Re: (finite)galois field

On 6 Apr., 15:27, oercim <oer...@xxxxxxxxx> wrote:
Hello, thanks to everyone for helping. Now i have a better
understanding. But i have still a lot of things that i cant
understand. I am sorry i coulndt replay since i havent any time last
days.

Galois showed that in general, given any prime p and any positive
integer n, there is one and only one field that can be constructed on
a set with p^n elements; and moreover, that if m is any positive
integer, then there is a field with exactly m elements if and only if
m=p^n for some prime p and positive integer n. These fields are
constructed by starting with F_p, and then "adding" the roots of some
polynomial equation of degree n that is irreducible over F_p. For
example, to get a field of order 3^2=9, you start with F_3, then take
a polynomial equation of degree 2 that is irreducible (x^2+1 = 0
works). Let a be an element such that a^2 + 1 = 0, that is, a
"solution" to this equation. Then the set

{0, 1, 2, a, 2a, 1+a, 1+2a, 2+a, 2+2a}

How did u construct this field?

You start with F_3 and add an element a which is not in F_3
and which we shall treat like an unkknown.
To have a field, we must have 1+a and 2+a as well, and of course
a+a =2a (note however that a+a+a = (1+1+1)*a = 0).
And of course we must have 1+2a and 2+2a.
That completes the list given above.
Note that all these expressions are of the form u+v*a with u,v in F_3
and in fact we have listed *all* possible such expressions.
Adding any two such expressions of course produces
again an expression of that form.
Also, the negative is of the given form.
Interestingly, formally multiplying two such expressions seems
to produce expressions of a different kind at first sight
because it involves an a^2 term.
But since we have the equation a^2 + 1 = 0, we can replace
a^2 = -1 = 2, we end up in our set of 9 elements again.

This makes our set a ring.

Finally we observe that the product of two elements in our list
never is 0, unless one of the factors is 0.
For assume that (u+v*a)*(r+s*a) = 0 in our ring,
then the polynomial (u+v*X)*(r+s*X) must be a multiple of X^2 +1
(otherwise we would simplify the expression to 0).
But X^2+1 was chosen because i is irreducible, hence
this happens only if one of the factors is 0.

As a consequence multiplication with a fixed non-zero element is an
injective mapping, hence (by finiteness) is bijective,
hence multiplicative inverses exist. We have a field.

Isnt there any other field?

You might start with a different polynomial than X^2+1.
As long as it is of degree 2 and irreducible,
the argument above applies and you obtain a different
field with 9 elements.
They are not too different, though: Any two fields with 9 elements
are isomorphic.

Similarly you can find - up to isomorphism - exactly one field
with p^n elements for any prime power p^n (n>=1).

why?
And for what do I use this field?
Can Galois fields be used in solutions of linear equations such
that the unknowns and coefficients are integers where number of
unknowns are less than the equations? Thanks a lot.

Finite fields have many aplications in math, but not immediately
in the way you suggest.
.

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