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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?

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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.

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