Is the following ring Cohen-Macaulay? Let $A=\mathbb{C}[a,b,c,d]/I$, where $I$ is the ideal generated by $$(a-b)(a-c),(d-b)(d-c),a^2-b^2,a^2-c^2,d^2-b^2,d^2-c^2.$$

Is $A$ a Cohen-Macaulay ring? What is the general method to check that whether a ring is Cohen-Macaulay? 

Thank you very much.
 A: Let me elaborate on the comments a bit:
First of all let us show that a noetherian one-dimensional reduced ring is Cohen-Macaulay. Of course we can reduce the local case. I will not use the homological characterization of depth to keep it elementary (i.e. I will not use that depth zero is equivalent to the maximal ideal being an associated prime, but I will show it on the way). 
If an one-dimensional local ring $(R,\mathfrak m)$ is not Cohen-Macaulay, it has no regular element in $\mathfrak m$. Now let $x \in \mathfrak m$ and $x \notin \mathfrak p$ for all the prime ideals of height zero (prime avoidance). $x$ is not regular, hence $xy=0$ for some $y$. $\operatorname{Ann}(y)$ contains $x$, hence is not contained in any prime ideal of height zero. This shows that a maximal element of $\{ \operatorname{Ann}(r) | r \in R, \operatorname{Ann}(r) \supset \operatorname{Ann}(x)\}$ must be equal to $\mathfrak m$. In particular $\mathfrak m = \operatorname{Ann}(r)$ for some $r \in \mathfrak m$, i.e. $r^2=0$. $\small\Box$
Now consider your ring $A$. To see $\dim A=1$, note that $A$ is a graded ring (because $I$ is homogeneous) and thus $\dim A = 1 + \dim \operatorname{Proj} A$. The latter can be computed on each affine cover, let me do the cover $d=1$:
$\mathbb C[a,b,c]/((a-b)(a-c),(b-1)(c-1),a^2-b^2,a^2-c^2,b^2-1,c^2-1)$ is equal to $\mathbb C[a,b,c]/((a-b)(a-c),(b-1)(c-1),a^2-1,b^2-1,c^2-1)$ and this is zero-dimensional because the solution set is finite: Clearly $a,b,c \in \{ \pm 1 \}$. Hence $\dim A=1+0=1$.
To check that $A$ is reduced, you cannot use that trick, because you also have to check the reduced property at the irrelevant maximal ideal. I currently do not see a trick to do it fast, but there are of course algorithms to have a computer check it for you.
