This is a problem in three parts, I managed to prove the first part, but the others I couldn't.
Let $R$ be a ring and let $X_1,X_2\subset Spec(R)$ be closed (in Zariski topology) and disjoint such that $X_1\cup X_2=Spec(R)$. Now define $$J_i=\bigcap_{P\in X_i}P$$
for $i=1,2$. The first problem is to show that $J_1+J_2 = R$, which I did. The second problem is to show that $R\cong R/J_1\oplus R/J_2$ and I'm stuck on this. The third is to show that if a ring is a direct sum of two other rings, then it's spectrum has two connected components (at least).
I appreciate any help in this two problems. Thank you very much.
The second part is false, if $R$ is not reduced (But dividing out the nil radical does not change the topology of $Spec(R)$). Actually we have $R/nil(R) \cong R/J_1 \times R/J_2$ by the Chinese Remainder Theorem, since $J_1 \cap J_2$ is the intersection of all primes of $R$, hence equal to $nil(R)$.
For the third part, you should take a ring $R \times S$ and consider the elements $e_1 = (1,0),e_2=(0,1)$. Any prime ideal contains either $e_1$ or $e_2$ (but never both of them). Along these lines, you can decompose the spectrum into two components, each corresponding to those primes who contain $e_1$ or $e_2$. More precisely speaking you should (and can easily) show $$Spec(R \times S) = V(e_1) \sqcup V(e_2) = D(e_2) \sqcup D(e_1)$$