Let be $R=\mathbb{C} \lbrace x,y,z \rbrace$ the formal series ring and let $f\_{1},f\_{2},f\_{3} \in R$ be nonzero elements of $R$.


(a) Consider the varieties $M:=V(f\_{1},f\_{2})$ and $N:=V(f\_{2},f\_{3})$ in $\mathbb{C}^{3}$ and suppose that $M$ and $N$ are curves with at least two irreducible components and exactly one irreducible component in common. Now, consider the intersection $Q=M \cap N$ given by $Q=V(f\_{1},f\_{2},f\_{3})$, the local ring of $Q$ is:


$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathcal{O}\_{Q}=\mathbb{C}\lbrace x,y,z \rbrace /(f\_{1},f\_{2},f\_{3}) $


I think that $\mathcal{O}\_{Q}$ is not Cohen-Macaulay, but I do not know how to prove. 


$\mathcal{O}\_{Q}$ is not Cohen-Macaulay?


(b) In general, consider two curves $M$ and $N$ in $\mathbb{C}^n$ with at least two irreducible components (that is, $M$ and $N$ are not irreducible) and exactly one irreducible component in common (which we call $Q$). Thus, the intersection of $M$ and $N$ is the curve $Q$. 


The curve $Q$ is not Cohen Macaulay?



Intersection of two curves is not Cohen Macaulay
The curve $Q$ can indeed be Cohen-Macaulay, and you can use the Hilbert-Burch-Schaps theorem to construct examples by considering the three $2\times 2$ minors of a $3\times 2$ matrix of elements of $R$. One simple example is $(f\_1,f\_2,f\_3) = (x-y,x^2-yz,x-z)$. The complete intersection ideal $\langle x-y,x^2-yz \rangle$ has two associated primes, $\mathfrak{p}\_1 = \langle x,y \rangle$ and $\mathfrak{p}\_2 = \langle x-y,x-z \rangle$. Similarly, the complete intersection ideal $\langle x^2-yz,x-z \rangle$ has two associated primes, $\mathfrak{p}\_2$ and $\mathfrak{p}\_3 = \langle x,z\rangle$. The ideal $\langle f\_1,f\_2,f\_3 \rangle$ equals $\langle x-y,x-z \rangle$ since $$x^2-yz = x(x-y)+y(x-z) = x(x-z)+z(x-y).$$ Thus, the ideal $\langle f\_1,f\_2,f\_3\rangle$ is a prime, complete intersection ideal $\mathfrak{p}\_2$. This example comes from the (non-minimal) realization of $\mathfrak{p}\_2$ via Hilbert-Burch-Schaps arising from the following $3\times 2$ matrix, $$ \left[ \begin{array}{cc} x & y \\ 1 & 1 \\ z & x \end{array} \right].$$ You can produce more examples in the same manner. 


