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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 ?

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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 $V(\mathfrak p)$.

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You know that there is a bijection between prime ideals of $A$ and irreducible closed subsets. Thus, you want to find a maximal irreducible subset. Clearly those must be of the form $V(p)$ for p minimal.

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