geometric interpretation of Cohen-Macaulay ring

This is my preparation for a small talk in commutative algebra conference in free discussing time, but it didn't happen because the organizers didn't have time for that (although my proof said we would have time to talk about our sections).


Introduction: 
 
When I was reading about COhen-Macaulay local ring, I found an example of a non-CM local ring, that is the ring $k[x,y,z,w]/(x,y) \cap (z,w)$. Because examples are important in mathematics, so I extend that example to a more general ring in light of a theorem in Eisenbud's book.   

Main talk:

My Non-CM rings that I've proved is:

$ k[x_1, x_2,..,x_n]/(x_1,...,x_k) \cap (x_k,x_{k+1},...,x_{2k})$ with $n \geq 4, k \geq 2, 2k \leq 2$.

To prove this, we need a theorem in Eisenbud's book "commutative algebra: a view toward to algebraic geometry".

Theorem:
 
Let $I, J$ are two ideals of a local ring $R$ whose radical ideals are incomparable. If $I \cap J = 0$, then grade($I+J)\leq 1$

proof:
 
Assume contrary that $grade(I+J) > 1$
Step 1: Consider an exact sequence $0 \to R \to R/I \oplus R/J \to R/(I+J)$. with $\mathrm{a} \mapsto \mathrm{ (a,-a)}$ and $(a,b) \mapsto a+b$. $grade(I+J) > 1$ implies $Ext^{1}_R(R/(I+J),R)=0$, then the sequence splxit.$\exists$ $(1,-1)$, $(a,b)$ such that $a+b \equiv 1 (mod I+J)$ and $R(1,-1) \cap R(a,b) = 0$.
Step 2: $J(a,b)=(Ja,0)=(J,0) \subset R(a,b)$ and $J(1,-1) = (J,0) \subset R(1,-1)$, then $(J,0)=0$, which means $J \subset I$, which is contradict to assumption.


Back to the proof the example.
$S=k[x_1, x_2,..,x_n]/(x_1,...,x_k) \cap (x_k,x_{k+1},...,x_{2k})$, $I=(x_1,...,x_k)/(x_1,...,x_k) \cap (x_k,x_{k+1},...,x_{2k})$
$(x_k,x_{k+1},...,x_{2k})/(x_1,...,x_k) \cap (x_k,x_{k+1},...,x_{2k}), P=(x_1, ..., x_2k)$
. $I_P \cap J_P = 0$, use theorem, we have $grade(I_p+J_p)=grade(P_P)=depth(S_P) \leq 1$. But $dim S_p \geq k \geq 2$ then $S_p$ is not CM local rings and so $S$ is not CM rings.


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