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If $(R,m)$ is a local Noetherian reduced ring of Krull dimension $1$ then $R$ is Cohen-Macaulay, since in a reduced Noetherian ring the set of zero divisors is the (finite) union $U$ of minimal prime ideals, so there exists an element $x\in m$ which is not a zero divisor (otherwise, $m$ lies in $U$ and equals to one of the minimal prime ideals by prime avoidance, which is a contradiction since the height of $m$ is $1$). Now, the singleton $\{x\}$ would be an $R$-sequence, and since the grade of $m$ is at most equal to the height of $m$ the equality occurs.

Now, if $R$ is not local what is the proof?

Thanks for any cooperation!

user26857
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karparvar
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  • Can I ask where you needed the reduced condition? Isn't it true that the set of zero divisors is the finite union of associated primes? – Kevin Sheng Dec 02 '16 at 16:55
  • We need to prove the set of zero divisors is in the union of minimal prime ideals. Then since dim$R$=1, $m$ is not minimal and there exists an element $x\in m$ which is not a zero divisor. https://commalg.subwiki.org/wiki/Reduced_Noetherian_implies_zero_divisor_in_minimal_prime –  Sep 11 '19 at 02:59

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Let $R$ be a noetherian reduced ring, $\dim R=1$, and $m$ a maximal ideal of $R$. Then $R_m$ is noetherian, reduced, and $\dim R_m\le 1$. If $\dim R_m=1$ you are in the local case, so $R_m$ is CM. If $\dim R_m=0$ there is nothing to prove since every zero-dimensional noetherian ring is CM.

user26857
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