Domain
A crone $$R$$ is a domain if the product of any two nonzero elements in $$R$$ is itself non-zero

Note that in other places they may call this an integral domain

Product Zero Implies Factor Zero Domain Equivalence
Suppose that $$R$$ is a crone. It is a domain if and only if for every $$a, b \in D$$, $$ab = 0$$ implies $$a = 0$$ or that $$b = 0$$
Since $$R$$ is a domain if and only if for any $$x \neq 0$$ and $$y \neq 0$$ it implies that $$x y \neq 0$$, this statement is equivalent to it's contrapositive: $$xy = 0$$ implies that $$x = 0$$ or $$y = 0$$ as needed.
Domain iff Cancellation
Let $$R$$ be a crone. $$R$$ is a domain if and only if given $$r, a, b \in R$$ we have $$ra = rb \land r \neq 0 \implies a = b$$

For the rightward implication we can use properties of a ring to see the following $\begin{gather} ra &= rb \iff \\ ra - rb &= 0 \iff \\ r \left( a - b \right) &= 0 \iff \\ \end{gather}$

Therefore we conclude that either $$a - b = 0$$ or $$r = 0$$, the latter of which is impossible through our assumption that $$r \neq 0$$, therefore $$a = b$$.

We move on to the leftward implication where our goal is to show that $$R$$ is a domain, we do so by chasing it's corollary. So let $$a, b \in R$$ and assume that $$ab = 0$$, note that at this point if $$a = 0$$, our proof is over, therefore we assume that $$a \neq 0$$, note that we can rewrite $$ab = 0$$ to $$ab = a0$$, then by our assumption we can conclude that $$b = 0$$, which proves it.

Since $$\mathbb{ Z } _ 4$$ forms a ring then we can then notice the following: \begin{align} \left( [ 2 ] x + [ 1 ] \right) ^ 2 &= [ 2 ] [ 2 ] x ^ 2 + [ 2 ] [ 1 ] x + [ 1 ] [ 2 ] x + [ 1 ] [ 1 ] \\ &= [ 4 ] x ^ 2 + [ 4 ] x + [ 1 ] \\ &= [ 0 ] x ^ 2 + [ 0 ] x + [ 1 ] \\ &= [ 1 ] \end{align} Where we're noticing that the degree of the product has decreased, this is occuring because in $$\mathbb{ Z } _ 4$$ there are non-zero elements which have a product of $$0$$.

Unit
Suppose that $$R$$ is a ring, then an element $$u \in R$$ is said to be a unit if there exists some element $$v \in R$$ such that $uv = vu = 1 _ R$
Zero Divisor
A non-zero element $$x$$ in a ring is said to be a zero divisor if there exists some other non-zero $$y$$ such that $$x y = 0$$
Nilpotent
Let $$R$$ be a crone, then an element $$x \in R$$ is said to be nilpotent if there is some $$n \in \mathbb{ N } _ 1$$ such that $$x ^ n = 0 _ R$$
Nilpotent Implies Zero or Zero Divisor
if $$x$$ is nilpotent then $$x = 0 _ R$$ or $$x$$ is a

If $$x = 0 _ R$$ then our proof is done, so instead let's assume that $$x \neq 0 _ R$$, we know that it's nilpotent, considering $$\left\{ a \in \mathbb{ N } _ 1: x ^ a = 0 _ R \right\} \subseteq \mathbb{ N } _ 0$$ it has a least element (as it's non-empty) say $$l$$ such that $$x ^ l = 0 _ R$$, note that $$l \gt 1$$ because $$x ^ 1 \neq 0 _ R$$

In that case we can see $$x ^ l = x ^ { l - 1 } x ^ 1$$, since $$l$$ was the smallest natural number such that $$x ^ l = 0 _ R$$ then we must have $$x ^ { l - 1 } \neq 0 _ R$$ since $$l - 1 \in \mathbb{ N } _ 0$$. Thus we can conclude that $$x x ^ { l - 1 } = 0 _ R$$ where $$x, x ^ { l - 1 } \neq 0 _ R$$ so that by definition $$x$$ is a zero divisor.

