Why is a root system irreducible iff its Dynkin diagram is connected? I think this shouldn't be a difficult proof, but I have problems in proving the implication $\Phi$ irreducible root system $\Longrightarrow$ Dynkin diagram associated $D_\Phi$ connected.
I tried arguing by contradiction. If $D_\Phi$ were disconnected, then there would exist a partition of a base $B$ of $\Phi$ $B=B_1 \cup B_2$ such that $(\beta_1,\beta_2)=0 \forall \beta_1 \in B_1, \beta_2 \in B_2$. 
Now I conjecture that the likely decomposition of the root system $\Phi$ is $\Phi=\Phi_1 \cup \Phi_2$ with $\Phi_1=\Phi \cap \text{span}_\mathbb{R} B_1$, $\Phi_2=\Phi \cap \text{span}_\mathbb{R} B_2$.
Am I right? If so, how can I prove that every element of $\Phi$ must be either a linear combination of the only elements of $B_1$ or a linear combination of the only elements of $B_2$?
Thanks! 
 A: Your reasoning is correct. You obtain an orthogonal decomposition of the vector space $V$ in direct sum $V_1\oplus V_2$ where $V_i$ is the span of roots in $D_i$. Now, you observe that the Weyl group will preserve this decomposition (as it is generated by reflections in simple roots) and argue that every root is the in the Weyl orbit of a simple root and, hence is either in $V_1$ or in $V_2$. That's all.
Edit. Here are some details. When I say that a group action preserves the decomposition I mean that every group element sends each  $V_i$ to itself. It suffices to check this form generators. The generators of the Weyl group are simple reflections corresponding to simple roots. Say, you have a simple root $\alpha$ in $D_1$. Then  $\alpha$ is orthogonal to all simple roots $\beta$ in $D_2$. Now it is immediate that the reflection $\tau_\alpha$ fixes the vector representing the root $\beta$. Thus,  $\tau_\alpha$ preserves $V_2$. By uniqueness of orthogonal complement, the isometry $\tau_\alpha$ also preserves $V_1$. 
