Intersection of ideals $(x,y),(y,z)$ and $(x,z)$ in $K[x,y,z]$

Question

I have to prove that the intersection of ideals $(x,y),(y,z)$ and $(x,z)$ is equal to the ideal generated by $xy ,yz$ and $xz$. I am unable to prove it using the definitions only.

Answer 1

Since $xy, yz, xz \in (x,y)\cap (y, z) \cap (x, z)$, clearly $(xy, yz, xz) \subset (x,y)\cap (y, z) \cap (x, z)$. Suppose now that $p(x,y,z)\in (x,y)\cap (y, z) \cap (x, z)$. Fix a monomial term, $m(x,y,z)$ of $p(x,y,z)$. Since $p(x,y,z)\in (x,y)$ we have $p(x,y,z)=xp_1(x,y,z)+yp_2(x,y,z)$ for some polynomials $p_1$ and $p_2$ which means that $m$ must either contain $x$ or $y$ as a factor. Suppose without loss of generality that it contains $x$ as a factor. Then we just observe that since $p(x,y,z)\in (y, z)$ also, a symmetric argument to the one above shows that $m$ either contains $y$ or $z$ as a factor. If the former is true, $m$ then contains $xy$ as a factor, if the latter is true it contains $xz$ as a factor. Since $m$ was an arbitrary term in $p$ this shows that each term in $p$ contains $xy$, $yz$ or $xz$ as a factor, i.e. $p(x,y,z)\in (xy, yz, xz)$