Characteristic of Field by Annihilator/Characteristic Zero

Theorem
Let $\struct {F, +, \times}$ be a field. Suppose that:
 * $\map {\mathrm {Ann} } F = \set 0$

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

Then:
 * $\Char F = 0$

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

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

By definition of characteristic, $\Char F = 0$ :
 * $\not \exists n \in \Z, n > 0: \forall r \in F: n \cdot r = 0_F$

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.

there exists a non-zero element of $\map {\mathrm {Ann} } F$.

From Non-Trivial Annihilator Contains Positive Integer, $\map {\mathrm {Ann} } F$ must contain a (strictly) positive integer.

But this would contradict the statement that $\Char F = 0$.

So it follows that:
 * $\map {\mathrm {Ann} } F = \set 0 \iff \Char F = 0$