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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?

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    $\begingroup$ Maybe you provide one of the proofs you found and tell us, where you see some flaws? $\endgroup$ – MooS Jul 14 '16 at 15:00
  • $\begingroup$ do all of them use Zorn's Lemma? $\endgroup$ – janmarqz Jul 14 '16 at 15:00
  • $\begingroup$ @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. $\endgroup$ – Amr Jul 14 '16 at 15:02
  • $\begingroup$ @Moos Hi. I added one proof that I think is flawed $\endgroup$ – Amr Jul 14 '16 at 15:09
  • $\begingroup$ This given proof is indeed flawed and thus plain wrong. $\endgroup$ – MooS Jul 14 '16 at 15:10
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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!

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  • $\begingroup$ Good job :) I liked that! $\endgroup$ – Amr Jul 14 '16 at 15:15
  • $\begingroup$ this miss the existence that there are maximals: ideals that has the property $\endgroup$ – janmarqz Jul 14 '16 at 20:52
  • $\begingroup$ Seriously: Please read the original question again. The OP never asked anyone to proof the existence of such maximal element, since he did so himself. He even told you so in the comments. $\endgroup$ – MooS Jul 15 '16 at 4:33
  • $\begingroup$ thanks for your patience xD $\endgroup$ – janmarqz Jul 15 '16 at 16:41

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