Finite Fields $\mathbb{F}_{p}$

Finite Fields ``sets of elements with addition" Finite fields are of the form

$\mathbb{F}_{p} = \overbrace{\{0, 1, 2, 3, ... p-1\}}^{p-elements}$
addition is a usual operation, except multiples of p should be left out of the set (modulo p). This type of field should not be confused with finite fields $\mathbb{F}_{p^n}$ because $\mathbb{F}_{p^n}$ also has multiplication as an operation. Vectors over $\mathbb{F}_{p}$ are denoted $\mathbb{F}_{p}^n$, where $n$ is the number of elements in the vector. The sum of any two vectors in $\FF[p]^n$ would be mod p. Ex: Adding Vectors Over $\mathbb{F}_{p}$

$\mathbb{F}_{5}={0, 1, 2, 3, 4}$

$\mathbb{F}_{5}^4$ has 4 elements in each vector

$$\left( \begin{array}{c}1 2 3 4 \end{array}\right) + \left(\begin{array}{c}0 1 2 3 \end{array}\right) \equiv \left( \begin{array}{c}1 3 5 7 \end{array}\right) \textrm{now take this vector mod 5}$$

$$\equiv \left( \begin{array}{c}1 3 0 2 \end{array}\right)$$

$\mathbb{F}_{5}^6$ has 6 elements

$$\left( \begin{array}{c}0 1 4 3 2 0 \end{array}\right) + \left( \begin{array}{c} 1 2 4 3 4 0 \end{array}\right) \equiv \left( \begin{array}{c} 1 3 8 6 6 0 \end{array}\right) \equiv \left( \begin{array}{c}1 3 3 1 1 0 \end{array}\right)$$

The additive identity is true for $\mathbb{F}_{p}$, so $x+0=0$. For every element $x$ there is another element $x^{-1}$ where $x+x^{-1}=0$.
Ex: Additive Identity
$3+(-3)\equiv$3mod(5) + (-3)mod(5)$\equiv$ 3 + 2$\equiv$ 0
$1+(-1)\equiv$1mod(5) + (-3)mod(5)$\equiv$ 1 + 4$\equiv$ 0
$2+(-3)\equiv$2mod(5) + (-3)mod(5)$\equiv$ 2 + 2$\neq$ 0