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Let's say $A=\{ x \mid x \text{ is a zero divisor}\}\cup\{ 0\}.$
Let $a,b\in A$.
Let $P$ be an ideal of $A$, i.e. $P\trianglelefteq A$.

Now it's necessary to show that: first, $P\ne A$. Second, $ab\in P\Rightarrow a\in P \vee b\in P$.

But the question is how? I've made some moves, but what's your idea? BTW, is the method correct at all?

Michael Hardy
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Mill
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  • The method is not correct. By letting $P$ be an ideal of $A$ and trying to show that $P$ is prime, you are trying to prove the statement "every ideal of $A$ is prime" instead of the statement "some ideal of $A$ is prime". The first statement is false. – RghtHndSd Jun 13 '14 at 13:14

1 Answers1

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Use Zorn's Lemma to construct a maximal ideal $M$ inside the $J$, where $J$ is the set of zero divisors and 0. And argue that the maximality of $M$ guarantees its primeness. In general an ideal maximal with respect to avoiding a multiplicative set is prime.

mez
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