Exercise 1.8

Question

Let $A$ be a ring $
e 0$. Show that the set of prime ideals of $A$ has minimal elements with respect to inclusion.

Amit Awasthi · Accepted Answer

\begin{proof}
  We order the set $\operatorname{Spec} A$ of prime ideals by reverse inclusion. $\operatorname{Spec} A 
e \emptyset$ by Thm.~$1.3$. By Zorn's lemma, it suffices to show for every descending chain of prime ideals $\{\mathfrak{p}_\alpha\}$ in $\operatorname{Spec} A$, the lower bound $\mathfrak{p} = \bigcap_\alpha \mathfrak{p}_\alpha$ is in $\operatorname{Spec} A$. If $xy \in \mathfrak{p}$, then $xy \in \mathfrak{p}_\alpha$ for all $\alpha$, and so for any $\alpha$, either $x \in \mathfrak{p}_\alpha$ or $y \in \mathfrak{p}_\alpha$. Since $\mathfrak{p}_\alpha$ are a descending chain, at least one of $x,y$ is in the intersection $\mathfrak{p}$. Thus, $\mathfrak{p} \in \operatorname{Spec} A$.
\end{proof}

Exercise 1.8

Answers

Comments

Exercise 1.8

Answers

Comments

Add answer