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It is written in chapter $4$, page $51$ of Atiyah-Mac Donald's $\textit{Introduction to Commutative Agebra}$:

"a prime power $\mathfrak{p}^n$ is not necessarily primary, although its radical is the prime ideal $\mathfrak{p}$. For example, let $A = k[x, y, z]/(xy - z^2)$ and let $\bar{x}, \bar{y}, \bar{z}$ denote the images of $x, y, z$ respectively in $A$. Then $\mathfrak{p} = (\bar{x}, \bar{z})$ is prime (since $A/\mathfrak{p}\sim k[y]$, an integral domain); we have $\bar{x}\bar{y} = \bar{z}^2\in \mathfrak{p}^2$ but $\bar{x} \notin \mathfrak{p}^2$ and $\bar{y} \notin r(\mathfrak{p}^2) = \mathfrak{p}$; hence $\mathfrak{p}^2$ is not primary"

By definition, $I$ is a primary ideal $\iff \forall a,b\in A:(ab\in I \implies a\text{ or }b\in\sqrt{I})$. I don't understand why $\bar{x}\bar{y}$ is valid for Atiyah's argument, since $\bar{y}\bar{x}\in\mathfrak{p}^2$ and $\bar{x}\in \sqrt{\mathfrak{p}^2}=\mathfrak{p}$, therefore the proposition $(ab\in I \implies a\text{ or }b\in\sqrt{I})$ is satisfied. Shouldn't he find $ab\in \mathfrak{p}^2$ such that this proposition is $\textit{refuted}$? What am I missing?

rmdmc89
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1 Answers1

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Your definition of primary ideal seems to be wrong. The definition is

$I$ is a primary ideal if and only if for all $a,b \in R$, we have $ab \in I \implies a \in I \text{ or } b\in\sqrt{I}$.

Here $\bar x \bar y \in \mathfrak p^2 = I$, but $\bar x \not \in I$ and $\bar y \not \in \sqrt{I} = \mathfrak p$, showing that $I$ is not a primary ideal.

To understand what is wrong with your definition, consider the ideal $I = (x^2,xy) \triangleleft k[x,y]=R$. Its radical is $(x)$, and $I$ is not primary. However, if $ab \in I$, then $x$ divide $ab$, so that $x \mid a$ or $x \mid b$, which implies that $a \in \sqrt I$ or $b \in \sqrt I$. Therefore $I$ satisfies your definition, without being primary.

The definition of "primary ideal" seems to be not symmetric. See this question to have more details.

Watson
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