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Let $R$ be a commutative nilpotent-free ring with unity and $\mathfrak{M}$ be a maximal $m$-system. I want to show the following.

For any nonzero $a\notin \mathfrak{M}$; there exists $b\in \mathfrak{M}$ with $ab=0$.

(proof:) Since $\mathfrak{M}$ is a maximal m-system, $R\setminus \mathfrak{M}$ is a prime ideal (see here). Now every prime ideal of $R$ contains a minimal prime ideal. Then by maximality of $\mathfrak{M}$, $R\setminus \mathfrak{M}$ is itself a minimal prime ideal. Every minimal prime ideal consists entirely of zero-divisors (see Lemma 1.1 of this paper). Therefore the statement follows.

Is my proof correct?

Arturo Magidin
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Dots_and_Arrows
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  • No, your proof is not correct. You know that every element not in $\mathfrak{M}$ is a zero divisor... ok. So if $a\notin \mathfrak{M}$, then there exists $b\in R$ such that $ab=0$ and $b\neq 0$. But what you need is a $b$ in $\mathfrak{M}$, not merely somewhere in $R$. – Arturo Magidin May 03 '24 at 19:46
  • See Lemma 1.1 of this article https://scholarship.claremont.edu/cgi/viewcontent.cgi?article=1335&context=hmc_fac_pub

    b happens to lie in $\mathfrak{M}$.

     – Dots_and_Arrows May 03 '24 at 19:50
  • So that Lemma says that there exists a $b\in \mathfrak{M}$ such that $ab$ is nilpotent. Last I checked, "nilpotent" is not equivalent to "is $0$". Your argument is still at best incomplete. – Arturo Magidin May 03 '24 at 19:52
  • See I have mentioned that my ring is nilpotent-free. – Dots_and_Arrows May 03 '24 at 19:53
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    So then your argument is missing the reference. Also, your opening two sentences should probably be "So $R\setminus\mathfrak{M}$ is a prime ideal", not "an ideal. Hence it is a prime ideal". – Arturo Magidin May 03 '24 at 19:54

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