Let $A$ be a ring. Let $I$, $J$ be two ideals of $A$.The following properties are ture.
(a) The radical $\sqrt[]{\mathstrut I}$ equals the intersection of the ideals $\rho$ $\in$ V(I).
(b) We have $V(I)$ $\supseteq$ $V(J)$ if and only if $J$ $\supseteq$ $\sqrt[]{\mathstrut I}$
Could some one help me to proof this with some details?
V(I) := {ρ ∈ Spec A ∣ I ⊆ ρ }
and the radical of I is the set of elements $a ∈ A$ such that $a^{n}$ ∈ I for some n ≥ 1. Thank you:)
(a): If $a \in \sqrt{I}$ then it's image in $A/I$ is nilpotent hence it is contained in all the prime ideals in $Spec(A/I)$ what are these prime ideals, they are the prime ideals in $A$ containing $I$ i.e $V(I)$. Which implies that if $a \in \sqrt(I)$ then $a \in P$ for all $P$ in $V(I).$ Conversely, assume that $a^n$ is not contained in $I$ for any $n$. We will construct a prime ideal containing $I$ but not $a$. Let $\Sigma$ be the set of ideals containing $I$ but not $a^n$ for any $n$, notice that $\Sigma$ is non empty as $I$ is in $\Sigma$. Use Zorn's lemma to show that $\Sigma$ contains a maximal element $P$ show that $P$ is a prime ideal (hint suppose that $xy \in P$ but neither $x$ nor $y$ are in $P$, show that then neither $(x) + P$ nor $(y) + P$ are in $\Sigma$ and use this to show that then we can't have $xy \in P$.)
(b): if $J$ doesn't contain the radical of $I$, then $\exists f$ such that $f^n \in I$ but $f \notin J$. Define $\Sigma$ as in (a), just as in (a) $\Sigma$ contains a prime ideal containing $J$ but not $f$, hence $P$ is in $V(J)$ but $P$ does not contain $f^n$ for any $n$ hence it does not contain $I$.