Constant times Nilpotent is Nilpotent
Suppose $$R$$ is a crone and that $$x$$ is nilpotent, then $$r x$$ is nilpotent for any $$r \in R$$
Since $$x$$ is nilpotent we have some $$n \in \mathbb{ N } _ 1$$ such that $$x ^ n = 0 _ R$$, since $$R$$ is commutative, then it can be shown by induction that an equation of the form $$\left( ab \right) ^ n = a ^ n b ^ n$$, in other words $$\left( r x \right) ^ n = r ^ n x ^ n = r ^ n 0 _ R = 0 _ R$$ which proves that $$rx$$ is nilpotent
One plus Nilpotent is Nilpotent
Suppose that $$R$$ is a crone, and $$x$$ nilpotent then $$1 _ R + x$$ is a unit

If $$x ^ 1 = 0 _ R$$, it clearly holds so lets assume that $$x ^ n = 0 _ R$$ for some $$n \gt 1$$.

We claim that $v := 1 - x + x ^ 2 - x ^ 3 + ... + \left( -1 \right) ^ { n - 1 } x ^ n - 1 = \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i$ is $$1 + x$$'s multiplicative inverse.

To see this true, we use distributivity, $\left( 1 + x \right) \left( \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i \right) = \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i + \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ { i + 1 }$ Define the following $S _ 1 \left( k \right) := \sum _ { i = 1 } ^ { k - 1 } \left( -1 \right) ^ i x ^ i = 1 - x + x ^ 2 - x ^ 3 + ... + \left( -1 \right) ^ { k - 1 } x ^ { k - 1 }$ and that $S _ 2 \left( k \right) := \sum _ { i = 1 } ^ { k - 1 } \left( -1 \right) ^ i x ^ { i + 1 } = x - x ^ 2 + x ^ 3 - x ^ 4 + \ldots \left( -1 \right) ^ { k - 2 } x ^ { k - 1 } + \left( -1 \right) ^ { k - 1 } x ^ { k - 1 }$

Note that $$S _ 2 \left( n \right) = S _ 2 \left( n - 1 \right)$$ because $$S _ 2 \left( n \right)$$'s last term is $$\left( -1 \right) ^ { n - 1 } x ^ { n } = 0 _ R$$ since $$x ^ n = 0 _ R$$ .

Through finite induction, we can show that for any $$j \in [ 2 \ldots n ], S _ 1 \left( j \right) + S _ 2 \left( j - 1 \right) = 1$$ which allows us to conclude that $$S _ 1 \left( n \right) + S _ 2 \left( n - 1 \right) = 1$$ and then $$S _ 1 \left( n \right) + S _ 2 \left( n \right) = 1$$ from the previous paragraph.

Finally this is good because now we continue the chain of equalities we mentioned at distributivity \begin{align} \left( 1 + x \right) \left( \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i \right) &= \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i + \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ { i + 1 }\\ &= S _ 1 \left( n \right) + S _ 2 \left( n \right) \\ &= 1 \end{align} Therefore $$1 + x$$ is a unit, as needed.

The Sum of a Unit and a Nilpotent Element is a Unit
Suppose $$R$$ is a crone and let $$u$$ be a unit and $$x$$ a nilpotent element in $$R$$ , then $$u + x$$ is a unit.

Since $$u$$ is a unit there exists some $$v$$ such that $$u v = v u = 1$$, if that's the case then $$\left( u + x \right) v = 1 + vx$$, we know that $$vx$$ is nilpotent and thus $$1 + vx$$ is a unit, so we have some $$a$$ such that $$\left( 1 + vx \right) a = 1$$.

But $$\left( u + x \right) v = 1 + vx$$ so in total we're saying $$\left( \left( u + x \right) v \right) a = 1$$ by associativity we have $$\left( u + x \right) \left( v a \right) = 1$$ and closure of multiplication shows us indeed $$u + x$$ has a multiplicative inverse so that $$u + x$$ is a unit.

Field is a Ring with Units
A field is a ring $$R$$ in which every non-zero element $$r \in R$$ is a unit