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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.

Mathematics Asked on November 12, 2021

1 Answers

One Answer

By definition, if $mathfrak psubseteqBbb Q[u,v,x,y]$ is a prime ideal and $abinmathfrak p$ then either $ainmathfrak p$ or $binmathfrak p$.

If $(ux,uy,vx,uv)inmathfrak p$ then $uxinmathfrak p$. We then get some possibilities:

  • If $uinmathfrak p$ then also $uy,uvinmathfrak p$. We only need to worry about $vx$. If $vinmathfrak p$ we get the prime ideal $P_1=(u,v)$.

  • If $xinmathfrak p$, we get the prime ideal $P_2=(u,x)$.

  • If $uxinmathfrak p$ and $xinmathfrak p$, then also $vxinmathfrak p$. We only have to worry about $uy,uvinmathfrak p$. $uyinmathfrak p$, we have already considered the case when $uinmathfrak p$, if $yinmathfrak p$ 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 $I=(u,v)cap (x,y)subset (u,v), (x,y)$.

Answered by cansomeonehelpmeout on November 12, 2021

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