How is the integers modulo p a finite field. 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?
 A: 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. 
A: 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 $0$. 
A: " 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 0\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.
