Let $K$ be a field and consider the ideal $I=(xz,yz)$ of $K[x,y,z]$. Find a primary decomposition of $I$.
Now I'd say the primary decomposition is $(z)\cap (x,y)$. To see that, we clearly have $(xz,yz)\subseteq (z)\cap (x,y)$ since each of the generators lies in each of the ideals. For the other way, I'm a little uncertain. My approach is negating the expression. So take an element $f$ which is not in $(xz,yz)$. This means that either $f$ consists of terms with $z$, or it consists of terms with $xy$, or it consists of constant terms. In either three cases, $f$ will also not be in $(z)\cap (x,y)$, so $(z)\cap (x,y)\subseteq (xz,yz)$. Is this correct?
Also I need to show that $(z)$ and $(x,y)$ are primary ideals. I do this by showing they are prime. $(x,y)$ is prime since $K[x,y,z]/(x,y)\cong K[z]$ is a integral domain, since $K$ is a field. For $(z)$, take $fg\in (z)$, so $fg=hz$ for some $h\in K[x,y,z]$. Assume $f\notin (z)$, so $f$ have a term, without $z$, but since $fg=hz$, then $g$ must have $z$ in all of it's terms, and so $g\in (z)$. Are this valid?