# Closure of a characteristic in an integral domain

To my understanding, one of the conditions of a characteristic is that the characteristic of a ring $R$ should also be in the ring $R$. However, that doesn't make quite sense for me when considering an integral domain. Let's consider $\mathbb{Z}_{5}$. Since 5 is prime, we know that $\mathbb{Z}_{5}$ is an integral domain. However, we also know that $\mathbb{Z}_{5}=\{0,1,2,3,4\}$. If the characteristic is inside the ring, then $\mathbb{Z}_{5}$ is of 0 characteristic. This is in conflict with my understanding of this theorem:

The smallest positive integer $n$ such that $n\cdot1=0$ is the characteristic of $D$, and if $n\cdot1\neq0$, then $D$ has characteristic 0.

Also, after a short google search, I found out that the characteristic of $\mathbb{Z}_{5}$ should be a prime. But no such integer in the ring satisfies the condition (I would pick $p=5$, since $5\equiv0\pmod 5$ but 5 is not inside the set so I think that's invalid). So what's the problem with my understanding of a characteristic?

The characteristic of a ring is not an element of the ring. It is the number of times that one has to add the identity to itself to get zero. In particular, the expression $n\cdot 1 = 0$ should be interpreted as "add $1$ to itself $n$ times and then the result is zero."
You are correct that in general, if $n$ is the characteristic of a ring, then "$n$" in the ring will have to be zero and so it "won't contain $n$".