3 Functoriality and Ring Homomorphisms

3.1 Functorial Behavior of Spec

Definition 25 Pushforward of Ideals

For a ring homomorphism \(\phi : R \to S\) and an ideal \(I \triangleleft R\), the pushforward is:

\[ \phi (I) := \phi (I) \cdot S = \{ \sum _j \phi (a_j) s_j : a_j \in I, s_j \in S\} \]

This is the ideal of \(S\) generated by \(\phi (I)\).

Definition 26 Induced Frame Homomorphism

For a ring homomorphism \(\phi : R \to S\), define:

\[ \phi ^*: \mathrm{Rad}(R) \to \mathrm{Rad}(S) \]

by:

\[ \phi ^*(I) := \sqrt{\phi (I) \cdot S} \]

for each radical ideal \(I \in \mathrm{Rad}(R)\).

Lemma 27 Image of Radical is Radical

If \(I \in \mathrm{Rad}(R)\), then \(\phi ^*(I) = \sqrt{\phi (I) \cdot S} \in \mathrm{Rad}(S)\).

Proof ▶

The radical of any ideal is radical, by definition.

Lemma 28 Preservation of Top

\(\phi ^*(R) = S\).

Proof ▶

\begin{align*} \phi ^*(R) & = \sqrt{\phi (R) \cdot S} \\ & = \sqrt{S} \quad \text{(since } \phi (R) \text{ generates } S \text{ as an ideal)} \\ & = S \quad \text{(since } S \text{ is radical: } \sqrt{S} = S \text{)} \end{align*}

Lemma 29 Preservation of Finite Meets

For \(I, J \in \mathrm{Rad}(R)\):

\[ \phi ^*(I \land J) = \phi ^*(I) \land \phi ^*(J) \]

Proof ▶

We have \(I \land J = I \cap J\) for radical ideals. Thus:

\begin{align*} \phi ^*(I \cap J) & = \sqrt{\phi (I \cap J) \cdot S} \\ & = \sqrt{(\phi (I) \cap \phi (J)) \cdot S} \end{align*}

On the other hand:

\begin{align*} \phi ^*(I) \land \phi ^*(J) & = \sqrt{\phi (I) \cdot S} \cap \sqrt{\phi (J) \cdot S} \end{align*}

By Lemma 15, \(\sqrt{(\phi (I) \cdot S) \cap (\phi (J) \cdot S)} = \sqrt{(\phi (I) \cdot S) \cdot (\phi (J) \cdot S)}\)...

Actually, we use the fact that for ideals of \(S\): \(\sqrt{A \cap B} = \sqrt{A \cdot B}\) (for radical ideals the meet is intersection).

So:

\begin{align*} \sqrt{(\phi (I) \cap \phi (J)) \cdot S} & = \sqrt{(\phi (I) \cdot S) \cap (\phi (J) \cdot S)} \\ & = \sqrt{(\phi (I) \cdot S) \cdot (\phi (J) \cdot S)} \\ & = \sqrt{\sqrt{\phi (I) \cdot S} \cdot \sqrt{\phi (J) \cdot S}} \\ & = \sqrt{\phi (I) \cdot S} \cap \sqrt{\phi (J) \cdot S} \end{align*}

Lemma 30 Preservation of Arbitrary Joins

For a family \((I_j)_{j \in J}\) of radical ideals:

\[ \phi ^*(\bigvee _j I_j) = \bigvee _j \phi ^*(I_j) \]

Proof ▶

We have:

\begin{align*} \phi ^*(\bigvee _j I_j) & = \phi ^*(\sqrt{\sum _j I_j}) \\ & = \sqrt{\phi (\sum _j I_j) \cdot S} \\ & = \sqrt{(\sum _j \phi (I_j)) \cdot S} \\ & = \sqrt{\sum _j (\phi (I_j) \cdot S)} \end{align*}

On the other hand:

\begin{align*} \bigvee _j \phi ^*(I_j) & = \bigvee _j \sqrt{\phi (I_j) \cdot S} \\ & = \sqrt{\sum _j \sqrt{\phi (I_j) \cdot S}} \\ & = \sqrt{\sum _j (\phi (I_j) \cdot S)} \end{align*}

(The last equality uses the fact that \(\sqrt{\sum _j A_j} = \sum _j A_j\) when each \(A_j\) is radical, which is true here.)

Theorem 31 \(\phi ^*\) is a Frame Homomorphism

For a ring homomorphism \(\phi : R \to S\), the map \(\phi ^*: \mathrm{Rad}(R) \to \mathrm{Rad}(S)\) is a frame homomorphism.

Proof ▶

A frame homomorphism must preserve arbitrary joins and finite meets, which are verified by Lemma 30 and Lemma 29. It must also preserve \(\top \), verified by Lemma 28.

3.2 Functorial Properties

Lemma 32 Identity Homomorphism

For the identity ring homomorphism \(\mathrm{id}_R : R \to R\):

\[ (\mathrm{id}_R)^* = \mathrm{id}_{\mathrm{Rad}(R)} \]

Proof ▶

For any \(I \in \mathrm{Rad}(R)\):

\begin{align*} (\mathrm{id}_R)^*(I) & = \sqrt{\mathrm{id}_R(I) \cdot R} \\ & = \sqrt{I \cdot R} \\ & = \sqrt{I} \\ & = I \end{align*}

Theorem 33 Functoriality of Spec

There exists a contravariant functor:

\[ \mathrm{Spec}: \mathbf{CRing}^{\mathrm{op}} \to \mathbf{Loc} \]

defined by:

Objects: \(\mathrm{Spec}(R) := \mathrm{Rad}(R)\) (viewed as a locale)
Morphisms: For \(\phi : R \to S\) in \(\mathbf{CRing}\), \(\mathrm{Spec}(\phi ) := (\phi ^*)^{\mathrm{op}}: \mathrm{Spec}(S) \to \mathrm{Spec}(R)\)

with:

\(\mathrm{Spec}(\mathrm{id}_R) = \mathrm{id}_{\mathrm{Spec}(R)}\)
For \(\phi : R \to S\) and \(\psi : S \to T\): \(\mathrm{Spec}(\psi \circ \phi ) = \mathrm{Spec}(\phi ) \circ \mathrm{Spec}(\psi )\)

Proof ▶

Objects: For each commutative ring \(R\), we assign the locale \(\mathrm{Spec}(R) := \mathrm{Rad}(R)\).

Morphisms: For each ring homomorphism \(\phi : R \to S\), the frame homomorphism \(\phi ^*: \mathrm{Rad}(R) \to \mathrm{Rad}(S)\) induces a locale morphism \(\mathrm{Spec}(\phi ): \mathrm{Spec}(S) \to \mathrm{Spec}(R)\) via the opposite functor.

Identity: Lemma 32 shows \(\mathrm{Spec}(\mathrm{id}_R) = \mathrm{id}_{\mathrm{Spec}(R)}\).

Composition: For \(\phi : R \to S\) and \(\psi : S \to T\):

\begin{align*} (\psi \circ \phi )^*(I) & = \sqrt{(\psi \circ \phi )(I) \cdot T} \\ & = \sqrt{\psi (\phi (I)) \cdot T} \end{align*}

We need to show this equals \(\phi ^*(\psi ^*(I))\). Working in the opposite category, we have:

\begin{align*} \mathrm{Spec}(\psi \circ \phi ) & = \mathrm{Spec}(\psi ) \circ \mathrm{Spec}(\phi ) \end{align*}

This follows from the composition law for frame homomorphisms.