Generators of the intersection of twelve ideals. I have to show that the intersection of the twelve ideals $$(X\pm 1, Y\pm 1), (X\pm 1, Z\pm 1), (Y\pm 1, Z\pm 1) \subset \mathbb{R}[X,Y,Z]$$ is the ideal $$\Big( (X^2 −1)(Y^2 −1),(X^2 −1)(Z^2 −1),(Y^2 −1)(Z^2 −1)\Big).$$ Is there an easy way to do it, maybe some combinatorial way or something?
Thank you.  
 A: Let $m$ be an element of the intersection of the twelve ideals. At first we have $m \in (x\pm1,y\pm1)$, so minimally we must have:
$y-1|m \lor x-1|m ,y+1|m \lor x-1|m  ,y-1|m \lor x+1|m  ,y+1|m \lor x+1|m$
There is a symmetry among these conditions, so assume WLOG that $y-1|m$, which satisfies the first and third minimal conditions. In order to satisfy the second and fourth, we must either have $y+1|m$ or $x^2-1|m$ by inspection. Overall if $m \in (x\pm1,y\pm1)$, then $x^2-1|m$ or $y^2-1|m$. By applying the same logic to $(x\pm1,z\pm1)$ and $(y\pm1,z\pm1)$ we must have the following:
$x^2-1|m \lor y^2-1|m, z^2-1|m \lor y^2-1|m, x^2-1|m \lor z^2-1|m$
Again we can assume WLOG that $x^2-1|m$, in which case we must also have $z^2-1|m \lor y^2-1|m$ in order to satisfy the second condition. Therefore we have shown that $(y^2-1)(x^2-1)|m $ or $(z^2-1)(x^2-1)|m $ or $(y^2-1)(z^2-1)|m $ by symmetry and so $m \in ((X^2−1)(Y^2−1),(X^2−1)(Z^2−1),(Y^2−1)(Z^2−1)).$
It remains to show that if $n \in ((X^2−1)(Y^2−1),(X^2−1)(Z^2−1),(Y^2−1)(Z^2−1))$, then $n$ lies in the intersection of the twelve ideals, which is straightforward. After this we have basically established a one to one relationship between the intersection and $((X^2−1)(Y^2−1),(X^2−1)(Z^2−1),(Y^2−1)(Z^2−1))$
