Irreducible components of the spectrum correspond to minimal prime ideals

Let $\mathfrak{p}$ be a minimal prime ideal. Since $\mathfrak{p}$ is prime, fact [L1] shows that $V(\mathfrak{p})$ is irreducible.

If $V(\mathfrak{p})\subseteq Y$ for an irreducible closed subset $Y$, then [L1] gives $Y=V(\mathfrak{q})$ for a prime ideal $\mathfrak{q}$. The inclusion $V(\mathfrak{p})\subseteq V(\mathfrak{q})$ means $\mathfrak{q}\subseteq \mathfrak{p}$ by [L2]. Minimality of $\mathfrak{p}$ forces $\mathfrak{q}=\mathfrak{p}$, so $Y=V(\mathfrak{p})$. Hence $V(\mathfrak{p})$ is maximal among irreducible closed subsets, that is, an irreducible component.

Let $Y$ be an irreducible component. By [L1], $Y=V(\mathfrak{p})$ for a prime ideal $\mathfrak{p}$. If $\mathfrak{q}\subseteq \mathfrak{p}$ is another prime, then $V(\mathfrak{p})\subseteq V(\mathfrak{q})$ by [L2]. Since $V(\mathfrak{q})$ is irreducible by [L1], maximality of the component $Y$ forces $V(\mathfrak{p})=V(\mathfrak{q})$, and [L1] then gives $\mathfrak{p}=\mathfrak{q}$. Thus $\mathfrak{p}$ is minimal.

The prime ideal in step 1.3 is unique because [L1] gives a unique generic point for every irreducible component.

Steps 1.1, 1.2, 1.3, and 2.1 give the claimed correspondence between irreducible components and minimal prime ideals.