Field has Prime Characteristic p iff exists Monomorphism from Field of Integers Modulo p

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $F$ be a field.

Then:

there exists some prime number $p$ such that $\Char F = p$

if and only if:

there exists a field monomorphism $\phi: \Z_p \to F$

where:

$\Char F$ denotes the characteristic of $F$.
$\Z_p$ denotes the field of integers modulo $p$.


Proof

Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.


Sufficient Condition

Let there exists some prime number $p$ such that $\Char F = p$.


Let us define the mapping $\phi: \Z_p \to F$ as:

$\forall \eqclass n p \in \Z_p: \map \phi {\eqclass n p} = n \cdot 1$

where $n \cdot 1$ denotes the power of $1$ in the context of the additive group $\struct {F, +}$:

$n \cdot 1 = \begin {cases}

0 & : n = 0 \\ \paren {\paren {n - 1} \cdot 1} + 1 & : n > 0 \end {cases}$


We show that $\phi$ is the field monomorphism required.


By definition, a field monomorphism is a field homomorphism which is also an injection.


We have:

\(\ds \map \phi {\eqclass a p} + \map \phi {\eqclass b p}\) \(=\) \(\ds \paren {a \cdot 1} + \paren {b \cdot 1}\)
\(\ds \) \(=\) \(\ds \paren {a + b} \cdot 1\) Powers of Group Elements: Additive Notation
\(\ds \) \(=\) \(\ds \map \phi {\eqclass {a + b} p}\) Definition of $\phi$


\(\ds \map \phi {\eqclass a p} \times \map \phi {\eqclass b p}\) \(=\) \(\ds \paren {a \cdot 1} \times \paren {b \cdot 1}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds \paren {a \times b} \cdot \paren {1 \times 1}\) Product of Integral Multiples
\(\ds \) \(=\) \(\ds \paren {a \times b} \cdot 1\) Definition of Multiplicative Identity
\(\ds \) \(=\) \(\ds \map \phi {\eqclass {a \times b} p}\) Definition of $\phi$

Then by definition $\phi$ is a (field) homomorphism.


It remains to be shown that $\phi$ is an injection.

Let $\eqclass a p, \eqclass b p \in \Z_p$ such that:

$\phi {\eqclass a p} = \phi {\eqclass b p}$




Necessary Condition

Let there exist a field monomorphism $\phi: \Z_p \to F$.

We require to show that $\Char F = p$.

It is sufficient to show that:

$\forall a \in F: n \circ a = 0 \iff n = p$




Sources