Define the finite field with 4 elements:
sage: F = GF(4)
sage: F
Finite Field in z2 of size 2^2
Failing any indication on how to name the generator of the finite field,
Sage names it z2 to indicate the extension is of degree two. So this z2
corresponds to $\omega$ in the question.
The polynomial used in constructing F is:
sage: F.polynomial()
z2^2 + z2 + 1
Define the polynomial ring over F in three variables x, y, z:
sage: R.<x, y, z> = PolynomialRing(F)
sage: R
Multivariate Polynomial Ring in x, y, z over Finite Field in z2 of size 2^2
then the two polynomials in the question:
sage: P = 1 + x + y + z
sage: Q = 1 + x*y + y*z + x*z
and finally the ideal I:
sage: I = R.ideal([P, Q])
sage: I
Ideal (x + y + z + 1, x*y + x*z + y*z + 1) of Multivariate Polynomial Ring in x, y, z over Finite Field in z2 of size 2^2
and check its dimension:
sage: I.dimension()
1
For ideals of dimension zero, the common zeros form a finite set of points,
obtained with I.variety().
Here the dimension is one, so we need to express solutions in terms of one parameter.
A useful tool for this is the resultant,
which can help eliminate one of the variables.
Compute the resultant of P and Q with respect to the variable z:
sage: res_P_Q = P.resultant(Q, z)
sage: res_P_Q
x^2 + x*y + y^2 + x + y + 1
From there either factor:
sage: res_P_Q.factor()
((z2 + 1)*x + y + (z2)) * ((z2)*x + y + (z2 + 1))
or notice that the change of variables x = 1 + s and y = 1 + t
brings the resultant to a simpler form:
sage: res_P_Q.subs({x: 1 + x, y: 1 + y})
x^2 + x*y + y^2
(where x and y in the output here stand for $s$ and $t$ from the question).
From either of these one can figure out that solutions are
$t = \omega s$, with $\omega$ satisfying $\omega^2 + \omega + 1 = 0$.
This translates to $x = 1 + s$, $y = 1 + \omega s$.
Then, solving for $z$ using $P = 0$ or $Q = 0$.
sage: 1 + (1 + x) + (1 + F.gen() * x)
(z2 + 1)*x + 1
so $z = 1 + (\omega + 1) s$, that is, $z = 1 + \omega^2 s$.
