Characteristic of Field by Annihilator

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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 $1 < a, b < p$.

\(\displaystyle 0_R\) \(=\) \(\displaystyle p \cdot 1_F\) $\quad$ Definition of Annihilator of Ring $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({a b}\right) \cdot 1_F\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({a \cdot 1_F}\right) \times \left({b \cdot 1_F}\right)\) $\quad$ Product of Integral Multiples $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({a \cdot 1_F}\right) = 0_F \lor \left({b \cdot 1_F}\right) = 0_F\) $\quad$ Field has no Proper Zero Divisors $\quad$

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:

\(\displaystyle \left({n p}\right) \cdot 1_F\) \(=\) \(\displaystyle n \cdot \left({p \cdot 1_F}\right)\) $\quad$ Integral Multiple of Integral Multiple $\quad$
\(\displaystyle \) \(=\) \(\displaystyle n \cdot 0_F\) $\quad$ as $p$ is in the annihilator of $F$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0_F\) $\quad$ Definition of Integral Multiple $\quad$

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:

\(\displaystyle 0_F\) \(=\) \(\displaystyle k \cdot 0_F\) $\quad$ Definition of Integral Multiple $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({q p + r}\right) \cdot 1_F\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({q p}\right) \cdot 1_F + r \cdot 1_F\) $\quad$ Integral Multiple Distributes over Ring Addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0_F + r \cdot 1_F\) $\quad$ from above: $q p$ is a multiple of $p$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle r \cdot 1_F\) $\quad$ Definition of Ring Zero $\quad$

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 $A$.

$\blacksquare$


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