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In Dummit & Foote, the definition of primary ideal says:

A proper ideal of a commutative ring is called primary if whenever $ab \in Q$ and $a \notin Q$, then $b \in {\rm rad}(Q)$.


Suppose $I$ is an ideal such that ${\rm rad}(I)$ is prime.

Suppose $ab \in I$. Then $ab \in {\rm rad}(I)$, hence $a \in {\rm rad}(I)$ or $b \in {\rm rad}(I)$.

Case 1: $a \notin I$, $a \notin {\rm rad}(I)$. Then $b \in {\rm rad}(I)$. So, $I$ is primary by definition.

Case 2: $a \notin I$, but $a, b \in {\rm rad}(I)$. Then $I$ is primary by definition.

Case 3: $a \notin I$, $b \notin {\rm rad}(I)$. Then, $a \in {\rm rad}(I)$. In this case, since $b \notin {\rm rad}(I)$, then $b \notin I$. Then we have $b \notin I$ but $a \in {\rm rad}(I)$. So, $I$ is primary.

What is wrong with this argument?

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  • $\begingroup$ In case 1, you say "$I$ is primary by definition". You can't conclude that until you have considered all cases. $\endgroup$ – Angina Seng Jun 16 '20 at 1:28
  • $\begingroup$ I don't see any contradiction in "case 3". $\endgroup$ – Angina Seng Jun 16 '20 at 1:31
  • $\begingroup$ Note: you should clean up this proof by assuming at the beginning that $ab \in I$ and $a \notin I$. Then you don't have to repeat this assumption in each case. $\endgroup$ – diracdeltafunk Jun 16 '20 at 1:34
  • $\begingroup$ @AnginaSeng is right that you shouldn't say "$I$ is primary" before you've proved this, but the fix is easy! Just say "we have the desired conclusion" (or something like that) instead. $\endgroup$ – diracdeltafunk Jun 16 '20 at 1:35
  • $\begingroup$ Related. $\endgroup$ – rschwieb Jun 16 '20 at 2:06
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Indeed Case 3 is incorrect. The issue is that you ended up with the wrong conclusion! You showed $b \notin I$ and $a \in \text{rad}(I)$, but you needed to conclude that $a \notin I $ and $b \in \text{rad}(I)$! In other words, in total you proved the following:

If $ab \in I$ and $a \notin I$, then either $b \in \text{rad}(I)$ or ($b \notin I$ and $a \in \text{rad}(I)$)

But this is not the definition of a primary ideal!

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  • $\begingroup$ Ah yes, I thought they were enumerating cases by possibilities for $a$, but they were enumerating cases by possibilities by $b$. $\endgroup$ – rschwieb Jun 16 '20 at 2:01
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    $\begingroup$ The problem is, as it always seems to be, rooted in this misconception about primary ideals. $\endgroup$ – rschwieb Jun 16 '20 at 2:03
  • $\begingroup$ @diracdeltafunk But to show primary, don't we have to show that whenever $ab \in I$, and one of $a$ or $b$ is not in $I$, then the element not in $I$ is in ${\rm rad}(I)$? $\endgroup$ – user46372819 Jun 16 '20 at 15:11
  • $\begingroup$ Well, yes, but that is not what you proved. Do you agree that you proved the statement I wrote in my answer? That statement is not equivalent to the one you wanted to show. To clarify: once you assume $a \notin I$, you must prove $b \in \text{rad}(I)$. It not enough to assume $a \notin I$ and deduce $a \in \text{rad}(I)$, even if you also deduced $b \notin I$ at the same time. $\endgroup$ – diracdeltafunk Jun 16 '20 at 20:36

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