Compare the definitions:

Page 136, Matsumura, Commutative Ring Theory:

A proper ideal $I$ in a Noetherian ring $A$ is said to be unmixed if the heights of its prime divisors are all equal.

This means that $\forall P\in {\rm Ass}(I) ,\quad \operatorname{ht} I = \operatorname{ht} P\quad (1) $

Page 59, Bruns&Herzog, Cohen-Macaulay Rings, 1998:

One says that an ideal $I$ is unmixed if $I$ has no embedded prime divisors, or in modern language if the associated prime ideals of $R/I$ are the minimal prime ideals of $I$.

This means that ${\rm Min}({\rm Ass}(I))={\rm Ass}(I)\quad \quad (2)$

Clearly $(1)$ implies $(2)$ but the reverse isn't true. How these definitions are equivalent? I mean while (2) doesn't imply (1) then how can we say the two definitions are the same?

  • $\begingroup$ maybe because: both Matsumura & Bruns Herzog use unmixedness to characterize Cohen-macaulay rings. and in Cohen-macaulay rings two definitions are equal $\endgroup$
    – user 1
    Commented Dec 5, 2014 at 16:16
  • 1
    $\begingroup$ As @11156 mentioned, this is not true without additional assumption. Take $k[x,y,z]$ and $I = (x) \cap (y,z)$. So, what you stated is correct. $\endgroup$
    – Youngsu
    Commented Dec 5, 2014 at 18:58

1 Answer 1


You are correct in observing this.

In the old literature, unmixedness means that all associated primes are of the same height. However, in modern literature, the definition of unmixedness is equidimensional and no embedded prime, that is, all its associated primes are minimal above it and are of the same height.

  • 1
    $\begingroup$ Maybe someone should update the wikipedia page for Cohen-Macaulay ring under unmixedness theorem then... $\endgroup$ Commented May 22, 2020 at 20:52

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