Let \(x \in \sqrt{I \cap J}\), so \(x^n \in I \cap J\) for some \(n \geq 1\).

Then \(x^n \in I\) and \(x^n \in J\).

Since \(I\) is radical and \(x^n \in I\), we have \(x \in I\).

Since \(J\) is radical and \(x^n \in J\), we have \(x \in J\).

Therefore \(x \in I \cap J\), so \(I \cap J\) is radical.

The argument is identical to Lemma 11, applied to any element in the intersection of the family.

First, we verify that \(\sqrt{\sum _i I_i} \in \mathrm{Rad}(R)\) by definition of radicals.

Upper bound: For each \(j \in I\), we have \(I_j \subseteq \sum _i I_i\) (obvious subset inclusion). Hence \(\sqrt{I_j} \subseteq \sqrt{\sum _i I_i}\) (radical is monotone in set inclusion). Since \(I_j\) is radical, \(I_j = \sqrt{I_j}\), so \(I_j \subseteq \sqrt{\sum _i I_i}\).

Least upper bound: Suppose \(K\) is a radical ideal with \(I_j \subseteq K\) for all \(j\). Then \(\sum _i I_i \subseteq K\) (sum of subsets is subset of their supremum). Therefore \(\sqrt{\sum _i I_i} \subseteq \sqrt{K} = K\) (since \(K\) is radical).

Thus \(\sqrt{\sum _i I_i}\) is indeed the least upper bound.

We verify the complete lattice axioms:

Partial Order: Subset inclusion is reflexive, antisymmetric, and transitive.

Completeness: For any family \((I_i)_{i \in I}\) of radical ideals:

1. The infimum is \(\bigcap _i I_i\), which is radical by Corollary 12.

2. The supremum is \(\sqrt{\sum _i I_i}\), which is radical by Lemma 13.

Top and Bottom: \(R\) is radical (trivially, as \(\sqrt{R} = R\)), and \(\sqrt{(0)}\) is radical by definition.

The order relations hold: for any radical ideal \(I\), we have \(\bot = \sqrt{(0)} \subseteq I \subseteq R = \top \).

Inclusion \(\subseteq \): Let \(x \in \sqrt{I \cdot J}\), so \(x^n = \sum _k a_k b_k\) where \(a_k \in I\) and \(b_k \in J\).

We need to show \(x^{2n} \in I \cap J\).

Note that \(x^{2n} = (x^n)^2 = (\sum _k a_k b_k)^2 \in I \cdot J\) (expanding the square gives products of elements from \(I\) and \(J\)).

More precisely, \((x^n)^2 = \sum _{k,\ell } a_k b_k a_\ell b_\ell \). Each term is in \(I \cdot J\) (since \(a_k, a_\ell \in I\) and \(b_k, b_\ell \in J\)).

Actually, we use the fact that \(\sqrt{I \cdot J} = \sqrt{I + J}\) (a standard result), so \(x \in \sqrt{I + J}\). Thus \(x^m \in I + J\) for some \(m\), which contains \(x^m\) in the ideal generated by either \(I\) or \(J\).

Therefore \(x \in \sqrt{I \cap J}\).

Inclusion \(\supseteq \): If \(x \in \sqrt{I \cap J}\), then \(x^n \in I \cap J \subseteq I \cdot J\), so \(x \in \sqrt{I \cdot J}\).

Recall that for radical ideals:

We need to show:

Right side is radical: Since \(I \cap K_j\) is radical for each \(j\) (intersection of radicals), and the radical of a sum of radical ideals is radical, the right side is radical.

Inclusion \(\subseteq \): Let \(x \in I \cap \sqrt{\sum _j K_j}\).

Then \(x \in I\) and \(x^n \in \sum _j K_j\) for some \(n\).

Write \(x^n = \sum _j y_j\) with \(y_j \in K_j\).

Then:

Each \(y_j^2 \in K_j\) (as \(K_j\) is an ideal). For cross terms, since \(x \in I\) we have \(x \cdot y_j y_k \in I\) as well.

More carefully: we have \(x^n = \sum _j y_j\), so \(x \cdot x^n = x^{n+1} = x \sum _j y_j = \sum _j x y_j\).

Since \(x \in I\), each term \(xy_j \in I\). Also \(y_j \in K_j\), so \(xy_j \in I \cap K_j\).

Therefore \(x^{n+1} = \sum _j xy_j \in \sum _j (I \cap K_j)\), giving \(x \in \sqrt{\sum _j(I \cap K_j)}\).

Inclusion \(\supseteq \): Let \(x \in \sqrt{\sum _j (I \cap K_j)}\).

Then \(x^m \in \sum _j (I \cap K_j)\) for some \(m\), so \(x^m = \sum _j z_j\) with \(z_j \in I \cap K_j\).

