My teacher stated the following result:

Let $R$ be a ring, and consider $I=\sqrt I\subseteq A$ a radical ideal. $I$ is equal to the intersection of the primes minimal over $I$, i.e. the minimal (with respect to inclusion) elements in the set of the prime ideals of $A$ containing $I$. Plus, this decomposition of $I$ is unique (among the decompositions made of minimal primes).

Now, I already know that $\sqrt I=I$ is equal to the intersection of all the prime ideals containing $I$, so $$I=\bigcap_{\mathfrak p \supseteq I}\mathfrak p\subseteq \bigcap_{\mathfrak p \supseteq I \ \mathrm {min.}}\mathfrak p;$$ however, I don't know how to proceed in order to prove the other inclusion; in particular, I don't know if I can prove that any prime ideal contains a minimal prime.

For the uniqueness of this decomposition, I would suppose that if there is another decomposition, say $\bigcap_{\mathfrak q\in S} \mathfrak q$, where $S$ is a subset of the set of the minimal primes over $I$, and $\bigcap_{\mathfrak q\in S} \mathfrak q=\bigcap_{\mathfrak p \supseteq I \ \mathrm {min}}\mathfrak p$. Then, for a minimal prime $\mathfrak p'\supseteq I$ not contained in $S$, we have $\bigcap_{\mathfrak q\in S} \mathfrak q\subseteq \mathfrak p'$ and so there is a $\mathfrak q'\in S$ such that $\mathfrak q'\subseteq \mathfrak p'$. By the minimality of $\mathfrak p'$ we obtain $\mathfrak q'= \mathfrak p'\in S$, which is absurd.

Can you give me a hint for the first part and check if the second part is correct? Thanks in advance.


1 Answer 1


Regarding the first point: claim: Given an ideal $I$ and a prime ideal $\mathfrak p$ such that $I\subseteq \mathfrak p$, there is a minimal prime ideal $\mathfrak q$ containing $I$ such that $\mathfrak q\subseteq\mathfrak p$.

To see this, employ Zorn's lemma. Let $V_\mathfrak p$=the set of all prime ideal $\mathfrak q$ such that $I\subseteq \mathfrak q\subseteq \mathfrak p$.

Consider the order on $V$ given by reverse inclusion i.e. $\mathfrak q_1\leq \mathfrak q_2$ if $\mathfrak q_1\supseteq \mathfrak q_2$. Now simply observe this set satisfies the assumptions of the Zorn's lemma. Now a maximal element of $V$ will give a minimal prime ideal contained in $\mathfrak p$ and containing $I$.

The argument for the 2nd part is fine.

  • $\begingroup$ I know that for any ideal there is a minimal prime containing it, but I don't see how this proves the first part of the theorem. My idea was to prove that every prime ideal, containing $I$, also contain a minimal prime over $I$. However I don't see if it follows from your argument. Thanks for the help $\endgroup$
    – Dr. Scotti
    Commented Nov 2, 2021 at 18:21
  • 1
    $\begingroup$ My argument was to answer your question. But I miswrote the 'claim'. See the updated claim and the proof again if you will. $\endgroup$ Commented Nov 3, 2021 at 5:47

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