Characteristic of Field by Annihilator

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.

Non-Zero 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 $p$:
 * $\operatorname{Ann} \left({F}\right) = p \Z$

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