Meet decompositions in the radical ideals lattice

Let $L$ be a lattice, then a semilattice decomposition (either meet or join) is described by a function from the lattice to the power set of the lattice: \[ f : L \to \mathcal{P}(L) \] Then there are numerous types of decompositions for lattice elements, but one particularly important type which is well-studied is the complete irreducible decompositions. Using this purely lattice-theoretic notion, we can define Spec(R) for any ring by the complete meet irreducible decompositions of the radical ideals lattice. \[ f : RadicalIdeals(R) \to Spec(R) \] We know from basic order-theory that this mapping $f$ is antitone, which means that the lattice Spec(R) is order-dual to the lattice of radical ideals. Additionally, Spec(R) is necessarily a Moore family which means it is union-closed. It does not follow, however, from basic order-theory that it is finite union-closed and therefore a cotopology.

To prove that it is union-closed requires us to drop back to commutative algebra to study properties of prime ideals, which are the meet irreducibles in the lattice of radical ideals. You can find this proof briefly outlined in Hatshorne's algebraic geometry, but I want to expand on it since it is fundamental to proving that Spec(R) is a cotopology. The only thing needed is a single lemma.

Spectrum-construction lemma: Let $R$ be a commutative ring with identity. Let $I,J$ be ideals and let $P$ be a prime ideal. Then $IJ \subseteq P$ implies that $I \subseteq P \lor J \subseteq P$.

(1) suppose that $I \subseteq P$ then we already have that one of the two ideals is a subset.
(2) if $I \not\subseteq P$ then there exists $a \in I - P$, that is there is some element which is a member of $I$ but not of $P$. We will now prove what we want by universal quantification on $J$. The first thing to realize is that $ab$ is in $P$. To see this, notice that $ab$ is in $IJ$ because $a \in I$ and $b \in J$ and the product of two ideals contains all products of members of the two ideals. Further, we had $IJ \subseteq P$ by supposition, so by the transitivty of inclusion $ab \in P$. \[ \forall b : ab \in (IJ \subseteq P) \implies ab \in P \] Once we have $ab \in P$ all it takes to see that $b \in P$ is that the definition of prime ideals states that if $ab \in P$ and $a not\in P$ then $b \in P$, and both of those two conditions are satisfied so $b \in P$. \[ ab \in P \implies b \in P \] Therefore, for all $b \in J$, we have that $b \in P$. Which means that $J \subseteq P$. Finally, by logical combination of the two cases (1) and (2) we know that either $I$ or $J$ is in $P$ which is what we wanted to show.

Theorem. Spec(R) is finite union closed.

Proof. let $IJ$ be two ideals then $I \subseteq I \cap J \subseteq P$ and $J \subseteq I \cap J \subseteq P$ means that $I \subseteq P$ or $J \subseteq P$ imply that $IJ \subseteq P$. The spectrum-construction lemma demonstrates the reverse direction so $ I \subseteq P \lor J \subseteq P \iff IJ \subseteq P$.

The elements of $Spec(R)$ are sets of prime ideals that are greater then a given ideal $I \subseteq P$, so the union of the them is all sets of the form $I \subseteq P \lor J \subseteq P$ for two ideals $I,J$ but we just showed that this is equal to all prime ideals greater then $IJ$ which is therefore a member of $Spec(R)$. Therefore, $Spec(R)$ is finite union closed.

Order-theoretic collaries:

• Spec(R) forms a cotopology
• The radical ideals lattice is distributive.
• Radical ideals form a locale in the sense of pointless topology.

The additional properties of $Spec(R)$ related to compactness will be discussed later, but for now I wanted to demonstrate and confirm that this is a valid topological construction. The nice thing is that $Spec(R)$ is entirely a lattice-theoretic construction, so it is fully determined by its order-type. Therefore, $Spec(R)$ is a fundamental construction in commutative algebra which can be entirely understood using lattice theory.