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Finite Fields $ \FF[p]$

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

$ \FF[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 $ \FF[p^n]$ because $ \FF[p^n]$ also has multiplication as an operation. Vectors over $ \FF[p]$ are denoted $ \FF[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 $ \FF[p]$

$ \FF[5]={0, 1, 2, 3, 4}$

$ \FF[5]^4$ has 4 elements in each vector


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


$ \FF[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 $ \FF[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

next up previous contents
Next: Vector Spaces Up: Spring 2004 Final Report Previous: Contents   Contents
Frederick Leitner 2004-05-12