13-00-EveryPrimeIdealIsRadical.tex

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\begin{document}
Let $\mathcal{R}$ be a commutative ring and let $\mathfrak{P}$ be
a prime ideal of $\mathcal{R}$.

\begin{prop}
Every prime ideal $\mathfrak{P}$ of $\mathcal{R}$ is a radical
ideal, i.e.
$$\mathfrak{P}=\operatorname{Rad(\mathfrak{P})}$$
\end{prop}
\begin{proof}
Recall that $\mathfrak{P}\subsetneq \mathcal{R} $ is a prime ideal
if and only if for any $a,b\in \mathcal{R}$ $$a\cdot b\in
\mathfrak{P} \Rightarrow a\in \mathfrak{P} \text{ or } b\in
\mathfrak{P}$$ Also, recall that
$$\operatorname{Rad}(\mathfrak{P})=\{ r\in \mathcal{R} \mid
\exists n\in \Nats \text{ such that } r^n\in \mathfrak{P} \}$$

Obviously, we have $\mathfrak{P}\subseteq
\operatorname{Rad}(\mathfrak{P})$ (just take $n=1$), so it remains
to show the reverse inclusion.

Suppose $r\in \operatorname{Rad}(\mathfrak{P})$, so there exists
some $n\in \Nats$ such that $r^n\in \mathfrak{P}$. We want to
prove that $r$ must be an element of the prime ideal
$\mathfrak{P}$. For this, we use induction on $n$ to prove the
following proposition:

For all $n\in\Nats$, for all $r\in \mathcal{R}$,
$r^n\in\mathfrak{P} \Rightarrow r\in \mathfrak{P}$.

{\bf Case $n=1$:} This is clear, $r\in \mathfrak{P}
\Rightarrow r\in \mathfrak{P}$.

{\bf Case $n$ $\Rightarrow$ Case $n+1$: } Suppose we have proved
the proposition for the case $n$, so our induction hypothesis is
$$\forall r\in\mathcal{R}, \quad r^n\in\mathfrak{P} \Rightarrow r\in
\mathfrak{P}$$ and suppose $r^{n+1}\in \mathfrak{P}$. Then
$$r\cdot r^n = r^{n+1} \in \mathfrak{P}$$
and since $\mathfrak{P}$ is a prime ideal we have
$$r\in \mathfrak{P} \text{ or } r^n\in \mathfrak{P}$$
Thus we conclude, either directly or using the induction
hypothesis, that $r\in \mathfrak{P}$ as desired.

\end{proof}
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