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In a commutative ring, prime ideals are radical


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.31

Solution:

We begin with a lemma.

Lemma: Let $R$ be a commutative ring and let $I \subseteq R$ be an ideal. Also let $J \subseteq R$ is an ideal containing $I$. Then $J/I$ is radical in $R/I$ if and only if $J$ is radical in $R$.

Proof: Suppose $J/I$ is radical in $R/I$, and let $x \in \mathsf{rad}(J)$. Then $x^m \in J$ for some $m \geq 1$. Now $$(x+I)^m = x^m+I \in J/I,$$ so that $$x+I \in \mathsf{rad}(J/I) = J/I.$$ Thus $x \in J$, and $J$ is a radical ideal in $R$. Conversely, suppose $J$ is radical in $R$ and let $x+I \in \mathsf{rad}(J/I)$. Then $(x+I)^m = x^m+I \in J/I$ for some $m \geq 1$, so that $x^m \in J$. Thus $x \in \mathsf{rad}(J) = J$, and we have $x+I \in J/I$; hence $J/I$ is radical. $\square$

Let $P \subseteq R$ be a prime ideal. Let $x \in \mathsf{rad}(P)$. Then for some $n \geq 1$, we have $x^n \in P$; suppose that $n$ is minimal with this property. If $n \geq 2$, then $xx^{n-1} \in P$. Since $P$ is prime, then either $x \in P$ or $x^{n-1} \in P$, a contradiction since $n$ is minimal. Thus $n = 1$, and we have $x \in P$.

Write the prime factorization of $n$ as $n = \prod p_i^{k_i}$. Suppose first that $0$ is radical in $\mathbb{Z}/(n)$, and suppose some $k_i$ is at least 2. Now $\prod p_i$ is nonzero in $\mathbb{Z}/(n)$, and $(\prod p_i)^{\max k_i} = 0$, a contradiction. Thus $n$ is squarefree. Conversely, suppose $n$ is squarefree and let $a \in \mathbb{Z}/(n)$ with $a^k = 0$ for some $k$. If some prime $p$ divides $n$ but not $a$, then no power of $a$ can be divisible by $n$, a contradiction. Thus $n$ divides $a$, and in fact $\mathsf{rad}(0) = 0$.

Consider $(n)$ as an ideal of $\mathbb{Z}$. By the lemma, $(n)$ is radical in $\mathbb{Z}$ if and only if 0 is radical in $\mathbb{Z}/(n)$, if and only if $n$ is squarefree.


Linearity

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