Field
A field is a set $$F$$ with two binary operations $$\oplus, \otimes$$ such that for any $$a, b, c \in F$$ we have
1. $$\oplus, \otimes$$ are both associative and commutative
2. Identities: There exist identity elements $$0 _ F \neq 1 _ F \in F$$ for $$\oplus$$ and $$\otimes$$ respectively
3. Additive Inverses: There exists an $$-a \in F$$ such that $$a \oplus -a = 0 _ F$$
4. Multiplicative Inverses: If $$a \neq 0$$ there exists an $$a ^ { -1 } \in F$$ such that $$a \otimes a ^ { -1 } = 1 _ F$$
5. Distributivity: $$\otimes$$ distributes into $$\oplus$$
The Complex Numbers Form a Field
The complex numbers with their standard addition and multiplication form a field
A Field is a Crone
Suppose that $$F$$ is a field, then $$F$$ is a crone
A Crone with Multiplicative Inverses is a Field
Suppose that $$\left( R, \oplus, \otimes \right)$$ is a non-zero crone then if for every $$a \neq 0$$, there is an $$a \otimes a ^ { -1 } = 1 _ R$$ then $$R$$ is a field
Field Implies Domain
Suppose that $$F$$ is a field, then $$F$$ is a domain

We already know that since $$F$$ is a field then $$F$$ is a crone, so one way would be just showing that it's a domain through this characterization of a domain

Suppose that $$x, y \in F$$ and assume that $$x \otimes y = 0 _ F$$, if $$x = 0 _ R$$, we've proven the statement, if $$x \neq 0 _ R$$, then since $$F$$ has multiplicative inverses for non-zero elements we know that $$x ^ { -1 } \otimes x \otimes y = x ^ { -1 } 0 _ R = 0 _ R$$, thus after cancellation and multiplication by one we have $$y = 0 _ R$$ as needed.

Finite Extension of the Rationals
We define the set for any $$x \in \mathbb{ R }$$ as $\mathbb{ Q } \left( x \right) := \left\{ a + bx : a, b \in \mathbb{ Q } \right\} \subseteq \mathbb{ R }$
Multiplicative Inverse in a Finite Extension of the Rationals
Suppose that $$p$$ is prime and that $$x + y \sqrt p \in \mathbb{ Q } \left( \sqrt p \right)$$, such that $$x + y \sqrt p \neq 0$$, then it's multiplicative inverse can be written as $$a + b \sqrt p$$ for some $$a, b \in \mathbb{ Q }$$
Since $$x + y \sqrt p \in \mathbb{ R }$$, and it's non-zero, then in $$\mathbb{ R }$$ we have an inverse: $$1 / \left( x + y \sqrt p \right)$$, we can then rationalize that as follows \begin{align} \frac{1}{x + y \sqrt p} &= \frac{1}{x + y \sqrt p} \cdot \frac{x - y \sqrt p}{x - y \sqrt p} \\ &= \frac{x - y \sqrt p}{x ^ 2 - y ^ 2 p} \end{align}

Now let's observe the denominator $$x ^ 2 - y ^ 2 p$$, if this equals zero, then we have the following $$\left( \frac{x}{y} \right) ^ 2 = p$$ note that division is justified since $$y \neq 0$$ from the fact that $$x + y \sqrt p$$ was non-zero. For this equation to make any sense we require that $$\left( \frac{x}{y} \right) ^ 2$$ be an integer, which isn't necessarily the case and if it's not the case we've reached a contradiction.

Perhaps $$\left( \frac{x}{y} \right) ^ 2$$ is an integer (non-zero), in that case we're saying that there exists some integer $$a$$ such that $$a ^ 2 = p$$, but this is a contradiction as clearly the square of any non-zero integer is not prime, thus our original assumption that $$x ^ 2 - y ^ 2 p = 0$$ is false, and thus we can safely write our multiplicative inverse as

$\frac{x}{x ^ 2 - y ^ 2 p} - \frac{y}{x ^ 2 - y ^ 2 p} \sqrt p$
The Integers mod a Prime Form a Field
$$\mathbb{ Z } / \mathbb{ Z } _ n$$ is a field if and only if $$n$$ is prime.

$$\impliedby$$ Suppose that $$n$$ is prime, we recall that $$\mathbb{ Z } / n \mathbb{ Z }$$ is a crone. Therefore to show it's a field we have to show it has multiplicative inverses, so let $$\overline{ a } \in \mathbb{ Z } / n \mathbb{ Z } \setminus \left\{ \overline{ 0 } \right\}$$. Since $$\overline{ a } \neq \overline{ 0 }$$ then $$a \not\equiv 0 \left( \operatorname{ mod } n \right)$$ therefore $$n \nmid a$$ thus $$\gcd \left( n, a \right) = 1$$ and we must have $$s, t \in \mathbb{ Z }$$ such that $$1 = n s + at$$, taking this equation mod $$n$$ we see that $$at \equiv 1 \left( \operatorname{ mod } n \right)$$ in otherwords $$\overline{ a } \overline{ t } = \overline{ 1 }$$ so that $$\overline{ t }$$ is $$\overline{ a }$$'s multiplicative inverse so that $$\mathbb{ Z } / n \mathbb{ Z }$$ is a field.

$$\implies$$ We use the contrapositive so assume that $$n$$ is not prime, therefore $$n$$ is composite, so we have $$n = a b$$ for some $$a, b \in { 1, ... n - 1 }$$ so clearly $$\overline{ a } \neq \overline{ 0 }$$ and $$\overline{ b } \neq \overline{ 0 }$$ but at the same time $$n | ab$$ therefore $$\overline{ a } \overline{ b } = \overline{ 0 }$$, which shows that $$\mathbb{ Z } / n \mathbb{ Z }$$ is not a domain, therefore it cannot be a field, as needed for the contrapositive, so the original implication holds true