Let $R$ be a commutative ring with identity. Let $X= \mathrm {Spec} (R).$ Let $A \subseteq X$ be a closed irreducible subset of $X.$ Then $A$ has a unique generic point.
I have proved the uniqueness part but can't prove the existence part of the above theorem. Uniqueness part is easy and I have proved it in the following way $:$
Let $p \in A$ be the generic point of $A.$ Then $A= \overline {\{p \}} = V(\mathcal {I} (\{p \})),$ (where $\mathcal {I} (A)$ is the vanishing ideal of $A$ for any subset $A$ of $\mathrm {Spec} (R)$ and for any ideal $I$ of $R,$ $V(I)$ is the zero set of the ideal $I$ in $\mathrm {Spec} (R)$). So we have $A = V(p) \implies \mathcal {I} (A) = \mathcal {I} (V (p)) = \sqrt {p} = p,$ since $p$ is a prime ideal of $R.$ This proves that $p = \bigcap\limits_{q \in A} q.$
So in order to show the existence of such generic point we need only to prove that $\bigcap\limits_{q \in A} q \in A.$ But how can I show that? Please help me in this regard.
Thank you very much.
$A=Spec(R/I)$ is irreducible is and only $0$ is a prime ideal of $R/I$, so $\cap_{p\in Spec(R/I)}p=0$. Another way to say that $A$ is irreducible $A=V(I)$ if $I$ is a prime since every prime in $V(I)$ contains $I$, we deduce that $\cap_{p\in V(I)}p=I$ and $I$ is the generic point.