Articles of cohen macaulay

Cohen-Macaulay and connected implies equidimensional?

I’m asking for a reality check. It seems to me that since Cohen-Macaulay rings are locally equidimensional, such a ring is either equidimensional or else disconnected (with different dimensions occurring on different connected components). But I have not slept enough, so I don’t trust my reasoning. Is it sound? Addendum: Having given this more (and […]

showing Cohen-Macaulay property is preserved under a ring extension

Let $R$ be an $\mathbb{N}$-graded Noetherian ring, with $R_0$ local Artinian. Assume also that $R$ is finitely generated over $R_0$ by elements of degree $1$. Let $M$ be a Cohen-Macaulay $R$-module. Let $Y$ be an indeterminate over $R_0$ and define the local ring $R_0(Y)=R_0[Y]_S$, where $S$ is the multiplicative set of $R_0[Y]$ which consists of […]

Cohen-Macaulay ring and saturated ideal

Let $A=\mathbb{C}[x_0,x_1,\dots,x_n]$ and $I$ an ideal of $A$. Is there any connection between “$A/I$ is Cohen-Macaulay ring” and “$I$ is a saturated ideal”? Does one of them imply another? Please give me any reply or reference.

Does maximal Cohen-Macaulay modules localize?

Let $A$ be a Noetherian local ring and $M$ a finitely generated $A$-module such that $$\operatorname{depth}M= \dim M=\dim A.$$ I can prove that $$\operatorname{depth}M_{\mathfrak{p}}= \dim M_{\mathfrak{p}}\leq\dim A_{\mathfrak{p}}\quad\forall\mathfrak{p}\in\operatorname{Supp}M.$$ I guess the inequality is in fact an equality, but can’t seem to be able to prove it. Does anyone have any idea? P.S. If $A$ is Cohen-Macaulay […]

$2$-dimensional Noetherian integrally closed domains are Cohen-Macaulay

Any 1-dimensional Noetherian domain is Cohen-Macaulay (C-M). For the $2$-dimensional case, a condition of being integrally closed is necessary to be added for a Noetherian domain to be C-M, which I could not prove it. Would anybody be so kind as to solve this? I also search for a non C-M $2$-dimensional Noetherian domain which […]

Examples of Cohen-Macaulay integral domains

Question 1 Could you find a non Cohen-Macaulay ring $A$ without zero divisors. I would like $A$ to be as simple as possible. For instance, I want $A$ to be finitely generated alegbra over $\mathbb{C}$. Comment I found plenty of examples of non Cohen-Macaulay rings in wiki and from other sources. But those rings are […]

Associated ideals of a principal ideal generated by a nonzero divisor

Let $R$ be a Noetherian ring and $b\in R$ a nonzerodivisor. Krull’s Principal Ideal Theorem implies that every minimal prime ideal over the ideal $(b)$ has codimension $1$. However, could it be possible that one of the associated ideals of $(b)$ is embedded, i.e. having codimension $>1$? If yes, could you give an example? Will […]

Prove that $ k/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle$ is not Cohen-Macaulay.

Prove that $ k[x_1,\ldots,x_4]/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle$ is not Cohen-Macaulay. We have $\langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle=\langle x_1,x_3 \rangle \cap \langle x_2,x_4\rangle$. Therefore $\dim k[x_1,\ldots,x_4]/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle=2$. How do I prove $ k[x_1,\ldots,x_4]/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle$ is not Cohen-Macaulay?

Why is this ring not Cohen-Macaulay?

I am stuck with exercise 18.8, page 466 of Eisenbud’s Commutative Algebra with a view towards Algebraic Geometry: Let $k$ be a field. The task is to prove that $R:=k[x^4, x^3y, xy^3, y^4] \subseteq k[x,y]$ is not a Cohen-Macaulay ring, i.e. the length of a maximal regular sequence (i.e. a sequence of elements $a_1, \dots […]

Are the rings $k]$ and $k]$ Gorenstein? (Matsumura, Exercise 18.8)

Here is question 18.8 of Matsumura’s Commutative Ring Theory. It asks whether the rings $k[[t^3,t^4,t^5]]$, $k[[t^4,t^5,t^6]]$ are Gorenstein. I got that 1) is not Gorenstein, but 2) is Gorenstein (by computing the socle). Just wanted to check if I am correct. I don’t need the answer necessarily, a yes or a no will suffice. Thanks.