Detached mathematics Bound fields

 25 June 05:18   Anamnesis from the antecedent area that we advised the case area F/ akin to modular addition but with polynomials, and that if we are searching at numbers modulo n, we accept a acreage iff Zn is a acreage if n is prime.

    Can we say something agnate about F/? Indeed, if m(x) is irreducible then F/ is a field.

    This area deals with these kinds of fields, accepted as a bound field.

    We accept the item F/ area this is the set of polynomials in F are disconnected by the polynomial m(x).

    Of the elements in F/ we can calmly ascertain addition, subtraction, multiplication, analysis and so on commonly but with a abridgement modulo m(x) to get the adapted remainder.

    We accept that F/ is a capricious ring with identity, and if m(x) is irreducible then F/ is a field.

    If m(x) has amount n, then

    :F/=

    If F is Zp (so p is prime) then |F/|=pⁿ

    Now bethink with circuitous numbers C, we accept invented the attribute i to angle for the basis of the band-aid x²+1=0. In fact, we accept C=R/2+1>.

    When we accept a accepted bound field, we can do this also. We address this generally as F/=F(α) area α is the basis of m(x) - we ascertain α to be the basis of m(x).

    F(α) in actuality is the aboriginal acreage which contains F and α.

    We accept a amount of theorems associated with bound fields.

    # If F is a bound field, |F|=q, then q=p^k for some k > 1 and p prime.

    # There then is a monic irreducible polynomial m(x) with amount k such that F=Zp/=Zp(α) with α a basis of m(x) over Zp

    # There is an aspect γ∈F such that the adjustment (the atomic aspect n such that γⁿ=1) of γ is q-1, so γ is archaic in F, and we can accomplish elements of F (not zero) from admiral of γ, ie F=

    # F is a with ambit k over Zp. It has base area n is the amount of m(x), so we accept F=

    # If a∈F, a+...+a p times (pa) is 0.

    # If m2(x) is any additional irreducible polynomial of amount k over Zp then F is isomorphic (meaning basically according to, except for a change in symbols) to Zp/ - so all means of autograph this acreage are basically the same. So there is about one acreage of admeasurement q=p^k and we denote it GF(p^k) (GF acceptation Evariste Galois Field).

    Lets attending at a few examples that go through these ideas.

    , briefly, are numbers in the form

    : $a\; +\; bi$

    where i is the band-aid to the blueprint x²+1=0

    These numbers in actuality anatomy a field, about it is not a bound field.

    Take m(x)=x²+1, with the acreage F getting R. Then we can anatomy the circuitous numbers as F/. Now F/ = because the remainders haveto be of amount beneath than m(x) - which is 2.

    So then (a+bx)(c+dx)=ac+bdx²+(ad+bc)x.

    But bethink that we are alive in F/. So x² modulo x²+1, can be accounting as (x²+1)-1=-1, and substituting -1 aloft yields a rather accustomed expression.

    If we let the attribute i to be the basis of x²+1, then i²+1=0 and i²=-1. The blow of the acreage axioms chase from here. We can then say the circuitous numbers C=R/2+1>=R(i).

    We can still do this for some acreage in general. Lets yield Z3 for example, and aces m(x)=x²+x+2. m(x) is irreducible - m(0)=2, m(1)=4=1, m(2)=4+2+2=8=2.

    So Z3/2+x+2> is a bound field. Accept α is a basis of m(x). Then Z3(α) = . Back Z3/2+x+2> is finite, we can account out all its elements. We accept the connected terms, then the α terms, then the α+constant terms, and so on. We accept .

    Now we accept α²+α+2=0, then

    : α²=-α-2

    : α²=2α-2=2α+1

    (Recall the coefficients are in Z3! We are alive in Z3/)

    We can verify multiplication works mod m(x) - for example

    : (1+2α)(2+α) = 2 + α+4α+2α²

    Reducing coefficients commonly mod 3 we get

    : (1+2α)(2+α) = 2 + 2α + 2α²

    Now using the blueprint aloft for α²

    : (1+2α)(2+α)

    :: = 2 + 2α + 2(2α+1)

    :: = 2 + 2α+4α+2

    :: = 2 + 6α+2

    :: = 2 + 2 = 4 = 1

    Verify for yourself that multiplication and additional operations plan too.

    Recall from that the adjustment of a amount a modulo n, in a field, is the atomic ability such that a^k-1=1 in Zn, with k the admeasurement of this field. Back the adjustment is authentic over a field, this leads us to accede whether we accept archaic elements in F/ - which we do. If we accept F(α), just like in Zn, α is archaic iff the adjustment of α is q-1 area q is the amount of elements in F/.

    Lets yield Z2/2+x+1>. Is α (root of x²+x+1) primitive?

    First, if α is a basis of x²+x+1,

    :α²+α+1=0

    :α²=-α-1=α+1

    Now, let us account admiral of α

    : 1, α

    : α²=α+1

    : α³=α(α²)=α(α+1)=α²+α=(α+1)+α=1

    Recall that the admeasurement of this acreage is 4 (the n in Zn, in this case, 2, aloft to the ability of the amount of the polynomial, in this case 2). Now we accept α³=α^4-1=1, and α is primitive.

    We about wish to attending at admiral of α in F(α), to see whether they are primitive, back we already understand about the orders of the constants in F(α) - which weve looked at in .