# Characteristic of Field by Annihilator

## Contents

## Theorem

Let $\left({F, +, \times}\right)$ be a field.

Then of the following two cases, exactly one applies:

### Characteristic Zero

- $\operatorname{Ann} \left({F}\right) = \left\{{0}\right\}$

That is, the annihilator of $F$ consists of the zero only.

If this is the case, then:

- $\operatorname{Char} \left({F}\right) = 0$

That is, the characteristic of $F$ is zero.

### Prime Characteristic

- $\exists n \in \operatorname{Ann} \left({F}\right): n \ne 0$

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:

- $\operatorname{Char} \left({F}\right) = p \ne 0$

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

- $\operatorname{Ann} \left({F}\right) = p \Z$

where $p$ is a prime number.

## Proof

Let the zero of $F$ be $0_F$ and the unity of $F$ be $1_F$.

### Proof for Characteristic Zero

By definition of characteristic, $\operatorname{Char} \left({F}\right) = 0$ iff:

- $\not \exists n \in \Z, n > 0: \forall r \in F: n \cdot r = 0_F$

That is, there exists no $n \in \Z, n > 0$ such that $n \cdot r = 0_F$ for all $r \in F$.

But note that $\forall r \in F: 0 \cdot r = 0_F$ by definition of integral multiple.

From Non-Trivial Annihilator Contains Positive Integer, however, if any element of $\operatorname{Ann} \left({F}\right)$ is non-zero, then $\operatorname{Ann} \left({F}\right)$ must contain a positive integer.

But this would contradict the statement that $\operatorname{Char} \left({F}\right) = 0$.

So it follows that:

- $\operatorname{Ann} \left({F}\right) = \left\{{0}\right\} \iff \operatorname{Char} \left({F}\right) = 0$

$\Box$

### Proof for Prime Characteristic

Let $A := \operatorname{Ann} \left({F}\right)$.

We are told that:

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

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

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

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

Suppose $p$ is not a prime number.

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

\(\displaystyle 0_R\) | \(=\) | \(\displaystyle p \cdot 1_F\) | Definition of Annihilator of Ring | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({a b}\right) \cdot 1_F\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({a \cdot 1_F}\right) \times \left({b \cdot 1_F}\right)\) | Product of Integral Multiples | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({a \cdot 1_F}\right) = 0_F \lor \left({b \cdot 1_F}\right) = 0_F\) | 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:

\(\displaystyle \left({n p}\right) \cdot 1_F\) | \(=\) | \(\displaystyle n \cdot \left({p \cdot 1_F}\right)\) | Integral Multiple of Integral Multiple | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n \cdot 0_F\) | as $p$ is in the annihilator of $F$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0_F\) | 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:

\(\displaystyle 0_F\) | \(=\) | \(\displaystyle k \cdot 0_F\) | Definition of Integral Multiple | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({q p + r}\right) \cdot 1_F\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left({q p}\right) \cdot 1_F + r \cdot 1_F\) | Integral Multiple Distributes over Ring Addition | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0_F + r \cdot 1_F\) | from above: $q p$ is a multiple of $p$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle r \cdot 1_F\) | Definition of Ring 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 $A$.

$\blacksquare$

## Sources

- 1964: Iain T. Adamson:
*Introduction to Field Theory*... (previous) ... (next): $\S 1.2$