Can you offer some examples of such rings, other than $\frac{k[x,y]}{(x^{2}, xy)}$. Thanks.
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1$\begingroup$ You can find some examples in part 1) of this: (mathoverflow.net/questions/7556/some-examples-of-depth/…) $\endgroup$– Hailong DaoCommented Feb 16, 2010 at 18:45
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2 Answers
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In dimension $1$, Cohen-Macaulay just mean unmixed, so all the associated primes have the same dimension. Thus the easiest way to cook up a non-CM ring of dimension $1$ is: Pick your favorite regular ring (say $A=k[x,y,z]$). Take an ideal of dimension $1$, say $I=(x,y)$. Take another ideal of dimension $0$, say $J=(x^3,y^4,z^5)$ such that $J$ is not contained in $I$. Now take $R=A/(I\cap J)$. Geometrically we just throw 2 things of pure dimensions $1$ and $0$ together. In some sense, all non-CM rings of dimension $1$ arise this way.
answered Feb 16, 2010 at 20:31
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$\begingroup$ Taking $R=A/(IJ)$ also works. $\endgroup$ Commented Feb 17, 2010 at 20:55
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You aren't going to find noetherian graded algebras $A$ much different from this. If $A$ is such an algebra with irrelevant ideal $A_+$ then there is a nonzero homogeneous element $x$ of positive degree such that $xA_+=0$. Similarly for local rings $R$ with maximal ideal $\mathfrak{m}$, such that $R$ has dimension one and in not Cohen-Macaulay, there is a nonzero $x\in\mathfrak{m}$ such that $x\mathfrak{m}=0$. See K. Baclawski and A. M. Garsia, Advances in Math. 39 (1981), 155--184 (Lemma 2.2) and I. Kaplansky, Commutative Rings, revised ed. (Theorem 82).
answered Feb 16, 2010 at 20:08