4 The Structure Sheaf

4.1 Sheaves on Locales

Definition 34 Sheaf on a Locale

Let \(L\) be a frame (viewed as a category with objects being elements of \(L\) and morphisms being the order relations \(u \leq v\)). A sheaf on \(L\) with values in a category \(\mathcal{C}\) is a functor \(\mathcal{F}: L^{\mathrm{op}} \to \mathcal{C}\) such that for every family \((u_i)_{i \in I}\) with \(\bigvee _i u_i = u\):

\[ \mathcal{F}(u) \to \prod _i \mathcal{F}(u_i) \rightrightarrows \prod _{i,j} \mathcal{F}(u_i \land u_j) \]

is an equalizer diagram in \(\mathcal{C}\).

Remark 35 Sheaf Exactness

The equalizer condition states that a section \(s \in \mathcal{F}(u)\) is uniquely determined by its restrictions to the basic opens \((u_i)_{i \in I}\), and any compatible family of sections on the opens glues to a unique global section.

4.2 The Structure Sheaf on Zariski Locales

Definition 36 Localization of a Ring at an Element

For \(f \in R\), define \(R_f\) as the localization of \(R\) at the multiplicative set \(\{ 1, f, f^2, \ldots \} \):

\[ R_f := \{ r/f^k : r \in R, k \in \mathbb {N}\} \]

with the obvious ring operations.

Lemma 37 Basic Properties of Localizations

There is a canonical ring homomorphism \(\iota _f: R \to R_f\) sending \(r \mapsto r/1\).
If \(f\) is a unit in \(R\), then \(R_f = R\).
If \(f\) is nilpotent, then \(R_f\) is the zero ring.

Proof ▶

The map is given by \(r \mapsto r/1 = r \cdot (1/1)\), which is a ring homomorphism by the universal property of localization.
If \(f\) is a unit with inverse \(f^{-1}\), then \(1/f^k = (f^{-1})^k\) is in \(R\), so \(R_f = R\).
If \(f^n = 0\), then for any \(r/f^k\), taking \(m \geq n\): \((r/f^k) \cdot (1/f^m) = r/f^{k+m}\) is defined, and we can kill any denominator. Actually, we need to show every element is zero. If \(f^n = 0\) and \(m \geq n\), then \(f^m = 0\), so \(r/f^m = r \cdot 0 = 0\) formally. Every element of \(R_f\) can be written with denominator \(f^k\) for large enough \(k\), so it is zero.

Definition 38 Structure Sheaf on Zariski Locale

Define \(\mathcal{O}_{\mathrm{Spec} R}\) on the basic opens by:

\[ \mathcal{O}_{\mathrm{Spec} R}(D(f)) := R_f \]

For a inclusion \(D(f) \leq D(g)\) (i.e., \(f \in \sqrt{(g)}\), so \(f^n = rg\) for some \(r\) and \(n \geq 1\)):

\[ \rho _{g,f}: R_g \to R_f, \quad \frac{a}{g^k} \mapsto \frac{a r^k}{f^{nk}} \]

Lemma 39 Restriction Maps are Well-Defined

For \(f \in \sqrt{(g)}\) (so \(f^n = rg\)), the map \(\rho _{g,f}\) is a well-defined ring homomorphism.

Proof ▶

Well-defined: If \(\frac{a}{g^k} = \frac{a'}{g^{k'}}\) in \(R_g\) (i.e., \(g^m(a g^{k'} - a' g^k) = 0\) for some \(m\)), we need to show \(\frac{ar^k}{f^{nk}} = \frac{a'r^k}{f^{nk'}}\) in \(R_f\).

From \(g^m(ag^{k'} - a'g^k) = 0\), we have \(g^{m+k'}a = g^{m+k}a'\).

