# Every non zero commutative ring with identity has a minimal prime.

Let $A$ be a non zero commutative ring with identity. Show that the set of prime ideals of $A$ has minimal elements with respect to inclusion.

I don´t know how to prove that, I can suppose that the ring is an integral domain, otherwise the ideal $(0)$ is a prime ideal , but I don´t know how to proceed. Probably it's a Zorn application.

• Zorn does seem promising. Jan 19, 2012 at 13:50
• What stops you from applying Zorn's lemma? Jan 19, 2012 at 13:51
• I don´t know how to use it D: Jan 19, 2012 at 13:53
• @August Do you know the proof that every non-zero ring has a maximal ideal? Jan 19, 2012 at 13:54
• Dear August: When you write "I can suppose that the ring is an integral domain", I think you mean "I can suppose that the ring is not an integral domain". --- +1 to Dylan's comments! Jan 19, 2012 at 13:55

Right, it's Zorn's lemma. Namely, show that the intersection of any downward chain of prime ideals is prime, and use Zorn's lemma to conclude that $\text{Spec}(A)$ has a minimal element.

Just in case you're having difficulty proving the statement about the intersections suppose that $\Omega$ is a downward chain of prime ideals and let $\mathfrak{P}$ be the intersection of all the members of $\Omega$.

Since the intersection of ideals are ideals, it suffices to show that $\mathfrak{P}$ is prime. To do this suppose that $ab\in\mathfrak{P}$ but neither $a$ nor $b$ was. Since $a$ nor $b$ is in $\mathfrak{P}$ we can find two prime ideals $\mathfrak{p},\mathfrak{p}'\in\Omega$ such that $a\notin\mathfrak{p}$ and $b\notin\mathfrak{p}'$. Since $\Omega$ is a downward chain we may assume without loss of generality that $\mathfrak{p}\subseteq\mathfrak{p}'$ so that $a,b\notin\mathfrak{p}$.

That said, since $ab\in\mathfrak{P}$ we know that $ab\in\mathfrak{p}$ which contradicts that $\mathfrak{p}$ is prime. Thus, we see that $ab\in\mathfrak{P}$ implies either $a\in\mathfrak{P}$ or $b\in\mathfrak{P}$ and so $\mathfrak{P}$ is prime. Since $\Omega$ was arbitrary it follows that $\text{Spec}(A)$ has a minimal element, by Zorn's lemma.

Remark: I left out a very small detail in the above proof that you should find and add.

• Dear Alex: What's a "downward chain"? Jan 19, 2012 at 13:58
• @Pierre-YvesGaillard Haha, it's just a habit of mine, when thinking about a poset with the opposite ordering. Of course, a chain is a chain, there is no 'directionality'--I probably have this linked when discussing chains in downward directed sets. Jan 19, 2012 at 14:01
• Dear Alex: Thanks for replying to my comment, and +1 for your nice answer! Jan 19, 2012 at 14:04
• @AlexYoucis, If I may ask you a question: how to show that the set $\{ J \in \textrm{Spec(R)} : I \subset J \}$ has minimal elements ? Appreciate any help.
– user358174
Nov 26, 2016 at 2:20
• @AlexYoucis,what is the detail left out in the proof? Sep 8, 2020 at 7:55

Below is a hint, with further remarks on the structure of the set of prime ideals, from Kaplansky's excellent textbook Commutative Rings. For a recent survey on the poset structure of prime ideals in commutative rings see R & S Wiegand, Prime ideals in Noetherian rings: a survey, in T. Albu, Ring and Module Theory, 2010.   