Basic question (part 2) about finite field example, p. 8, of Golan, Linear Algebra a Beginning Graduate Student Ought to Know

Question

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 (really it is two questions) is:

1. How does one prove the claim: "that $[n]_p \in \mathbb{Z}/(p)$ for each nonnegative integer $n$"?

2. How does one prove the claim: "that $[i]_p=i$ for all $i \in \mathbb{Z}/(p)$"?

At this point, I can only demonstrate to myself numerical examples that hold, but not construct an actual proof for either of them.

Answer 1

1. The remainder of the division by $p > 0$ is by construction an $r$, with $0 \le r < p$.
2. If $0 \le r < p$, the remainder of the division of $r$ by $p$ is $r$ itself, as \begin{cases} r = 0 \cdot p + r\\ 0 \le r < p\\ \end{cases} and quotient and remainder are unique.

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Reminder. One can prove that if $p > 0$, then for each $a$ there exist unique $q, r$ (the quotient and the remainder) such that \begin{cases} a = q \cdot p + r\\ 0 \le r < p\\ \end{cases}