Vtyjkyuk

This is inspired by an exercise in Eisenbud's text on commutative algebra. In the text, he gives the following alternative characterization of localization: for any commutative ring $R$ with unity and any subset $U$ of $R$ (not necessarily multiplicatively closed as the text defines $R[U^-1] = R[barU^-1]$), there is a natural isomorphism

beginalign
R[U^-1] cong R[x_u_u in U]/langleux_u-1_u in Urangle.
endalign

Applying this to $R = mathbbZ$ and $U = 2,3,6$, this isomorphism becomes

beginalign
mathbbZ[U^-1] cong mathbbZ[x,y,z]/langle 2x-1, 3y-1, 6z-1rangle.
endalign

Intuitively, $x = 1/2$, $y = 1/3$, and $z = 1/6$. It seems natural, then, that $z-xy$ should be in the ideal $I = langle 2x-1, 3y-1, 6z-1rangle$ of $mathbbZ[x,y,z]$. I used Sage to verify this, but I can't think of a way to express $z-xy$ in terms of the generators of $I$ (note carefully, that I'm considering $I$ as an ideal of $mathbbZ[x,y,z]$)

beginalign
z-xy&=z-6xyz+xy(6z-1)\
&=z-(2x)(3y)z+xy(6z-1)\
&=z-3(2x-1)yz-3yz+xy(6z-1)\
&=-(3y-1)z-3(2x-1)yz+xy(6z-1)
endalign

Note that $(6z-1)-2x(3y-1)-(2x-1)=6z-6xy$ is in the ideal

Now using the intuition $z=frac 16$ we multiply by $z$ to find also $6z^2-6xyz$ in the ideal and this is $z(6z-1)-xy(6z-1)+z-xy$

You should be able to work backwards to the expression you need.