I realize that $K= \mathbb Z /2\mathbb Z$ is simply the set of equivalence classes. But I recently came across $K^{2}$ given $K= \mathbb Z /2\mathbb Z$
It is then stated that $K^{2}=\{\{0,0\},\{1,0\},\{0,1\},\{1,1\}\}$
I have no idea what those vectors have to do with $\mathbb Z /2\mathbb Z$. Any ideas as to what I am missing here?
On
If you follow the definition, $K= \mathbb Z /2\mathbb Z$ consists of two classes: even and odd integers. It is natural to map the class of even integers into $0$ and the class of odd integers into $1$. This bijection in turn becomes a ring isomorphism between $\mathbb Z /2\mathbb Z$ and $\{0,1\}$. Strictly speaking, these sets are isomorphic, but more loosely they are often identified. $K^2=K \times K$ is the Cartesian product of $K$ with itself: $K^{2}=\{(0,0),(1,0),(0,1),(1,1)\}$
Yes, $K = \mathbb{Z}/2\mathbb{Z}$ contains two equivalences classes:
$$ \mathbb{Z}/2\mathbb{Z} = \{[0],[1]\}$$
where $0$ and $1$ are used to act as the representative of those equivlances classes. As you know, $[0]$ is the set of all even integers, and $[1]$ is the set of all odd integers. So the Cartesian product $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ is the set
$$\{([0],[0]),([0],[1]),([1],[0]),([1],[1])\}$$
This is pretty cluttered. Since there is a canonical isomorphism between $\mathbb{Z}/2\mathbb{Z}$ and the subgroup $\{0,1\}$ of $\mathbb{Z}$, there is then a canonical isomorphism between the groups $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ and $\{0,1\}\times\{0,1\}\subseteq \mathbb{Z}\times\mathbb{Z}$. Since $\mathbb{Z}/2\mathbb{Z}$ is usually identified with $\{0,1\}$, the group $K^2 = K \times K =(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ is usually just identified with $\{0,1\}\times\{0,1\}$: $$\{(0,0),(0,1),(1,0),(1,1)\}$$
Note that you used sets rather than 2-tuples in your expression, which isn't correct, since, for example, $\{1,1\} = \{1\}$ by Extensionality.