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 $\struct {\Q, +, \times}$.
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:
\(\ds \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) | \(=\) | \(\ds \frac {\paren {\paren {m_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} } + \paren {\paren {m_2 \cdot 1_F} \circ \paren {n_1 \cdot 1_F} } } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }\) | Addition of Division Products | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\paren {m_1 n_2} \cdot \paren {1_F \circ 1_F} } + \paren {\paren {m_2 n_1} \cdot \paren {1_F \circ 1_F} } } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }\) | Product of Integral Multiples | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {\paren {m_1 n_2} \cdot 1_F} + \paren {\paren {m_2 n_1} \cdot 1_F} } {\paren {n_1 n_2} \cdot 1_F}\) | as $1_F$ is the unity of $F$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) | Integral Multiple Distributes over Ring Addition | |||||||||||
\(\ds \) | \(\in\) | \(\ds P'\) | by definition of $P'$ |
So $P'$ is closed under $+$.
Next:
\(\ds \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \circ \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) | \(=\) | \(\ds \frac {\paren {m_1 \cdot 1_F} \circ \paren {m_2 \cdot 1_F} } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }\) | Product of Division Products | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {m_1 m_2} \cdot \paren {1_F \circ 1_F} } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }\) | Product of Integral Multiples | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) | as $1_F$ is the unity of $F$ | |||||||||||
\(\ds \) | \(\in\) | \(\ds P'\) | by definition of $P'$ |
So $P'$ is closed under $\circ$.
Next:
\(\ds -\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F}\) | \(=\) | \(\ds \frac {-\paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}\) | Negative of Division Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1 \cdot \paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}\) | Definition of Integral Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-m_1 \cdot 1_F} {n_1 \cdot 1_F}\) | Integral Multiple of Integral Multiple | |||||||||||
\(\ds \) | \(\in\) | \(\ds P'\) | Definition of $P'$, as $-m_1 \in \Z$ |
So $P'$ is closed under taking inverses of $+$.
Next, assuming that $m \ne 0$:
\(\ds \paren {\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} }^{-1}\) | \(=\) | \(\ds \frac {n_1 \cdot 1_F} {m_1 \cdot 1_F}\) | Inverse of Division Product | |||||||||||
\(\ds \) | \(\in\) | \(\ds P'\) | Definition of $P'$ |
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:
\(\ds \map \phi {\frac {m_1} {n_1} } + \paren {-\map \phi {\frac {m_2} {n_2} } }\) | \(=\) | \(\ds \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \paren {-\frac {m_2 \cdot 1_F} {n_2 \cdot 1_F} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {-m_2 \cdot 1_F} {n_2 \cdot 1_F}\) | from above: negative of element of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {m_1 n_2 - m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) | from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {0 \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) | as $m_1 n_2 = m_2 n_1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F\) |
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:
\(\ds \map \phi {\frac {m_1} {n_1} } + \map \phi {\frac {m_2} {n_2} }\) | \(=\) | \(\ds \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) | from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\frac {m_1 n_2 + m_2 n_1} {n_1 n_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\frac {m_1} {n_1} + \frac {m_2} {n_2} }\) |
and:
\(\ds \map \phi {\frac{m_1} {n_1} } \circ \map \phi {\frac {m_2} {n_2} }\) | \(=\) | \(\ds \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \times \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}\) | from above: product of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\frac {m_1 m_2} {n_1 n_2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {\frac {m_1} {n_1} \circ \frac {m_2} {n_2} }\) |
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 $1_F \in K$.
\(\ds 1_F\) | \(\in\) | \(\ds K\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall k \in \Z: \, \) | \(\ds k \cdot 1_F\) | \(\in\) | \(\ds K\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall m, n \in \Z: n \ne 0: \, \) | \(\ds \paren {m \cdot 1_F} \circ \paren {n \cdot 1_F}^{-1}\) | \(\in\) | \(\ds K\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds P\) | \(\subseteq\) | \(\ds K\) |
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 $P_1 = P_2$.
$\blacksquare$
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Theorem $3.2$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 61$. Characteristic of an integral domain or field