Q:

Prove that $\mathbb{Z}_3$ is a field

We define $\mathbb{Z}_3=\{0,1,2\}$, the set of integers modulo $3$
Prove it is a field.
Here I have made two tables, showing the relationship by $+,\cdot$
$$
\begin{array}{|c|c|c|c|}\hline
+ & 0 & 1 & 2 \\ \hline
0 & 0 & 1 & 2 \\ \hline
1 & 1 & 2 & 0 \\ \hline
2 & 2 & 0 & 1\end{array}$$
$$
\begin{array}{|c|c|c|c|}\hline
\cdot & 0 & 1 & 2 \\ \hline
0 & 0 & 0 & 0 \\ \hline
1 & 0 & 1 & 2 \\ \hline
2 & 0 & 2 & 1\end{array}$$
Read it as, $2+1=0$, since $2+1=3$, and $3\mod 3 = 0$.
I want to prove the property that any $d\in\mathbb{Z}_3$, there is $y$ such that $d\cdot y = 1$, where $d,y\in\{0,1,2\}$.
Consider all cases.
If $d=1$, let $y=1$, then $d\cdot y = 1$
If $d=2$, let $y=2$, then $d\cdot y = 1$
If $d=0$, let $y=??$, then $d\cdot y = 1$

In the final case, no matter what we choose $y$ as, $d\cdot y$ will never be $1$. So then $0$ does not have a multiplicative identity, but my book still says this is a field. How?

A:

The definition of a field requires that the NONzero elements have inverses.
Good exercise:  prove from ring axioms alone that if $R$ is a ring with identity and $0$ is invertible, then $R=\{0\}$.  So, if you want more than one element in your field, you better not let $0$ have an inverse. 

