Field of Characteristic Zero has Unique Prime Subfield

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 1

Follows directly from:

• Subring Generated by Unity of Ring with Unity
• Quotient Theorem for Monomorphisms

$\blacksquare$

Proof 2

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:

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

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

Next:

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

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

Next:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle - \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F}\)	\(=\)	\(\displaystyle \frac {-\left({m_1 \cdot 1_F}\right)} {n_1 \cdot 1_F}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Negative of Division Product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {-1 \cdot \left({m_1 \cdot 1_F}\right)} {n_1 \cdot 1_F}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of integral multiple
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {-m_1 \cdot 1_F} {n_1 \cdot 1_F}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Integral Multiple of Integral Multiple
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\in\)	\(\displaystyle P'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $P'$, as $-m_1 \in \Z$

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

Next, assuming that $m \ne 0$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} }\right)^{-1}\)	\(=\)	\(\displaystyle \frac {n_1 \cdot 1_F} {m_1 \cdot 1_F}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Inverse of Division Product
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\in\)	\(\displaystyle P'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $P'$

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

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

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:

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

and:

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

thus proving that $\phi$ is a ring homomorphism.

From Homomorphism from Field Either 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 $1_F \in K$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 1_F\)	\(\in\)	\(\displaystyle K\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \forall k \in \Z: k \cdot 1_F\)	\(\in\)	\(\displaystyle K\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \forall m, n \in \Z: n \ne 0: \left({m \cdot 1_F}\right) \circ \left({n \cdot 1_F}\right)^{-1}\)	\(\in\)	\(\displaystyle K\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle P\)	\(\subseteq\)	\(\displaystyle K\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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 $P_1 = P_2$.

$\blacksquare$