$ from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {0 \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) $\quad$ as $m_1 n_2 = m_2 n_1$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0_F\) $\quad$ $\quad$

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:

\(\displaystyle \map \phi {\frac {m_1} {n_1} } + \map \phi {\frac {m_2} {n_2} }\) \(=\) \(\displaystyle \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) $\quad$ from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\frac {m_1 n_2 + m_2 n_1} {n_1 n_2} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\frac {m_1} {n_1} + \frac {m_2} {n_2} }\) $\quad$ $\quad$

and:

\(\displaystyle \map \phi {\frac{m_1} {n_1} } \circ \map \phi {\frac {m_2} {n_2} }\) \(=\) \(\displaystyle \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \times \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) $\quad$ from above: product of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\frac {m_1 m_2} {n_1 n_2} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {\frac {m_1} {n_1} \circ \frac {m_2} {n_2} }\) $\quad$ $\quad$

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 $1_F \in K$.

\(\displaystyle 1_F\) \(\in\) \(\displaystyle K\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall k \in \Z: k \cdot 1_F\) \(\in\) \(\displaystyle K\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall m, n \in \Z: n \ne 0: \paren {m \cdot 1_F} \circ \paren {n \cdot 1_F}^{-1}\) \(\in\) \(\displaystyle K\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle P\) \(\subseteq\) \(\displaystyle K\) $\quad$ $\quad$


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 $P_1 = P_2$.

$\blacksquare$


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