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.