Fields

Field

A field is a set

F

with two binary operations

\oplus︎, \otimes

such that for any

a, b, c \in F

we have

$\oplus︎, \otimes$ are both associative and commutative
Identities: There exist identity elements $0_{F} \neq 1_{F} \in F$ for $\oplus︎$ and $\otimes$ respectively
Additive Inverses: There exists an $- a \in F$ such that $a \oplus︎ - a = 0_{F}$
Multiplicative Inverses: If $a \neq 0$ there exists an $a^{- 1} \in F$ such that $a \otimes a^{- 1} = 1_{F}$
Distributivity: $\otimes$ distributes into $\oplus︎$

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

(R, \oplus︎, \otimes)

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 ℝ

ℚ (x) : = {a + b x : a, b \in ℚ} \subseteq ℝ

Multiplicative Inverse in a Finite Extension of the Rationals

Suppose that

p

is prime and that

x + y \sqrt{p} \in ℚ (\sqrt{p})

, 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 ℚ

Since

x + y \sqrt{p} \in ℝ

, and it's non-zero, then in

ℝ

we have an inverse:

1 / (x + y \sqrt{p})

, we can then rationalize that as follows

\begin{matrix} \frac{1}{x + y \sqrt{p}} & a m p; = \frac{1}{x + y \sqrt{p}} \cdot \frac{x - y \sqrt{p}}{x - y \sqrt{p}} \\ a m p; = \frac{x - y \sqrt{p}}{x^{2} - y^{2} p} \end{matrix}

Now let's observe the denominator $x^{2} - y^{2} p$ , if this equals zero, then we have the following ${(\frac{x}{y})}^{2} = p$ 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 ${(\frac{x}{y})}^{2}$ be an integer, which isn't necessarily the case and if it's not the case we've reached a contradiction.

Perhaps ${(\frac{x}{y})}^{2}$ 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

ℤ / ℤ_{n}

is a field if and only if

n

is prime.

$⟸$ Suppose that $n$ is prime, we recall that $ℤ / n ℤ$ is a crone. Therefore to show it's a field we have to show it has multiplicative inverses, so let $a \in ℤ / n ℤ ∖ {0}$ . Since $a \neq 0$ then $a \equiv̸ 0 (\mod n)$ therefore $n ∤ a$ thus $\gcd (n, a) = 1$ and we must have $s, t \in ℤ$ such that $1 = n s + a t$ , taking this equation mod $n$ we see that $a t \equiv 1 (\mod n)$ in otherwords $a t = 1$ so that $t$ is $a$ 's multiplicative inverse so that $ℤ / n ℤ$ is a field.

$⟹$ 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 $a \neq 0$ and $b \neq 0$ but at the same time $n | a b$ therefore $a b = 0$ , which shows that $ℤ / n ℤ$ is not a domain, therefore it cannot be a field, as needed for the contrapositive, so the original implication holds true