Hi!

I need help for solve this problem:

Construct a field with four elements by adjoining a suitable root of x^{4}-x to Z_{2}

Tks..

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- Dec 2nd 2012, 11:15 AMcharles01Finite fields
Hi!

I need help for solve this problem:

Construct a field with four elements by adjoining a suitable root of x^{4}-x to Z_{2}

Tks.. - Dec 2nd 2012, 12:06 PMtopsquarkRe: Finite fields
- Dec 2nd 2012, 12:19 PMcharles01Re: Finite fields
The new possibilities are $\displaystyle (-1+i\sqrt 3)/2$ and $\displaystyle (-1-i\sqrt 3)/2$?? but their product is zero in Z

_{2}.

Tks - Dec 2nd 2012, 12:30 PMPlatoRe: Finite fields
- Dec 2nd 2012, 08:54 PMDevenoRe: Finite fields
this makes no sense at all: 1/2 is undefined in a field of characteristic 2 (such as Z

_{2}).

instead, let's factor x^{4}- x into irreducible factors:

x^{4}- x = x(x - 1)(x^{2}+ x + 1)

it is obvious that the linear factors x and x - 1 are irreducible (and have the roots x = 0 and x = 1 in Z_{2}), to show that x^{2}+ x + 1 is irreducible over Z_{2}, it is sufficent to show it has no linear factors, that is: no roots in Z_{2}:

0^{2}+ 0 + 1 = 0 + 1 = 1 ≠ 0

1^{2}+ 1 + 1 = (1 + 1) + 1 = 0 + 1 = 1 ≠ 0.

let u be a root of x^{2}+ x + 1. just to be clear, what we are REALLY doing is taking the quotient ring:

Z_{2}[x]/(x^{2}+ x + 1), the ring of ALL polynomials in Z_{2}, modulo the ideal generated by the polynomial x^{2}+ x + 1.

we embed the field Z_{2}in this quotient by mapping a to the coset a + I (where we are taking I = (x^{2}+ x + 1) because i'm lazy and hate typing a polynomial over and over again), for any element a in Z_{2}(there's only 2 of them)).

in this quotient ring u is the coset of x. it is clear that this actually gives a root:

u^{2}+ u + 1 = (x + I)^{2}+ (x + I) + (1 + 1) = (x^{2}+ I) + (x + I) + (1 + I) = x^{2}+ x + 1 + I = I = 0 + I, since x^{2}+ x + 1 is IN I.

since it's tiresome to keep running "+I" after everything, we'll drop it (just remember that "a" means a+I, and "u" means x+I).

now in our new quotient ring, we have just 4 distinct cosets:

I = 0 + I (the identity of the ring Z_{2}[x]/I), which we are just going to call "0".

1 + I , which we are just going to call "1" (this is indeed the multiplicative identity of Z_{2}[x]/I, verify this!).

x + I, which we are going to call "u"

(x+1) + I, which we are going to call "u+1".

note that for any polynomial of degree 2 or more, say f(x), we can turn it into one of the 4 cosets above by writing:

f(x) = q(x)(x^{2}+ x + 1) + r(x), where deg(r) < 2, or r = 0 whence:

f(x) + I = (q(x) + I)(I) + (r(x) + I) = r(x) + I (since I is the "0" of the ring Z_{2}/I).

and the polynomials 0,1,x, and x+1 are the only elements of Z_{2}[x] of degree < 2, together with 0.

now, we can use some "high-powered" ring theory, to establish that {0,1,u,u+1} is a field by showing that (x^{2}+x+1) is a maximal ideal, because x^{2}+x+1 is irreducible over Z_{2}, but we can just show that our 4-element set is a field DIRECTLY.

it is immediate that Z_{2}[x]/I is a ring, being a quotient ring. so we only need to verify that the set {1,u,u+1} forms a group under multiplication.

since we already know (since we're in a ring) the multiplication is associative, we only need to show closure, and inverses.

before we start:

since u is a root of x^{2}+ x + 1, we know that u^{2}+ u + 1 = 0, hence u^{2}= -(u + 1) = u + 1 (since we are in a ring of characteristic 2).

now:

(1)(1) = 1

(1)(u) = u

(1)(u+1) = u+1

(u)(u) = u^{2}= u+1

(u)(u+1) = u^{2}+u = (u+1)+u = (u+u)+1 = 0+1 = 1

(these are all the products we need to check, since Z_{2}[x] is a commutative ring, hence so is Z_{2}[x]/I).

thus we have closure.

finally, we see that:

(1)(1) = 1, so 1 has an inverse (itself).

(u)(u+1) = 1, so u and u+1 are inverses of each other.

note that we could, if we desired, write {1,u,u+1} as {1,u,u^{2}}, that is u is a generator for this group and:

u^{3}= (u)(u^{2}) = (u)(u+1) = 1, so u is of order 3, while:

(u+1)^{3}= (u+1)(u+1)^{2}= (u+1)(u^{2}+1) = (u+1)((u+1)+1) = (u+1)(u) = 1, so u+1 is ALSO of order 3.

this is no accident, the non-zero elements of a finite field ALWAYS form a cyclic group. in this case, u is a "primitive element" of our field of four elements. - Dec 4th 2012, 02:21 AMcharles01Re: Finite fields
Thank you so much for the help Deveno! :)