The example I reference in the subject line reads:
Fields do not have to be infinite. Let $p$ be a positive integer and let $\mathbb{Z}/(p) = {0, 1, . . . , p − 1}$. For each nonnegative integer $n$, let us, for the purposes of this example, denote the remainder after dividing $n$ by $p$ as $[n]_p$ . Thus we note that $[n]_p \in \mathbb{Z}/(p)$ for each nonnegative integer $n$ and that $[i]_p = i$ for all $i \in \mathbb{Z}/(p)$. We now define operations on $\mathbb{Z}/(p)$ by setting $[n]_p + [k]_p = [n + k]_p$ and $[n]_p \cdot [k]_p = [nk]_p$ . It is easy to check that if the integer $p$ is prime then $\mathbb{Z}/(p)$, together with these two operations, is again a field, known as the Galois field of order $p$. This field is usually denoted by $\operatorname{GF}(p)$.
My question is: For the claim $[i]_p=i$ for all $i \in \mathbb{Z}/(p)$ to hold, should the author have specified that $p \in $ prime numbers rather than just that $p \in $ positive integers? Because if we chose only the latter, as the author does, wouldn't $[1]_1=0$ be a counterexample to the claim?
Did I miss something?
Update: I see now that someone else already caught this here
If you choose $p = 1$, then $\mathbb{Z}/(1) = \{ 0 \}$ (in general, $\mathbb{Z}/(p)$ has $p$ elements, so it has only $p = 1$ element in this case), so $[1]_{1} = 0$ is not a counterexample, as $1 \notin \mathbb{Z}/(1)$.