For n=1, fields are of the formDenote , where :

is a prime

is a positive integer

has elements

multiplication and addition are usual operations, except multiples of p should be left out of the set (modulo p).

__Ex__

We use the sign because the result is modulo p. That means that we are dividing the product by and the answer is the remainder of that division. All of the multiplication in the example is mod 5 because .

__Multiplicative Inverse__ - for every
*non- zero*
element, ,

there is a multiplicative inverse, .

All of the operations below are (mod 5).

__Additive Identity__ -same as subtraction

Having a prime {0, 1, 2, 3...p-1}

guarantees a multiplicative inverse.

__No Inverse Example__ ``" ``", 6 is
not prime.

{0, 1, 2, 3, 4, 5}

-Is there an inverse of 2?
A zero divisor is not what we want because the product of two positive numbers should not be 0

We tested the product of 2 and every number in the field, but we never got a product of 1. This means that there is no multiplicative inverse.

( elements)

The reason cannot have those elements is because not all elements will have multiplicative inverses.

__Explanation:__

,

In this field
3 is a
zero divisor = ``bad"

Construction of omplex numbers from eal numbers

The field of complex numbers is denoted and

This field has the multiplication rule:

Another way is to look at the polynomials in , denoted , modulo the quadratic .

which is the same as the multiplication
rule

0 | |||

__ Example for the Goal__ : Figure out
, the field
with 4 elements.

Note: sits inside of

This is a list of all the quadratic polynomials ``over = {0, 1}''Start with - look at polynomials in

in fact look for just quadratic polynomials,

with the condition:

Condition- has no zeros in .

f(x) | f(0) | f(1) |

0 | 1 | |

1 | 0 | |

0 | 0 | |

1 | 1 |

construct =

a+c= 0 c= 0

b+d= b d= 0

0 is still the additive identity.

(a+bx) has the additive inverse -a-bx.

this seems to be a problem because only has linear polynomials but we ended up with a quadratic one.

replacement rule, anytime you see , replace with because

0 | |||

When constructing the elements of remember had multiplication modulo p = 2, so has polynomial multiplication modulo . Also sits inside of as the ''constant'' polynomials.

We end up with the multiplication table of :

If we name and , then the table will look like this:

0 | 1 | |||

0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | ||

0 | 1 | |||

0 | 1 |

As a vector space,

To find the multiplication table we need a monic-quadratic that has no zeros in

A monic-quadratic will have a coefficient of 1 on the highest degree term.

Also note that we only need one polynomial that works.

Multiplication table for ,

0 | 1 | 2 | x | x+1 | x+2 | 2x | 2x+1 | 2x+2 | |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 2 | x | x+1 | x+2 | 2x | 2x+1 | 2x+2 |

2 | 0 | 2 | 1 | 2x | 2x+2 | 2x+1 | x | x+2 | x+1 |

x | 0 | x | 2x | 2 | x+2 | 2x+2 | 1 | x+1 | 2x+1 |

x+1 | 0 | x+1 | 2x+2 | x+2 | 2x | 1 | 2x+1 | 2 | x |

x+2 | 0 | x+2 | 2x+1 | 2x+2 | 1 | x | x+1 | 2x | 2 |

2x | 0 | 2x | x | 1 | 2x+1 | x+1 | 2 | 2x+2 | x+2 |

2x+1 | 0 | 2x+1 | x+2 | x+1 | 2 | 2x | 2x+2 | x | 1 |

2x+2 | 0 | 2x+2 | x+1 | 2x+1 | x | 2 | x+2 | 1 | 2x |

__Construct__
out of
.

We need a monic-cubic polynomial with no zeros/roots in
. The polynomial will be of the form
, but with no
zeros.

__Try__ a=1, b=0.
, has no zeros over
.
.

(
as vector spaces)

Multiplication Table of
(0,1 columns and rows excluded)

x | x+1 | |||||

x | 1 | |||||

x+1 | 1 | x | ||||

1 | x+1 | x | ||||

x | x+1 | 1 | ||||

1 | x | x+1 | ||||

1 | x |

__Sample multiplication__