Ideal intersection in Boolean polynomial ring

Question

I'm trying to calculate

$ \langle x,y,z\rangle \cap \langle x+1,y+1,z+1\rangle $

in the ring of Boolean polynomials. In CoCoA initially I set the ring as

Use XYZ ::= QQ[x,y,z];

(which is the default ring anyway). Then,

I := ideal(x,y,z);
J:=ideal(x+1,y+1,z+1);
intersection(I,J);

outputs

ideal(y -z, x -z, z^2 +z)

Now, in a Boolean ring multiplication is idempotent and addition is cancellative, so I'm expecting z^2+z gone. Yet, setting

Use BXYZ := Z/(2)(x,y,z);

or

Use BXYZ := ZZ/(2)(x,y,z);

didn't change the ideal intersection. What is correct way to set the ring so that the answer is

ideal(y+z, x+z)

?

Answer 1

It looks a little bit like you believe that $z^2$ ought to be $z$, and then $z^2+z=0$.

It appears you are working in the polynomial ring $(\Bbb Z/2\Bbb Z)[x,y,z]$. This is certainly a ring of characteristic $2$, but it is not a Boolean ring at all. All the powers of $x$, all the powers of $y$ and all the powers of $z$ are distinct.

Calling these "Boolean polynomials" seems somewhat double edged for this reason. The polynomials are not themselves boolean. In fact $(\Bbb Z/2\Bbb Z)[x,y,z]$ is a domain and the only idempotent elements are $0$ and $1$.