[proofplan] We use the definition of characteristic as the least positive integer $n$ for which $n\cdot 1_k=0_k$, with characteristic $0$ when no such positive integer exists. If such an $n$ exists, we first rule out $n=1$ using the nontriviality of fields. Then we show that any nontrivial factorization $n=ab$ would make $a\cdot 1_k$ and $b\cdot 1_k$ nonzero elements whose product is zero, contradicting the fact that [every field is an integral domain](/theorems/8301). [/proofplan]

[step:Split according to the definition of characteristic] For each $m\in \mathbb{N}$, let $m\cdot 1_k$ denote the sum of $m$ copies of the multiplicative identity $1_k$ in the additive group of $k$. By definition, $\operatorname{char}(k)=0$ precisely when there is no positive integer $m$ such that $m\cdot 1_k=0_k$. In this case the theorem holds. It remains to consider the case where such a positive integer exists. Let \begin{align*} n := \operatorname{char}(k). \end{align*} Then $n\in \mathbb{N}$, $n\cdot 1_k=0_k$, and $n$ is minimal among positive integers with this property. [/step]

[step:Rule out characteristic one] Since $k$ is a field, it is nontrivial, so $1_k\neq 0_k$. If $n=1$, then \begin{align*} n\cdot 1_k = 1\cdot 1_k = 1_k \neq 0_k, \end{align*} contradicting $n\cdot 1_k=0_k$. Therefore $n>1$. [/step]

[step:Use a nontrivial factorization to produce zero divisors]Suppose, for contradiction, that $n$ is not prime. Since $n>1$, there exist integers $a,b\in\mathbb{N}$ such that \begin{align*} n = ab \end{align*} and \begin{align*} 1<a<n, \qquad 1<b<n. \end{align*} By the minimality of $n$, neither $a\cdot 1_k$ nor $b\cdot 1_k$ is zero. Indeed, if $a\cdot 1_k=0_k$, then $a$ would be a positive integer smaller than $n$ annihilating $1_k$; the same argument applies to $b$. Using distributivity in the field $k$, the product of these two elements is \begin{align*} (a\cdot 1_k)(b\cdot 1_k)=ab\cdot 1_k. \end{align*} Since $ab=n$, this gives \begin{align*} (a\cdot 1_k)(b\cdot 1_k)=n\cdot 1_k=0_k. \end{align*} By [citetheorem:8301], the field $k$ is an [integral domain](/page/Integral%20Domain), so it has no zero divisors. This contradicts the fact that $a\cdot 1_k$ and $b\cdot 1_k$ are nonzero while their product is zero.[/step]

[guided]Assume, toward a contradiction, that the positive characteristic $n$ is composite. Since $n>1$, this means that there are positive integers $a$ and $b$ with \begin{align*} n = ab \end{align*} and with both factors strictly between $1$ and $n$: \begin{align*} 1<a<n, \qquad 1<b<n. \end{align*} The reason this factorization is useful is that it turns an additive equation about $1_k$ into a multiplicative equation in $k$. First, the minimality built into the definition of characteristic tells us that \begin{align*} a\cdot 1_k\neq 0_k \end{align*} and \begin{align*} b\cdot 1_k\neq 0_k. \end{align*} For example, if $a\cdot 1_k=0_k$, then $a$ would be a positive integer smaller than $n$ with $a\cdot 1_k=0_k$, contradicting that $n$ is the least such positive integer. The same argument applies to $b$. Now multiply the two nonzero elements $a\cdot 1_k$ and $b\cdot 1_k$. By repeated distributivity in the ring structure of the field, multiplying a sum of $a$ copies of $1_k$ by a sum of $b$ copies of $1_k$ gives a sum of $ab$ copies of $1_k$. Hence \begin{align*} (a\cdot 1_k)(b\cdot 1_k)=ab\cdot 1_k. \end{align*} Since $ab=n$, we obtain \begin{align*} (a\cdot 1_k)(b\cdot 1_k)=n\cdot 1_k=0_k. \end{align*} Thus two nonzero elements of $k$ have product zero. This is impossible because, by [citetheorem:8301], every field is an integral domain, and an integral domain has no zero divisors. The contradiction shows that $n$ cannot be composite.[/guided]

[step:Conclude that the characteristic is zero or prime] We have shown that if no positive integer annihilates $1_k$, then $\operatorname{char}(k)=0$. In the remaining case, $n=\operatorname{char}(k)$ is a positive integer greater than $1$ and is not composite. Therefore $n$ is prime. Hence $\operatorname{char}(k)$ is either $0$ or a prime number. [/step]