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?


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.

  • 1
    $\begingroup$ +1 Yes, it seems to happen a lot that students fall prey to this misconception about the definition of primary ideals. $\endgroup$ – rschwieb Apr 17 '17 at 14:29
  • $\begingroup$ Great answer, thank you! $\endgroup$ – rmdmc89 Apr 17 '17 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.