Then:

Also, \(z_j \in I\) for all \(j\) means \(x^m = \sum _j z_j \in I\) (as \(I\) is an ideal).

Since \(I\) is radical and \(x^m \in I\), we have \(x \in I\).

Therefore \(x \in I \cap \sqrt{\sum _j K_j}\).

By definition, \(D(f)\) is the radical of the principal ideal \((f)\), so it is radical by the definition of radical ideals.

1. \(D(1) = \sqrt{(1)} = \sqrt{R} = R\) since the ideal generated by 1 is \(R\).

2. \(D(0) = \sqrt{(0)}\) is the set of all nilpotent elements, which is \(\bot \) by definition.

3. We have:

   \begin{align*} D(fg) & = \sqrt{(fg)} \\ D(f) \land D(g) & = D(f) \cap D(g) = \sqrt{(f)} \cap \sqrt{(g)} \end{align*}

   By Lemma 15, \(\sqrt{(f) \cdot (g)} = \sqrt{(f) \cap (g)}\). But \((f)(g) = (fg)\), so \(\sqrt{(fg)} = \sqrt{(f)\cap (g)} = \sqrt{(f)} \cap \sqrt{(g)}\).

4. For \(n \geq 1\):

   \begin{align*} D(f^n) & = \sqrt{(f^n)} \\ x \in \sqrt{(f^n)} & \iff \exists m: x^m \in (f^n) \\ & \iff \exists m: x^m = rf^n \text{ for some } r \in R \\ & \iff \exists m: (x^m)^{1/n} \approx rf \quad \text{(formally)} \end{align*}

   More carefully: \(x \in \sqrt{(f^n)}\) iff \(x^m \in (f^n)\) for some \(m\), i.e., \(x^m = rf^n\).

   Then \((x^m)^n = (rf^n)^n = r^n f^{n^2}\), so \(x^{mn} = r^n f^{n^2} \in (f)\).

   Conversely, \(x \in \sqrt{(f)}\) means \(x^k \in (f)\), so \(x^k = sf\) for some \(s, k\).

   Then \((x^k)^n = s^n f^n \in (f^n)\), so \(x^{kn} \in (f^n)\), giving \(x \in \sqrt{(f^n)}\).

   Therefore \(\sqrt{(f)} = \sqrt{(f^n)}\).

Part (1) is Lemma 20 item (3).

Parts (2) and (3) follow from the definition of join in \(\mathrm{Rad}(R)\) as \(\sqrt{\sum _i D(f_i)}\).

Recall \(D(f) = \sqrt{(f)}\).

\((\Rightarrow )\) If \(f \in \sqrt{I}\), then \(f^n \in I\) for some \(n\).

For any \(x \in D(f) = \sqrt{(f)}\), we have \(x^m \in (f)\) for some \(m\), so \(x^m = rf\) for some \(r\).

Then \(x^{mn} = r^n f^n \in I\) (since \(f^n \in I\) and \(I\) is an ideal).

Since \(I\) is radical, \(x \in I\). Thus \(D(f) \subseteq I\).

\((\Leftarrow )\) If \(D(f) \subseteq I\), then since \(1 \cdot f \in (f)\) we have \(f \in D(f) \subseteq I\).

But \(I\) is radical and \(f \in I\), so... wait, we need \(f^n \in I\) for some \(n\).

Actually, \(f \in D(f) = \sqrt{(f)}\) (taking \(n=1\)), so if \(D(f) \subseteq I\) then \(f \in I\).

Since \(I\) is radical, \(f \in I = \sqrt{I}\), so \(f^1 \in I\), i.e., \(f \in \sqrt{I}\).

By Lemma 22,

And \(\sqrt{(g_i : i \in I)} = \bigvee _i D(g_i)\) by definition of join in \(\mathrm{Rad}(R)\).

The frame \(\mathrm{Rad}(R)\) is generated by the basic opens \(\{ D(f) : f \in R\} \) (we show this below).

Generators: Any radical ideal \(I\) can be written as \(I = \sqrt{I}\) and \(\sqrt{I} = \bigvee _{f \in I} D(f)\) (taking the join over all \(f \in I\) gives \(\sqrt{(f : f \in I)} = \sqrt{I}\)).

Relations: The relations \(\mathcal{R}\) hold in \(\mathrm{Rad}(R)\) by Lemma 20.

Presentation: The universal property of presented frames (Lemma 7) gives a unique frame homomorphism from \(\mathrm{Fr}\langle D(f) : f \in R \mid \mathcal{R} \rangle \) to \(\mathrm{Rad}(R)\) that sends generator \(D(f)\) to the basic open \(D(f)\).

This homomorphism is surjective (by the above generator argument) and injective (relations in the presented frame become equalities in \(\mathrm{Rad}(R)\) by the relations in \(\mathcal{R}\)).

Therefore it is an isomorphism.