Ideals in a UFD Consider the ideal $I=(ux,uy,vx,uv)$ in the polynomial Ring $\mathbb Q[u,v,x,y]$, where $u,v,x,y$ are indeterminates. Prove that every prime Ideal containing I contains the Ideal $(x,y)$ or the Ideal $(u,v)$.
I am not able to choose the correct combinations of products of the four indeterminates to arrive at the answer.
 A: By definition, if $\mathfrak p\subseteq\Bbb Q[u,v,x,y]$ is a prime ideal and $ab\in\mathfrak p$ then either $a\in\mathfrak p$ or $b\in\mathfrak p$.
If $(ux,uy,vx,uv)\in\mathfrak p$ then $ux\in\mathfrak p$. We then get some possibilities:

*

*If $u\in\mathfrak p$ then also $uy,uv\in\mathfrak p$. We only need to worry about $vx$. If $v\in\mathfrak p$ we get the prime ideal $P_1=(u,v)$.


*If $x\in\mathfrak p$, we get the prime ideal $P_2=(u,x)$.


*If $ux\in\mathfrak p$ and $x\in\mathfrak p$, then also $vx\in\mathfrak p$. We only have to worry about $uy,uv\in\mathfrak p$. $uy\in\mathfrak p$, we have already considered the case when $u\in\mathfrak p$, if $y\in\mathfrak p$ we get the prime idel $P_3=(x,y)$. We get the last case when considering $uv$.


*$P_4=(x,v)$.
All of the ideals $P_1,\ldots,P_4$ are prime, since $\Bbb Q[u,v,x,y]/P_i$ is an integral domain.

It looks like your proposition is not correct. It may be, as suggested in the comments that the ideal is in fact $$I=(ux,uy,vx,vy)$$ Then we would also have $I=(u,v)\cap (x,y)\subset (u,v), (x,y)$.
