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Let $I$ be a radical ideal (i.e. $\sqrt I=I$) in a commutative ring with unity. Then is it true that $I$ is a prime ideal if and only if it is not an intersection of two radical ideals properly containing $I$ (i.e. not an intersection of two properly larger (w.r.t. inclusion) radical ideals ) ?

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Yes, it is.

If $I$ is prime, then the conclusion follows easily.

For the converse, write $I=\bigcap_{j\in J} P_j$ with $P_j$ prime ideals. (This can be done since $I$ is radical.) Let's prove that $I$ is prime. Let $ab\in I$ such that $b\notin I$. If $b\notin P_j$ for all $j\in J$, then $a\in I$ and we are done. Otherwise, set $J_1=\{j\in J:b\in P_j\}$. We have $I=I_1\cap I_2$ where $I_1=\bigcap_{j\in J_1} P_j$ and $I_2=\bigcap_{j\in J-J_1} P_j$. Note that $I_1$ and $I_2$ are radical ideals, and moreover $I\subsetneq I_1$ (why?). Since $I$ can not be an intersection of two radical ideals properly containing it, it follows $I=I_2$, so $a\in I$ (why?).

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  • $\begingroup$ Why if $I$ is prime , the conclusion follows easily ? $\endgroup$ – user228168 May 4 '15 at 4:53
  • $\begingroup$ @SaunDev Do you know the definition of prime ideals using ideals? $\endgroup$ – user26857 May 4 '15 at 8:34
  • $\begingroup$ Do you mean that $P$ is a prime ideal if for ideals $A,B$ such that $AB \subseteq P \implies A \subseteq P$ or $B \subseteq P$ ? I still don't see how that helps ... $\endgroup$ – user228168 May 4 '15 at 14:46
  • $\begingroup$ @SaunDev If $I=I_1\cap I_2$ then $I\supseteq I_1I_2$, so $I=I_1$ or $I=I_2$. (Any more questions?) $\endgroup$ – user26857 May 4 '15 at 20:13

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