Let $p$ be a prime number. Let $F$={0,1,...,p-1}.
+= addition mod p
.= multiplication mod p
The only nontrivial thing to check is the existence of multiplicative inverses.
Proof. Let x $\in F\setminus\{0\}$. Then x and p are relatively prime. So there are a,b $\in$ $\mathbb{Z}$ such that ax+bp=1 (Bézout). xxx Taking ''mod p'' yields ax$\equiv$1 (mod p).xx
Note that I did not understand, from ''xxx''to ''xx''. Can you be explained in detail?
One way of looking at this is that $ax\equiv 1\mod p$ means simply that $ax$ and $1$ leave the same remainder when divided by $p$, so that their difference must be divisible by $p$. Thus $ax - 1 = kp$ for some integer $k$ and then $ax - kp = 1$.
Alternatively, you have in $\mathbb{Z}$ that $ax + bp = 1$. Now consider the map $\varphi:\mathbb{Z}\to F:n\mapsto \overline{n}$, the "remainder $\mod p$" map. Then $\varphi(1) = 1$ and $\varphi(ax+bp) = \overline{ax} + \overline{bp} = \overline{ax} + \overline{b}\cdot\overline{p}$. But $\overline{p}$ is the image of $p$ in $F$ is $0$, so $\overline{ax} = \overline{1}$, which says that $\overline{ax}$ and $1$ are the same element of $F$.