Substituting \(g = f^n/r\):

\begin{align*} (f^n/r)^{m+k'} a & = (f^n/r)^{m+k} a’ \\ \frac{f^{n(m+k')}}{r^{m+k'}} a & = \frac{f^{n(m+k)}}{r^{m+k}} a’ \end{align*}

Multiply both sides by \(r^{m+k'}\) and divide by \(f^{n(m+k')}\):

\begin{align*} a & = \frac{r^{m+k'}}{r^{m+k}} \cdot \frac{f^{n(m+k)}}{f^{n(m+k')}} a’ = r \cdot f^{-n} a’ \end{align*}

So \(ar^k f^{-nk} = a' r^k f^{-nk'}\) in \(R_f\).

Ring homomorphism: \(\rho _{g,f}\) preserves addition, multiplication, and the unit by the homomorphism properties of localization.

Lemma 40 Restriction Maps Compose

For \(D(f) \leq D(g) \leq D(h)\), the restriction maps satisfy:

\[ \rho _{h,f} = \rho _{g,f} \circ \rho _{h,g} \]

Proof ▶

If \(f^n = rg\) and \(g^m = sh\), then:

\begin{align*} f^{nm} = r^m g^m = r^m s h \end{align*}

Taking \(N = nm\) and \(R_0 = r^m s\), we have \(f^N = R_0 h\).

For \(\frac{a}{h^k} \in R_h\):

\begin{align*} \rho _{g,f}(\rho _{h,g}(\frac{a}{h^k})) & = \rho _{g,f}(\frac{as^k}{g^{mk}}) \\ & = \frac{as^k r^{mk}}{f^{nmk}} \\ & = \frac{a(s^k r^{mk})}{f^{nmk}} \\ & = \frac{a(R_0)^k}{f^{Nk}}\\ & = \rho _{h,f}(\frac{a}{h^k}) \end{align*}

Definition 41 Extension to Arbitrary Radical Ideals

For a radical ideal \(I \in \mathrm{Rad}(R)\), extend the structure sheaf by:

\[ \mathcal{O}_{\mathrm{Spec} R}(I) := \lim _{f \in I} R_f \]

the inverse limit of localizations as \(f\) ranges over \(I\).

For \(I \leq J\) (i.e., \(I \subseteq J\)), the restriction map is induced by the universal property of limits.

Theorem 42 Sheaf Property of the Structure Sheaf

The structure sheaf \(\mathcal{O}_{\mathrm{Spec} R}\) is a sheaf on the Zariski locale \(\mathrm{Rad}(R)\).

Proof ▶

We verify the equalizer condition. Let \((D(f_i))_{i \in I}\) be a family of basic opens with \(\bigvee _i D(f_i) = D(f)\), i.e., \(f \in \sqrt{(f_i : i \in I)}\), so \(f^n = \sum _i r_i f_i\) for some \(r_i \in R\) and \(n \geq 1\).

Injectivity: If \(\alpha /f^k \in R_f\) restricts to zero in each \(R_{f_i}\), then there exists \(m_i\) with \(f_i^{m_i} \alpha = 0\) in \(R\).

Since the \((f_i)\) generate the unit ideal in \(R_f\) (i.e., \(1 = \sum _i (r_i/f^n)(f_i/1)\) in \(R_f\)), we can use this to show \(\alpha = 0\) in \(R_f\). Specifically, multiply by a high power to clear denominators.

Surjectivity (Gluing): Given compatible elements \((\alpha _i/f_i^{k_i}) \in \prod _i R_{f_i}\) (compatible means they agree on overlaps \(D(f_i) \land D(f_j) = D(f_i f_j)\)), we must construct a preimage in \(R_f\).

The compatibility says that on \(D(f_i f_j)\), we have \(\alpha _i/f_i^{k_i} = \alpha _j/f_j^{k_j}\), i.e., there exists \(m_{ij}\) with \((f_i f_j)^{m_{ij}} (f_j^{k_j} \alpha _i - f_i^{k_i} \alpha _j) = 0\).

Use a partition of unity in \(R_f\) (obtained from \(f^n = \sum _i r_i f_i\)) to glue:

\[ \alpha /f^N := \sum _i (r_i/f^n) \cdot (\alpha _i/f_i^{k_i}) \quad \text{(in } R_f\text{)} \]

This element restricts to \((\alpha _i/f_i^{k_i})\) on each \(D(f_i)\) by the partition of unity property.