Exercise with localization and minimal primes

Solution 1:

Here is a better way to prove this. The containment $\sqrt{I} \subseteq p$ is trivial once you have $I \subseteq p$ (by taking radical on both sides). By definition, $\sqrt{I} \supseteq p$ if an only if for all $x \in p$ there is $n>0$ and $t \in A \setminus p$ such that $tx^n=0$ if and only if $x/1 \in \mathfrak{N}(A_p)$ (for all $x \in p$). Recall that $\mathfrak{N}(A_p)$ is the intersection of all prime ideals of $A_p$.

If $p$ is minimal, then $x/1 \in pA_p = \mathfrak{N}(A_p)$ (there is only one prime ideal of $A_p$). Conversely, suppose $p$ is not minimal. Then we can find a minimal prime $q \subset p$. Assume, in contrary, that $x/1 \in \mathfrak{N}(A_p)$. In particular, $x/1 \in qA_p$. Thus, there exists $y \in q$, $t \in A \setminus q$ such that $x/1 = y/t$. Hence, for some $u \in A \setminus p$, $utx = uy \in q$ and so $x \in q$ by primality (since $u,t \not\in q$). Since $x \in p$ is arbitrary, we have $p \subseteq q$, a contradiction.

Solution 2:

Here is an equivalent statement and proof, which do not involve localisation. Thus you get a different perspective, and can see what is happening in the ring $A$ more explicitly.

Let $A$ be a commutative unital ring and let $P$ be a prime ideal. Let $I$ be the ideal of $i\in A$ such that $ix=0$ for some $x\notin P$.

Then $\sqrt I\subseteq P$ as $a^nx=0$ with $x\notin P$ implies $a\in P$. Similarly, for any prime ideal $P'\subsetneq P$, we have $a^nx=0$ with $x\notin P$ implies $x\notin P'$, so $a\in P'$. Thus $\sqrt I \subseteq P'\subsetneq P$ and $\sqrt I\neq P$.

Conversely, suppose $\sqrt I \neq P$. Pick $a\in P \backslash \sqrt I$. Then $a$ and $A\backslash P$ generate a multiplicatively closed set $M$, disjoint from $\sqrt I$ (if $a^nx\in \sqrt I$ with $x\notin P$, then $(a^nx)^ky=0$, with $y\notin P$, which implies $a\in \sqrt I$ - a contradiction). By Zorn's lemma, one may find an ideal $J\supseteq \sqrt I$, maximal amongst ideals disjoint from $M$. Then $J$ is a prime ideal and $J\subsetneq P$, as it is disjoint from $M$.

Note the following lemma is used explicitly at the end of the above argument, and implicitly in the corresponding argument by @Ray (as it is the key ingredient of the proof that only nilpotent elements lie in the intersection of all primes):

Lemma Let $A$ be a commutative unital ring, let $M\subseteq A$ be a multiplicatively closed subset, and let $J$ be maximal among ideals disjoint from $M$. Then $J$ is prime.

Proof: If $u,v\notin J$ then by the maximality property we have: $$\lambda u=j_1+m_1, \qquad \mu v=j_2+m_2,$$ with $\lambda,\mu\in A$, and $j_1,j_2\in J$, and $m_1,m_2\in M$. However in that case: $$\lambda\mu uv=(j_1j_2+j_1m_2+j_2m_1)+m_1m_2,$$ so $uv\notin J$ (otherwise $m_1m_2\in J \cap M$ - a contradiction). $\qquad\qquad\qquad\Box$