i have trouble proving this assertion:
Let $R=\oplus_{n\in\mathbb{Z}_{\geq 0}}R_n$ be a positively graded ring, $R_+:=\oplus_{n\in\mathbb{Z}>0}R_n$, $\mathrm{Spec}(R):=\{p: p\text{ prime ideal of }R\}$ and $\mathrm{Proj}(R):=\{p\in \mathrm{Spec}:p \text{ is homogeneous and }p\nsupseteq R_+\}$.
Every irreducible subset of $\mathrm{Proj}(R)$ respect to the topology induced by the Zarisky topology on $\mathrm{Spec}(R)$ is of the form $V(p)\cap\mathrm{Proj}(R)$ with $p\in\mathrm{Proj}(R)$ (the empty set is considered reducible).
($V$ is the function that brings an ideal $I$ of $R$ in the subset of $\mathrm{Spec}(R)$ composed by all the prime ideals that contains $I$).
I've tried to approach the problem directly:
It is clear that if $p\in\mathrm{Proj}(R)$ than $V(p)\cap\mathrm{Proj(R)}$ is irreducible.
For the first implication let $V(a)\cap\mathrm{Proj}(R)$ be irreducible with $a$ not prime and homogeneous than $V(a)$ is reducible in $\mathrm{Spec}(R)$, so $V(a)=V(b)\cup V(c)$ with $a$ and $b$ ideals of $R$.
Now I don't know how to show that $V(c)\cap\mathrm{Proj}(R)$ and $V(b)\cap \mathrm{Proj}(R)$ are non empty and strictly contained in $V(a)\cap\mathrm{Proj}(R)$.
I also tried to use a proof that looks like the one in $\mathrm{Spec(R)}$ so suppose that $a$ is radical, homogeneous and not prime, take $x,y$ homogeneous not in $a$ with $xy\in a$. Now one has that $V(a+(x))\cup V(a+(y))=V(a)$ and the same holds passing to $\mathrm{Proj(R)}$ but i can't prove that $R_+\nsubseteq a+(x)$ and $R_+\nsubseteq b+(y)$ and also that $V(a+(x))\cap \mathrm{Proj(R)}\subsetneq V(a) \cap \mathrm{Proj(R)}$ and $V(a+(y))\cap \mathrm{Proj(R)}\subsetneq V(a) \cap \mathrm{Proj(R)}$.
I know that for every ideal $I$ of $R$ exists an homogeneous ideal $J$ such that $V(I)\cap\mathrm{Proj}(R)=V(J)\cap\mathrm{Proj}(R)$.
I am a bit surprised by the fact that this question is not a duplicate, if it is so i sorry, and please link me where is.
Can please someone help me?
Thanks.