How is the integers modulo p a finite field.

Question

Let's a assume we have integers Z := {..-3,-2,-1,0,1,2,3..}, if this is modded by p, how does it satisfy the inverse axiom of a field? Assume p is 3

:. p = {0,1,2}

The identity 0,1 and exist for both operations but the field lacks an inverse for both operations, how is this still a field?

Answer 1

Note that in mod (3) we have $2+1=0$ thus $2$and $1$ are additive inverse of each other.

Also $2*2=1$ Thus $2$ is its own multiplicative inverse.

Similarly you can check out other elements for field properties.

Answer 2

An inverse exists because it's modulo $3$, so going past $2$ takes you back to $0$.

$2+1$mod$3 = 3$mod$3=0$

$2\times2$mod$3 = 4$mod$3=1$

Similarly for $1$ and $0$.

Answer 3

" but the field lacks an inverse for both operations"

quite the opposite!

It is because $p $ is prime that we are assured every element does have multiplicative inverses.

You seem to forget that when "modded" the all the $kp+j $ are all consider the same as $j$. So $2+1=3=3+0\equiv 0$ so $1$ and $2$ are additive inverses. And $2*2=4=3+1\equiv 1$ so $2$ is its own multiplicative inverse.

In general. For $k $, the $p-k $ is the additive inverse (because $k +(p-k)=p+0\equiv 0 \mod p$.

And because $p$ is prime, then for any $k\not \equiv 0\mod p$ there exist $a,b $ so that $ak +bp=1$ (Bezout's lemma). Thus $a $ is the multiplicative inverse of $k $ (because $ka=1-bp\equiv 1\mod p $)

===

Example: take $p=7$

The $1*1+0*7=1$

$2*4-1*7=1$

$3*5-2*7=1$

$4*2-1*7=1$

$5*3-2*7=1$

$6*6-5*7=1$

So in $\mod 7$ the multiplicative inverses of $1,2,3,4,5,6$ are respectively $1,4,5,2,3,6$.

And the additive inverses are $6,5,4,3,2,1$ respectively.