Basic Arithmetic in Finite Fields

There’s really no big deal. You are really working in integers (never in the reals), and whenever you get an answer that’s too big or too negative, subtract or add a multiple of your prime number $n$ (conventionally, we call the modulus $p$ in these cases). So, if you’re working modulo $7$, to add $4+1$, you do it in the integers $\Bbb Z$. Answer $5$. Is it at least $7$? No, so just leave it be. But to add $4+5$, the integer sum is $9$, so you subtract $7$ to get $2$, and in the system of integers modulo $7$, you have $4+5=2$. A standard way of writing this is $4+5\equiv2\pmod7$, which you read, “four plus five is congruent to two modulo seven”. This notation and terminology goes back to Gauss (1801), maybe even farther.

This actually brings up a subtle point. What do we mean by $5$ in a finite field? Or if you choose to define $5$ in terms of $1 ~(5=1+1+1+1+1)$, then what do we mean by $1$?

One answer is to define $5$ in terms of equivalence classes. Say that two integers $m$ and $n$ are equivalent if $p \vert (m-n).$ First, you prove this really is an equivalence relation on the integers. Then you define $[m]+[n]=[m+n]$ and $[m][n]= [mn]$. So by $5$ we actually mean the equivalence class $[5]$.

You need to prove that your field operations are well-defined (you get the same answer no matter which representative of an equivalence class you choose) and that $[0]$ and $[1]$ really are the additive and multiplicative identities, as you'd expect. But once you've done that, you can see that $[4]+[1]=[5]$ (and usually we abuse notation by dropping the brackets) because we've defined it that way.