Union of Cohen-Macauley components Let $X=X_1\cup X_2\cup X_3$ be a variety with three irreducible equidimensional components which are normal and Cohen-Macauley. Suppose $X_1$ and $X_3$ intersects trasversally and $X_2$ and $X_3$ intersects trasversally but $X_1$ and $X_2$ touches only at a point $P$. Also $X_1\cap X_2\cap X_3=\{P\}$.
Is $X$ Cohen-Macauley?
Transverse intersection means the analytic local ring is formally smooth over $\mathbb{C}[[x,y]]/(xy)$.
 A: This is true in your case, mainly because your hypotheses imply several restrictions. 
Restriction 1.  Each of $X_1,$ $X_2,$ and $X_3$ is smooth at $P$.
Your definition of "transverse intersection" implies that both branches are smooth at every point of the intersection.  In particular, all three branches are smooth at $P$.  
Restriction 2. Each of $X_1,$ $X_2,$ and $X_3$ has dimension equal to $2.$ Also by your definition of transverse intersection, both $X_1\cap X_3$ and $X_2\cap X_3$ are smooth Cartier divisors in $X_3$, which is smooth at every point of these divisors.  Yet these divisors intersect in an isolated point.  By Krull's Hauptidealsatz, every irreducible component of the intersection has codimension $\leq 2$ in $X_3$.  Thus, $X_3$ is $2$-dimensional.  By your equidimensional hypothesis, also $X_1$ and $X_2$ are $2$-dimensional.
Restriction 3. The embedding dimension equals $4.$ When you write, "but $X_1$ and $X_2$ touches only at a point $P$", I assume that you mean that the scheme-theoretic intersection of $X_1$ and $X_2$ equals one reduced point $P$.  So the embedding dimension is at least $4$.  Also, the tangent directions of $X_1\cap X_3$ and $X_2\cap X_3$ at $P$ are linearly independent.  So these tangent directions span the $2$-dimensional tangent space of $X_3$ at $P$.  Thus, the direct sum of the Zariski tangent spaces of $X_1$ and $X_2$ at $P$ already contains the Zarisk tangent space of $X_3$ at $P$.  So the embedding dimension is exactly $4$.
Restriction 4. A special system of parameters.  The previous point deserves more attention.  Denote by $R$ the local ring $\mathcal{O}_{X,p}$ with maximal ideal $\mathfrak{m}_{X,p}.$  Denote by $I_1,$ resp. $I_2,$ $I_3,$ the ideal in $R$ of the germ at $P$ of $X_1,$ resp. $X_2,$ $X_3.$
Then $I_3/(I_3+I_1)$ in $R/I_1$ is a principal ideal.  So there exists an element $s_1\in I_3$ whose image in $R/I_1$ is a generator of $I_3/(I_3+I_1)$.  Similarly, there exists an element $s_2\in I_3$ whose image in $R/I_2$ is a generator of $I_3/(I_3+I_2)$.  
Since $I_2/(I_1+I_2)$ is the maximal ideal in $R/I_1$, there are elements in $I_2$ mapping to a generating set of this maximal ideal.  Since $R/I_1$ is a regular ring of dimension $2$, the maximal ideal is generated by a regular sequence.  Since $X_1\cap X_3$ is smooth at $P$, the element $s_1$ is part of a regular sequence of generators.  Thus, there exists an element $t_1\in I_2$ such that $(s_1,t_1)$ is a regular sequence of generators of the maximal ideal of $R/I_1.$  Similarly, there exists an element $t_2\in I_1$ such that $(s_2,t_2)$ is a regular sequence of generators of the maximal ideal of $R/I_2.$  Since $s_1s_2,$ $s_1t_2,$ and $s_2t_1,$ are in $I_1\cap I_2\cap I_3,$ these products are zero.
Restriction 5. Characterization of the completion.
Denote by $S$ the image of the following local homomorphism of local rings,
$$k[s_1,s_2,t_1,t_2]_{\langle s_1,s_2,t_1,t_2\rangle}/\langle s_1s_2,s_1t_2,s_2t_1 \rangle \rightarrow \mathcal{O}_{X,P}.$$ The domain ring is already a Cohen-Macaulay ring of dimension $2$.  Indeed, by the Hilbert-Burch(-Schaps) theorem, it is equivalent to prove that the ideal $\langle s_1s_2,s_1t_2,s_2t_1\rangle$ is determinantal.  This is the ideal of $2\times 2$ minors of the following matrix, $$\left[\begin{array}{ccc} s_1 & t_1 & 0 \\ 0 & t_2 & s_2 \end{array}\right].$$
Since the homomorphism of rings is injective at generic points, the kernel is a torsion ideal.  Since the domain ring is Cohen-Macaulay and generically reduced, it is reduced.  Thus the kernel is zero.  Thus $S$ is the isomorphic image of the domain ring.  
Denote by $J_1,$ resp. $J_2,$ $J_3,$ the ideal in $S$ generated by $\langle s_2,t_2\rangle,$ resp. $\langle s_1,t_1\rangle,$ $\langle s_1,s_2\rangle.$  Denote by $S_i$, resp. $S_{i,j}$, $S_{1,2,3}$, the quotient ring $S/J_i,$ resp. $S/(J_i+I_j),$ $S/(J_1+J_2+J_3).$  Similarly, denote by $R_i$, resp. $R_{i,j},$ $R_{1,2,3},$ the quotient ring $R/I_i,$ resp. $R/(I_i+I_j),$ $R/(I_1+I_2+I_3).$
Consider the following commutative diagram with exact rows, 
$$
\begin{array}{ccccccccccc}
0 & \rightarrow & S & \rightarrow & S_1 \oplus S_2 \oplus S_3 & \rightarrow & S_{1,2} \oplus S_{1,3} \oplus S_{2,3} & \rightarrow & S_{1,2,3} & \rightarrow & 0 \\
& & \downarrow & & \downarrow & & \downarrow & & \downarrow & &\\
0 & \rightarrow & R & \rightarrow & R_1 \oplus R_2 \oplus R_3 & \rightarrow & R_{1,2} \oplus R_{1,3} \oplus R_{2,3} & \rightarrow & R_{1,2,3} & \rightarrow & 0 \\
\end{array}.
$$
After passing to the completions, every vertical arrow is an isomorphism, except possibly the first.  Thus, by the Snake Lemma, also the first vertical arrow is also an isomorphism after passing to the completion.  Thus the completion of $R$ is isomorphic to the completion of $S$.  
A local Noetherian ring is Cohen-Macaulay if and only if the completion is Cohen-Macaulay, cf. Theorem 17.5, p. 136 of Commutative ring theory by Matsumura.  Thus, the completion of $S$ is Cohen-Macaulay.  So the completion of $R$ is Cohen-Macaulay.  Therefore, also $R$ is Cohen-Macaulay.
