A topological space $X$ is called irreducible if given $A_{1}, A_{2} $ open sets $ \neq \emptyset $ then $A_{1} \cap A_{2} \neq \emptyset$.
The maximal irreducible topological subspaces of $X$ are called irreducible components.
Let $A$ be a commutative ring with unit, $X = Spec(A) $ with the Zariski topology. I have to prove that the irreducible components are $\lbrace V(p) : p\subset A \ \text{minimal prime ideal} \rbrace$ where $V(P) =\lbrace q \ \text{prime ideal } \mid p\subset q\rbrace$.
Any hint ?
Best Answer
You should start by showing that $V(I)$ is irreducible if and only if the radical $\sqrt{I}$ of $I$ is a prime ideal.
This tells you that each $V(\mathfrak p)$ is irreducible, and using that $\mathfrak p$ is a minimal prime you will quickly get that each $V(\mathfrak p)$ is an irreducible component.
Finally for the converse take a $V(I)$ which is irreducible and show that it's contained in, hence equal to, one of the $V(\mathfrak p)$.