Why is a null field also primary, in CFT? I do not understand how to prove the statement, mentioned in pag 205 of Di Francesco's book "Conformal Field Theory":
Let $|\chi\rangle$ be a null vector, i.e. such that $L_n |\chi\rangle = 0$ for $n>0$, let $\chi$ be the corresponding field, via the operator-vector correspondance:
$$|\chi\rangle = \lim_{z->0}\ \chi |0\rangle,$$
then $\chi$ is a primary field, that is $(L_n \chi)(z)=0$ for all $n>0$.
I can see that $(L_n \chi)(0)=0$ for all $n>0$, but I cannot see why the descendent field is zero at every $z$.
 A: The state $L_n\chi$ is zero. The state-field correspondence is linear, therefore the corresponding field is identically zero.
A: I think your problem comes from a misunderstanding of the conclusion: 

I cannot see why the descendent field is zero at every $z$.

This is not what is said in the book. Rather, the claim is that all the descendants of $|\chi \rangle$ are orthogonal to all states of the Verma module having the same level. 
Let me recall how the story goes. You construct the Verma module $V(c,h)$ starting from a highest-weight state $| h \rangle$ which satisfies $L_0 | h \rangle = h | h \rangle$ and $L_n | h \rangle = 0$ for $n>0$. By definition, the descendants of $| h \rangle$ are the states obtained by acting on it with the operators $L_{-n}$, $n>0$. 
Now it may happen that one of these descendants (call it $|\chi \rangle$) is null, i.e. satisfies $\langle \chi |\chi \rangle = 0$. For concreteness let's take $c=1$ and $h=1/4$. Then you can see that $$|\chi \rangle = (L_{-2} - L_{-1}^2) |h \rangle$$ is null. This is a state at level 2. The statement is that any descendent of $|\chi \rangle$ of level $k$ is orthogonal to any state of $V(c,h)$ at level $k$. 
The proof is given around equation (7.18), but maybe it's easier to see an example. For instance $L_{-6} |\chi \rangle $ is a descendent of $|\chi \rangle $ at level $8$, so it has to be orthogonal to $L_{-8} | h \rangle$. I leave it to you to check that, using the Virasoro algebra. 
A: We need to verify is that if we mod out $|\chi\rangle$ (i.e. we declare it is the zero vector), then $(L_n\chi)(z)$ is zero for all $z$ and not just in $0$.
In the same page (205) Di Francesco says if $|\chi\rangle$ is orthogonal to the entire Verma moduel then all its descendants $L_{-k_1}\cdots L_{-k_n}|\chi\rangle$ are also orthogonal. So we can mod them out too, thus implying
$$
(L_{-k_1}\cdots L_{-k_n}\chi)(0) = 0\,,\label{1}\tag{1}
$$
through the operator-state correspondence. Now we need to introduce the $z$ dependence. Since $L_{-1} = \partial_z$ is the generator of translations
$$
\chi(z) = (e^{z\,L_{-1}}\chi)(0) = \sum_{k=1}^\infty \frac{z!}{n!} (L_{-1}^k\chi)(0)
\,.$$
Apply $L_{n}$ to this and you'll see that each term will eventually be either $(L_{m_1}\cdots L_{m_j}\chi)(0)$, for $m_j>0$, or \eqref{1}.
