# every prime ideal is radical

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

###### Proposition 1.

Every prime ideal $\mathfrak{P}$ of $\mathcal{R}$ is a radical ideal, i.e.

 $\mathfrak{P}=\operatorname{Rad(\mathfrak{P})}$
###### 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

 $\operatorname{Rad}(\mathfrak{P})=\{r\in\mathcal{R}\mid\exists n\in\mathbb{N}% \text{ such that }r^{n}\in\mathfrak{P}\}$

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

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

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.

Title every prime ideal is radical EveryPrimeIdealIsRadical 2013-03-22 13:56:54 2013-03-22 13:56:54 alozano (2414) alozano (2414) 5 alozano (2414) Theorem msc 13-00 msc 14A05 prime ideal is radical RadicalOfAnIdeal PrimeIdeal HilbertsNullstellensatz