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 donΒ΄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 donΒ΄t know how to proceed. Probably it's a Zorn application.
Best Answer
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.