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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


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


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


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

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


Subsections
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Next: Finite Fields Up: Summer 2004 Final Report Previous: Contents   Contents
Frederick Leitner 2004-09-01