How can a field have a finite characteristic $p$, given that a field has no zero divisors?

The characteristic of a field is defined to be the smallest positive integer $p$ such that $$p \cdot 1 = 0.$$

But I have learned that field has no zero divisors. How is this possible?

Solution 1:

The notation $p \cdot 1$ just means $1+1+\cdots+1$ ($p$ times), where $1$ is the unit of the field and $+$ is the addition in the field.

You can also interpret $p$ as this same element in the field, but in this case you'll have $p=0$.

Solution 2:

For any unital ring $R$, fields included, and for any $n \in \Bbb Z_{\geq 0}$, we denote $$n \cdot 1 := \underbrace{1 + \cdots + 1}_n.$$ Now, $n \mapsto n \cdot 1$ is by construction a group homomorphism $$\phi: (\mathbb{Z}, +) \to (R, +) ,$$ (to achieve this we need to declare for $-m < 0$ that $(-m) \cdot 1 = -(m \cdot 1)$) and we define the characteristic of $R$ to be the additive order $c$ of the multiplicative unit $\phi(1) = 1_R$ of $R$, or more informally, "$c = 0$ in $R$". (If $n \cdot 1 \neq 0$ for all positive integers $n$, we usually say by convention that the field has characteristic $0$; some older sources say such fields have characteristic $\infty$ instead.)

Example Consider the field $\mathbb{F}_2$ with underlying set $\{0, 1\}$, with elements just an additive and multiplicative identity (respectively) and satisfying $1 + 1 := 0$. Since $0 \neq 1$, the characteristic of this field is $\operatorname{char} \mathbb{F}_2 = 2$.

Remark If the characteristic $c$ of a ring is positive and composite, say, $c = kl$ for $k, l > 1$, then $$\phi(k) \phi(l) = \phi(kl) = \phi(c) = 0 ,$$ so $\phi(k)$ and $\phi(l)$ (which are nonzero, because $k, l$ are both smaller than $c$) are zero divisors. In particular, the characteristic of any field must be $0$ or prime.