What is wrong with this proof that every ideal whose radical is prime is a primary ideal?

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?

• In case 1, you say "$I$ is primary by definition". You can't conclude that until you have considered all cases. – Angina Seng Jun 16 '20 at 1:28
• I don't see any contradiction in "case 3". – Angina Seng Jun 16 '20 at 1:31
• 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. – diracdeltafunk Jun 16 '20 at 1:34
• @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. – diracdeltafunk Jun 16 '20 at 1:35
• – rschwieb Jun 16 '20 at 2:06

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 $$a \in \text{rad}(I)$$)
• Ah yes, I thought they were enumerating cases by possibilities for $a$, but they were enumerating cases by possibilities by $b$. – rschwieb Jun 16 '20 at 2:01
• @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 ${\rm rad}(I)$? – user46372819 Jun 16 '20 at 15:11
• 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. – diracdeltafunk Jun 16 '20 at 20:36