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.

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 $$I$$.
• 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 Commented Nov 2, 2021 at 18:21