every prime ideal is radical

Let $ℛ$ be a commutative ring and let $𝔓$ be a prime ideal of $ℛ$.

Proposition 1.

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

$$𝔓=\mathrm{Rad}(𝔓)$$

Proof.

Recall that $𝔓⊊ℛ$ is a prime ideal if and only if for any $a,b\in ℛ$

$$a\cdot b\in 𝔓\Rightarrow a\in 𝔓\text{ or }b\in 𝔓$$

Also, recall that

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

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

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

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

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

Case $n$ $\mathrm{\Rightarrow }$ Case $n\mathrm{+}\mathrm{1}$: Suppose we have proved the proposition for the case $n$, so our induction hypothesis is

$$\forall r\in ℛ,r^n\in 𝔓\Rightarrow r\in 𝔓$$

and suppose $r^{n+1}\in 𝔓$. Then

$$r\cdot r^n=r^{n+1}\in 𝔓$$

and since $𝔓$ is a prime ideal we have

$$r\in 𝔓\text{ or }{r}^n\in 𝔓$$

Thus we conclude, either directly or using the induction hypothesis, that $r\in 𝔓$ as desired.

∎

Title	every prime ideal is radical
Canonical name	EveryPrimeIdealIsRadical
Date of creation	2013-03-22 13:56:54
Last modified on	2013-03-22 13:56:54
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	5
Author	alozano (2414)
Entry type	Theorem
Classification	msc 13-00
Classification	msc 14A05
Synonym	prime ideal is radical
Related topic	RadicalOfAnIdeal
Related topic	PrimeIdeal
Related topic	HilbertsNullstellensatz