Exercise 1.9

Question

Let $\mathfrak{a}$ be an ideal $
e (1)$ in a ring $A$. Show that $\mathfrak{a} = r(\mathfrak{a}) \Leftrightarrow \mathfrak{a}$ is an intersection of prime ideals.

Amit Awasthi · Accepted Answer

\begin{proof}
  $\Rightarrow$ follows by Prop.~1.14.
  \par $\Leftarrow$. Suppose $\mathfrak{a} = \bigcap \mathfrak{p}_i$. $\mathfrak{a} \subseteq r(\mathfrak{a})$ is true by definition. Since $\{\mathfrak{p}_i\} \subseteq \{\mathfrak{q}_k \mid \mathfrak{q}_k \supseteq \mathfrak{a}\}$ and by Prop.~$1.14$ $r(\mathfrak{a}) = \bigcap \mathfrak{q}_k$, we have $\mathfrak{a} \supseteq r(\mathfrak{a})$. Thus, $\mathfrak{a} = r(\mathfrak{a})$.
\end{proof}

Exercise 1.9

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Exercise 1.9

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