# 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

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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