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, $\Char F = 0$ if and only if:
- $\not \exists n \in \Z, n > 0: \forall r \in F: n \cdot r = 0_F$
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.
Aiming for a contradiction, suppose 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:
- $\map {\mathrm {Ann} } F = \set 0 \iff \Char F = 0$
$\Box$
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties