The set of radical ideals forms a distributive lattice

Question

Let $R$ be a commutative ring with unity. For radicals ideals $I, J \subseteq K$, define $I \cdot_{\operatorname{rad}} J = \sqrt{I \cdot J}$, $I \cap_{\operatorname{rad}} J = \sqrt{I \cap J}$ and $I +_{\operatorname{rad}} J = \sqrt{I + J}$. Indeed, it is well-known that $I \cdot_{\operatorname{rad}} J = I \cap_{\operatorname{rad}} J$, so we might as well refer to both operations by a common name $\circ_{\operatorname{rad}}$. I'm trying to show that $+_{\operatorname{rad}}$ and $\circ_{\operatorname{rad}}$ distribute over each other, i.e. for radical ideals $I, J, K \subseteq R$: $$ (I +_{\operatorname{rad}} J) \circ_{\operatorname{rad}} K = (I \circ_{\operatorname{rad}} K) +_{\operatorname{rad}} (J \circ_{\operatorname{rad}} K) \tag{1} $$ and $$ (I \circ_{\operatorname{rad}} J) +_{\operatorname{rad}} K = (I +_{\operatorname{rad}} K) \circ_{\operatorname{rad}} (J +_{\operatorname{rad}} K) \tag{2} $$ Let's start by showing (1). Let's show "$\supseteq$" (the easy direction): $$ \begin{array}{crc} & & I + J \supseteq I, J \\ & \Rightarrow & I +_{\operatorname{rad}} J \supseteq I, J \\ & \Rightarrow & (I +_{\operatorname{rad}} J) \circ_{\operatorname{rad}} K \supseteq I \circ_{\operatorname{rad}} K, J \circ_{\operatorname{rad}} K \\ & \Rightarrow & (I +_{\operatorname{rad}} J) \circ_{\operatorname{rad}} K \supseteq (I \circ_{\operatorname{rad}} K) + (J \circ_{\operatorname{rad}} K) \\ \text{(the LHS is radical)} & \Rightarrow & (I +_{\operatorname{rad}} J) \circ_{\operatorname{rad}} K \supseteq (I \circ_{\operatorname{rad}} K) +_{\operatorname{rad}} (J \circ_{\operatorname{rad}} K) \\ \end{array} $$ Let's show "$\subseteq$" (the hard direction): $$ \begin{array}{crc} \text{(well known)} & & (I + J) \cdot K = I \cdot K + J \cdot K \\ & \Rightarrow & (I + J) \cdot K \subseteq (I \circ_{\operatorname{rad}} K) +_{\operatorname{rad}} (J \circ_{\operatorname{rad}} K) \\ \text{(the RHS is radical)} & \Rightarrow & (I + J) \circ_{\operatorname{rad}} K \subseteq (I \circ_{\operatorname{rad}} K) +_{\operatorname{rad}} (J \circ_{\operatorname{rad}} K) \\ \left( A \circ_{\operatorname{rad}} B = \sqrt{A} \circ_{\operatorname{rad}} \sqrt{B} \right) & \Rightarrow & (I +_{\operatorname{rad}} J) \circ_{\operatorname{rad}} K \subseteq (I \circ_{\operatorname{rad}} K) +_{\operatorname{rad}} (J \circ_{\operatorname{rad}} K) \\ \end{array} $$ Now let's show (2). Let's show "$\subseteq$" (the easy direction): $$ \begin{array}{crc} & & I \cap J \subseteq I, J \\ \text{(the RHS is radical)} & \Rightarrow & I \circ_{\operatorname{rad}} J \subseteq I, J \\ & \Rightarrow & (I \circ_{\operatorname{rad}} J) +_{\operatorname{rad}} K \subseteq I +_{\operatorname{rad}} K, J +_{\operatorname{rad}} K \\ & \Rightarrow & (I \circ_{\operatorname{rad}} J) +_{\operatorname{rad}} K \subseteq (I +_{\operatorname{rad}} K) \cap (J +_{\operatorname{rad}} K) \\ & \Rightarrow & (I \circ_{\operatorname{rad}} J) +_{\operatorname{rad}} K \subseteq (I +_{\operatorname{rad}} K) \circ_{\operatorname{rad}} (J +_{\operatorname{rad}} K) \\ \end{array} $$ Let's show "$\supseteq$" (the hard direction): $$ \begin{array}{crc} & & (I \cdot J) + K \supseteq (I \cdot J) + (I \cdot K) + (J \cdot K) + K^2 = (I + K) \cdot (J + K) \\ & \Rightarrow & (I \cdot J) +_{\operatorname{rad}} K \supseteq (I + K) \circ_{\operatorname{rad}} (J + K) \\ \left( A +_{\operatorname{rad}} B = \sqrt{A} +_{\operatorname{rad}} \sqrt{B} \right) & \Rightarrow & (I \circ_{\operatorname{rad}} J) +_{\operatorname{rad}} K \supseteq (I + K) \circ_{\operatorname{rad}} (J + K) \\ \left( A \circ_{\operatorname{rad}} B = \sqrt{A} \circ_{\operatorname{rad}} \sqrt{B} \right) & \Rightarrow & (I \circ_{\operatorname{rad}} J) +_{\operatorname{rad}} K \supseteq (I +_{\operatorname{rad}} K) \circ_{\operatorname{rad}} (J +_{\operatorname{rad}} K) \\ \end{array} $$

My questions are:

1. Is there a better and more insightful proof of facts (1) and (2)?

2. If I understand it correctly, this implies that the set of radical ideals of $R$ forms a distributive lattice. Is this correct? Can we gain any further insight from this?

Answer 1

It suffices to prove (1) by the general characterization of distributive lattices.

Radical ideals are closed under arbitrary intersections, so the meet is the usual intersection - you don't need to take the radical of it anymore. Also, we can take sums for arbitary index sets, and prove the general distributive law

$$\textstyle \sqrt{\sum_{s \in S} I_s} \cap K = \sqrt{\sum_{s \in S} (I_s \cap K)}.$$

In other words, the radical ideals even form a frame. The inclusion $\supseteq$ is immediate from the definitions, and in fact the inequality $$\textstyle (\bigvee_{s \in S} x_s) \wedge y \geq \bigvee_{s \in S} (x_s \wedge y)$$ holds in every lattice (with $S$-indexed joins), for obvious reasons. So there is no necessity to prove it again in this special case.

Let me show you two conceptual proofs that the distributive law holds, and also an equational proof.

Proof 1 (with adjunctions). We need to prove that for a fixed radical ideal $K$ the map $I \mapsto I \cap K$ preserves joins of radical ideals. Since left adjoints preserve colimits, it will be enough (and in this situation, namely partial orders, it is actually equivalent!) to provide a right adjoint. Indeed, for another radical ideal $J$ we find $$I \cap K \subseteq J \iff \sqrt{IK} \subseteq J \iff IK \subseteq J \iff I \subseteq [J:K],$$ where $[J:K] = \{r \in R : rK \subseteq J\}$ is the ideal quotient. An easy argument shows that $[J:K]$ is a radical ideal (using that $J$ is one). Thus, $I \mapsto I \cap K$ is indeed left adjoint to $J \mapsto [J:K]$.

Proof 2 (with affine schemes). From algebraic geometry we have a bijection $$\{\text{radical ideals of } R\} \xrightarrow{\sim} \{\text{closed subsets of } \mathrm{Spec}(R)\},\quad I \mapsto V(I).$$ It is an anti-isomorphism of lattices because of the well-known equations $$V(I + J) = V(I) \cap V(J),\quad V(I \cap J) = V(I) \cup V(J).$$ Since the closed subsets form a distributive lattice, the same is true for the radical ideals. Also the generalized law above, because we have $$\textstyle (\bigcap_{s \in S} A_s) \cup B = \bigcap_{s \in S} (A_s \cup B)$$ for closed subsets. The failure of the distributive law for general ideals means that, somewhat unconveniently, the closed subschemes do not form a distributive lattice.

Notice that the approach with affine scheme needs the axiom of choice, otherwise there might not be enough prime ideals. It must be said that in constructive commutative algebra the story goes backwards. We start with the distributive lattice of radical ideals, which is even a frame as mentioned, and treat them as the closed subsets of a "generalized space", namely, of a locale. We can do this also when no prime ideals exist.

Proof 3 (with elements). Let $x \in \sqrt{\sum_{s \in S} I_s} \cap K$. This means that $x \in K$ and there is some $n \in \mathbb{N}$ such that $x^n = \sum_{s \in S} u_s$ with $u_s \in I_s$ (finite sum). Then $x^{n+1} = \sum_{s \in S} u_s x \in \sum_{s \in S} I_s K \subseteq \sum_{s \in S} (I_s \cap K)$. Thus, $x \in \sqrt{\sum_{s \in S} (I_s \cap K)}$.