Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Zorn does seem promising. –  Dylan Moreland Jan 19 '12 at 13:50
1  
What stops you from applying Zorn's lemma? –  Tom Bachmann Jan 19 '12 at 13:51
1  
I don´t know how to use it D: –  August Jan 19 '12 at 13:53
1  
@August Do you know the proof that every non-zero ring has a maximal ideal? –  Dylan Moreland Jan 19 '12 at 13:54
1  
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! –  Pierre-Yves Gaillard Jan 19 '12 at 13:55

2 Answers 2

up vote 10 down vote accepted

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 minimial element, by Zorn's lemma.

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

share|improve this answer
    
Dear Alex: What's a "downward chain"? –  Pierre-Yves Gaillard Jan 19 '12 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. –  Alex Youcis Jan 19 '12 at 14:01
    
Dear Alex: Thanks for replying to my comment, and +1 for your nice answer! –  Pierre-Yves Gaillard Jan 19 '12 at 14:04

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. enter image description here enter image description here enter image description here

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.