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