# Is there a 'maximal prime ideal' contained in a given ideal?

It is well known, that every ring (commutative and unital) which is not the zero ring has at least one maximal (consequently prime) ideal.

Let $R$ be a commutative and unital ring and $I\neq 0$ an ideal. Does there exist at least one maximal prime ideal $P$ with $P\subseteq I$?

More precisely, does there exist a prime ideal $P$ such that for every other prime ideal $Q$ with $P\subseteq Q\subseteq I$ one has $P=Q$ or $Q=I$?

How about the following modification (which is the same if $I=R$): Does there exist a prime ideal $P$ such that for every other prime ideal $Q$ with $P\subseteq Q\subseteq I$ one has $P=Q$ or $Q=I$?

Any nested union of prime ideals bounded above by $I$ is a prime ideal bounded above by $I$. So, if there are any prime ideals at all in the poset of prime ideals bounded by $I$, then by Zorn's lemma there is a maximal one.
• Thank you Hurkyl. Is it possible that there are no prime ideals at all contained in $I\neq 0$? – user8463524 Sep 28 '17 at 19:28
• @user8463524: Yes. Note for this to happen we need a ring where $(0)$ is not prime. Let $F$ be a field. Then you can take $R = F[x]/(x^3)$ and $I = (x^2)$. The only prime ideal is $(x)$, and $(x) \not\subseteq (x^2)$. – user14972 Sep 28 '17 at 19:39
In the ring $\mathbb Q[x,y,z]:=\mathbb Q[X,Y,Z]/\langle XY,XZ,YZ\rangle$ the ideal $I=\langle z\rangle$ does not contain any prime ideal.