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Given a ring $A$, By the chain of prime ideals of $A$, we mean a sequence of prime ideals of $A$ such that $P_0 \subsetneq P_1 \dots \subsetneq P_{n-1}\subsetneq P_n$. Further this chain has a length $n$.

Define the dimension of a ring $A$ by $\operatorname{Dim}(A)$ to the supremum the lengths of all chains of prime ideals of $A$.

Show that if $\operatorname{Dim}(A)=0$ then every prime ideal is maximal.

Attempt: Suppose that there exists a prime ideal $X$ which is not maximal, then there exists an ideal $Y\subsetneq A$ properly containing $X$. Now my aim is to show that there exist a sequence of prime ideals $P_0 \subsetneq P_1 \dots \subsetneq P_{n-1}\subsetneq P_n$ with $n \geq 1$. I am getting $Y$ as an ideal. I am not getting any prime ideal after that. How should I proceed from here?

Also I cannot use $0=\{0\}$ since $0$ is not a prime ideal.

GraduateStudent
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    Choose $Y$ to be maximal. – user6 Jun 03 '20 at 12:21
  • By definition I am getting $Y$ to be a proper ideal of $A$. How can I make it a maximal ideal? – GraduateStudent Jun 03 '20 at 12:27
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    Assuming your ring has the multiplicative identity, every proper ideal is contained in a maximal ideal(use Zorn's lemma). – cqfd Jun 03 '20 at 12:28
  • When I searched, I immediately found someone had the exact same problem as you. And I mean that they experienced the exact same sticking point, not just the same question. Please search first next time. – rschwieb Jun 03 '20 at 13:25

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