Let $A$ be a ring and $p\subset A$ be a prime ideal. Call $f$ the canonical map $A\to A_p$, and set $I:=\operatorname {ker }f$. Show that $I\subseteq p$ and that $\sqrt I=p$ if and only if $p$ is minimal among the prime ideals of $A$.
$I\subseteq p$ is easy: $x\in I $ iff exists $t\in A\setminus p$ such that $tx=0$; since $0\in p$ and $t\notin p$, be $p$ prime $x\in p$.
Now I try to show that, if $p$ is a minimal prime, $\sqrt I=p$. Since $I\subseteq \sqrt I\subseteq p$, we can look at what happens in $A_p$. Here we have that the radical of $IA_p=0$ is $pA_p$, being the only prime containing it. Am I finished?
Now suppose that $\sqrt I=p$. Again, passing to $A_p$, we are in the situation that there is only a prime ideal, namely $\sqrt 0=pA_p$. Since any prime contained in $p$ would survive in $A_p$, can I conclude that $p$ is minimal?
These arguments actually seem defective to me, but I have not clear why. It seems to me that sometimes I reasoned like if the prime ideals of $A_p$ were in bijection with the primes of $A$ contained in $p$, other times like if the bijection was with the prime ideals contained in $p$ and containing $I$ (for example in saying that $\sqrt I A_p$ is the nilradical of $A_p$). Could you clear my ideas? Thanks
Best Answer
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.