# Characteristic of Field by Annihilator/Prime Characteristic

## Theorem

Let $\struct {F, +, \times}$ be a field.

Suppose that:

$\exists n \in \map {\mathrm {Ann} } F: n \ne 0$

That is, there exists (at least one) non-zero integer in the annihilator of $F$.

If this is the case, then the characteristic of $F$ is non-zero:

$\Char F = p \ne 0$

and the annihilator of $F$ consists of the set of integer multiples of $p$:

$\map {\mathrm {Ann} } F = p \Z$

where $p$ is a prime number.

## Proof

Let $A := \map {\mathrm {Ann} } F$.

We are told that:

$\exists n \in A: n \ne 0$

Consider the set $A^+ \set {n \in A: n > 0}$.

From Non-Trivial Annihilator Contains Positive Integer we have that $A^+ \ne \O$.

As $A^+ \subseteq \N$ it follows from the well-ordering principle that $A^+$ has a least value $p$, say.

Aiming for a contradiction, suppose $p$ is not a prime number.

Then $p$ can be expressed as $p = a b$ where $1 < a, b < p$.

 $\ds 0_R$ $=$ $\ds p \cdot 1_F$ Definition of Annihilator of Ring $\ds$ $=$ $\ds \paren {a b} \cdot 1_F$ $\ds$ $=$ $\ds \paren {a \cdot 1_F} \times \paren {b \cdot 1_F}$ Product of Integral Multiples $\ds$ $=$ $\ds \paren {a \cdot 1_F} = 0_F \lor \paren {b \cdot 1_F} = 0_F$ Field has no Proper Zero Divisors

But then either $a \in A$ or $b \in A$, and so $p$ is not the minimal positive element of $A$ after all.

So from this contradiction it follows that $p$ is necessarily prime.

Next let $n \in \Z$.

Then:

 $\ds \paren {n p} \cdot 1_F$ $=$ $\ds n \cdot \paren {p \cdot 1_F}$ Integral Multiple of Integral Multiple $\ds$ $=$ $\ds n \cdot 0_F$ as $p$ is in the annihilator of $F$ $\ds$ $=$ $\ds 0_F$ Definition of Integral Multiple

So all multiples of $p$ are in $A$.

Finally, suppose $k \in A$.

By the Division Theorem $k = q p + r$ where $0 \le r < p$.

Then:

 $\ds 0_F$ $=$ $\ds k \cdot 0_F$ Definition of Integral Multiple $\ds$ $=$ $\ds \paren {q p + r} \cdot 1_F$ $\ds$ $=$ $\ds \paren {q p} \cdot 1_F + r \cdot 1_F$ Integral Multiple Distributes over Ring Addition $\ds$ $=$ $\ds 0_F + r \cdot 1_F$ from above: $q p$ is a multiple of $p$ $\ds$ $=$ $\ds r \cdot 1_F$ Definition of Field Zero

So $r \in A$ contradicting the stipulation that $p$ is the smallest positive element of $A$.

Hence all and only multiples of $p$ are in $A$.

$\blacksquare$