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So here is the part of exercise 14 of chapter 1 that has been bothering me:

Let $A$ be a commutaive ring with identity. Let $\Sigma $ be the set of ideals with the property that every element in them is a zero divisor. Show that maximal elements of $\Sigma $ are prime.

I saw many online solutions, but I found them all to be flawed proofs. It would be great to hear a valid proof from somebody here.

Thank you a lot.

Here is one proof that I think is flawed :

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Criticism: why is it true that all elements of $(m,x) $ are zero divisors?

Amr
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    Maybe you provide one of the proofs you found and tell us, where you see some flaws? – MooS Jul 14 '16 at 15:00
  • do all of them use Zorn's Lemma? – janmarqz Jul 14 '16 at 15:00
  • @janmarqz Proving Existence of maximal elements requires zorns lemma. This was straightforward so I didn't ask abt it. Primality of maximal elements is what I couldn't prove. – Amr Jul 14 '16 at 15:02
  • @Moos Hi. I added one proof that I think is flawed – Amr Jul 14 '16 at 15:09
  • This given proof is indeed flawed and thus plain wrong. – MooS Jul 14 '16 at 15:10
  • For the sake of completeness: The given proof is wrong because the sum of zero divisors can be a non-zero divisor (even a unit). Take 2 and 3 modulo 6. – MooS Jul 14 '16 at 15:32

1 Answers1

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Here is a correct proof:

Let $I$ be a maximal element of $\Sigma$. Let $xy \in I$ but $x \notin I, y \notin I$.

By the maximality of $I$ there is a non-zero divisor in $(I,x)$ and a non-zero divisor in $(I,y)$. The product of those two guys is a non-zero divisor in $(I,x)(I,y) \subset (I,xy) = I$, contradiction!

MooS
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