Compute the arithmetic genus of the union of the three coordinate axes $Z(x_1x_2,x_1x_3,x_2x_3) \subseteq \mathbb{P}^3$ 
Compute the arithmetic genus of the union of the three coordinate axes $Z(x_1x_2,x_1x_3,x_2x_3) \subseteq \mathbb{P}^3$

The arithmetic genus of $X \subseteq \mathbb{P}^n$ is $g(X)=(-1)^{dim(X)}(\chi(0)-1)$, where $\chi$ is the Hilbert polynomial of $X$.
Let 
\begin{equation*}
X = Z(x_1x_2,x_1x_3,x_2x_3) \subseteq \mathbb{P}^3.
\end{equation*}
I guess that $dim(X)=1$, since the coordinate $x_0$ is free.
I showed that if $X$ is a hypersurface of degree $d$ in $\mathbb{P}^n$, then $g(X)=\binom{d-1}{n}$. I don't think that this can be applied to this problem but it's a general result that I know.
I showed also that if $X$ is a complete intersection of two surfaces of degree $a$ and $b$ in $\mathbb{P}^3$, then $g(X)=\frac{1}{2}ab(a+b-4)+1$.
Is there any solution of this problem that does not involve computing the Hilbert polynomial of $X$? However, if computing the Hilbert polynomial is simplier, that's ok for me.
Thanks!
 A: Here is an elementary, easy, self-contained  proof which, as a friend reminded me, results from a discussion we had a few months ago.  I could kick myself for forgetting about it yesterday ...
Consider two subschemes $X_1,X_2\subset \mathbb P_k^n=\operatorname {Proj}(S)\quad (S=k[T_0,\cdots,T_n]) $.
From the exact sequence $$0\to S/I(X_1)\cap I(X_2)\to S/I(X_1)\oplus S/I(X_2)\to S/(I(X_1)+ I(X_2))\to 0$$ we get  the relation between Hilbert polynomials $$ H_{X_1}(t)+H_{X_2}(t)= H_{X_1\cup X_2}(t)+ H_{X_1\cap X_2}(t) \quad (\bigstar )$$  Applying this to $$X_1=V(x_3,x_1x_2),\: X_2=V(x_1,x_2),\: X_1\cap X_2=V(x_1,x_2,x_3)=\{[1:0:0:0]\}, X=X_1\cup X_2$$ 
we get, remembering that $X_1$ is just a reducible  plane conic so that $H_{X_1}(t)=2t+1$, that $X_2$ is a line so that  $H_{X_2}(t)=t+1$, and that $X_1\cap X_2$ is a reduced point with Hilbert polynomial the constant $1$: 
             $$(2t+1)+(t+1)=H_X(t)+(1) \quad (\bigstar \bigstar)$$ Hence $H_X(t)=3t+1$ and thus $p_a(X)=0$.  
WARNING
One might wonder why (fortunately!) the above calculation  doesn't work  when $X'_2$ is a line in the same plane as $X_1$, going through the singularity $P$ of $X_1$.
The answer is that in this case the intersection $X_1\cap X'_2$ is no longer a simple point but the double point at  $P$ embedded in $X'_2$.
In that case the analogous  calculation yields for $X_0=X_1\cup X'_2$ the Hilbert polynomial $H_{X_0}(t)= 3t$, as already mentioned in my other answer.
One must  very carefully note that the formula $(\bigstar)$ is correct only if one interprets union and intersection in the scheme-theoretic sense: never ever forget about nilpotents!
A: There is a nice general method for calculating the arithmetic genus of a reducible curve like this where a bunch of curves are being glued transversely. It uses the (equivalent) description of the arithmetic genus of a curve $X$ over a field as the dimension of the cohomology group $H^1(\mathcal O_X)$.
So suppose $C_1$ and $C_2$ are curves over a field. Let $X$ denote the curve we get by glueing a point $p_1 \in C_1$ transversely to a point $p_2 \in C_2$. Call the image of the glueing points $p \in X$. Then we have an exact sequence of sheaves on $X$ of the form
$$ 0 \rightarrow O_X \rightarrow \mathcal O_{C_1} \oplus \mathcal O_{C_2} \rightarrow k_p \rightarrow 0 $$
where $k_p$ is the skyscraper sheaf supported at $p$. (The first map is pullback of functions; the second is zero away from $p$, and $(f_1,f_2) \mapsto f_1(p)-f_2(p)$ on open sets containing $p$.)
Now take cohomology of this sequence: you get
$$0 \rightarrow H^0(O_X) \rightarrow H^0(\mathcal O_{C_1}) \oplus H^0(\mathcal O_{C_2}) \rightarrow H^0(k_p) \rightarrow H^1(O_X) \rightarrow H^1(\mathcal O_{C_1}) \oplus H^1(\mathcal O_{C_2}) \rightarrow 0$$
where the final zero comes from the fact that $k_p$ is supported at a single point. 
Now, I leave it to you to check that the first three nonzero terms give a short exact sequence. That implies that $$H^1(O_X)=H^1(\mathcal O_{C_1}) \oplus H^1(\mathcal O_{C_2})$$
In other words, arithmetic genus is additive when we glue curves transversely at a point! Notice that we didn't require anything to be smooth or irreducible, so you can apply this in your situation in two steps by glueing two of the axes together, then glueing on the third. 
Exercise: Modify the above argument to show that if $Y$ is a curve obtained by transverse glueing of two points on the same connected curve $X$, then 
$$\operatorname{dim} H^1(\mathcal O_Y) = \operatorname{dim} H^1(\mathcal O_X)+1$$
Use this to figure out what happens when you glue any number of curves transversely at any number of points!
A: The required arithmetic genus is $$p_a(X)=0$$ You can see it by using a formula published in 1957 by  Hironaka in the article:
On the arithmetic genera and the effective genera of algebraic curves  (Theorem 2, page 190).
The formula is $$ p_a(C)=\pi(C)+\sum_P\delta (P)  -(r-1)$$
The sum is over the singular points $P\in C$ and $\delta (P)=\dim (\tilde {\mathcal O} _P/\mathcal O_P) $ .
The number $r$ is the number of irreducible  components of $C$ and $\pi(C)$ is the sum of the genera of those irreducible components of $C$
In our case $\pi(X)=3, \sum_P\delta (P)=3-1, r=3$ and we  thus get, as announced, $p_a(X)=0$.  
Here is a related article by Sánchez  while  this article by Hartshorne also leads to our result (cf. Theorem 3.1 ).  
Note that as a consequence (!) the Hilbert polynomial of $X$ is $H_X(t)=3t+1$.
Note also that the Hilbert polynomial of the union $X_0$ in $\mathbb P^2$ of three lines passing through a point is $H_{X_0}(t)=3t$, so that  $X$ can't be flatly deformed into $X_0$, contrary to what one might naïvely conjecture.
Edit
Browsing the Internet I have come across this paper on Hironaka's work
It puts the result in context and explains that Hironaka's article was his first one, and originated with his Master Thesis. 
Not bad for a beginner!  
