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Let $R\subset R'$ be an extension of integral domains. So we have an inclusion map $i:R\hookrightarrow R'$. Let $\mathfrak{p}\subset R$ be a prime ideal. We know that $\mathfrak{p}^e$ (generated by the image of $\mathfrak{p}$ under $i$) need not be a prime ideal. But is it always a radical ideal? proof or counter-example?

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$k[x]\rightarrow k[x^{1/p}]$ is a counterexample. –  PVAL Jan 20 '14 at 0:55

2 Answers 2

up vote 4 down vote accepted

No. If you accept the characterization of radical ideals in Dedekind domains provided here and are familiar with some algebraic number theory, then there are a plethora of counter-examples. More precisely, given an extension of number fields $K/k$, let $\mathfrak{O}$ and $\mathfrak{o}$ denote the rings of integers of $K$ and $k$, respectively. Then if $\mathfrak{p}$ is an ideal in $\mathfrak{o}$ the ideal $\mathfrak{p}\mathfrak{O}$ in the ring of integers of $K$ lying over $\mathfrak{p}$ factors into a product

$$ \mathfrak{p}\mathfrak{O} = \prod_{i=1}^n \mathfrak{P}_i^{e_i} $$ where $\mathfrak{P}_i$ is a prime in $\mathfrak{O}$ and $e_i \geq 0$ for all $i$, and it can happen that $e_i \geq 2$ for any number of the $i$'s. When $e_i \geq 2$ for some $i$, the prime $\mathfrak{p}$ is said to ramify in $\mathfrak{O}$, and the concept has been extensively studied.

As a simple example, take $k = \mathbb{Q}$ and $K = \mathbb{Q}(i)$ so that $\mathfrak{o} = \mathbb{Z}$ and $\mathfrak{O} = \mathbb{Z}[i]$. Then $\mathbb{p} = (2)$ is prime in $\mathbb{Z}$, but it follows from the factorization $2 = (1+i)(1-i)$ of $2$ in $\mathbb{Z}[i]$ that

$$ (2) = (1+i)(1-i) = (1+i)^2 $$

in $\mathbb{Z}[i]$. Note that this ideal does not contain $1+i$.

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It is not always radical. Consider the prime ideal $(x) \subseteq k[x]$, and the inclusion of domains $k[x] \subseteq k[x,y]/(y^2 - x)$. Notice that $k[x,y]/(y^2 - x)$ is a domain, as $y^2 - x$ is an irreducible polynomial. However, the extension of $(x)$ is not radical, as $y^2 \in (x)^e$, but $y \not \in (x)^e$.

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