research30 min studyLesson 561 in Mathematics

Sheaves and locally ringed spaces

Organize compatible local functions and require geometric morphisms to preserve the maximal ideals of their stalks.

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What you’ll understand

  • Separate a sheaf's gluing condition from its presheaf restriction maps.
  • Construct stalks as germs and identify their residue fields.
  • Test the local-ring condition on a proposed geometric pullback.

Schemes retain both a space and the functions that can be used near its points. A collection of rings attached independently to open sets would lose the relationship between neighboring descriptions. Sheaves supply that relationship, while local rings remember which germs vanish at the chosen point. Throughout this module rings are commutative with identity and ring homomorphisms preserve identity. The direction of a geometric map and the direction of its pullback on functions are opposite, a convention that becomes essential when checking any proposed morphism.

Restriction maps become geometry through gluing

A presheaf assigns an object to every open set and a restriction map to every inclusion, with identity and composition laws. For a sheaf, sections that agree on every pairwise overlap glue to exactly one section on the union. Existence and uniqueness are separate requirements: agreement prevents contradictory local instructions, while uniqueness prevents hidden global differences invisible on the cover. The empty open has the terminal object, which is the zero ring for sheaves of rings.

The condition concerns an actual open covering and restrictions to its overlaps. It does not say that any arbitrary list of local functions glues, nor that a section on a small open extends to the entire space. A rational expression may be regular on one chart while having a pole elsewhere. For an abelian sheaf, compatibility can be written as the kernel of a difference map; for rings, the more intrinsic formulation is an equalizer of two restriction maps.

F(U)≅Eq⁡ ⁣(∏iF(Ui)⇉∏i,jF(Ui∩Uj)),U=⋃iUi\mathcal F(U)\cong\operatorname{Eq}\!\left(\prod_i\mathcal F(U_i)\rightrightarrows\prod_{i,j}\mathcal F(U_i\cap U_j)\right),\qquad U=\bigcup_i U_i
Each arrow restricts one of the two indexed sections to an overlap. The equalizer is exactly the compatible families, including both existence and uniqueness of gluing.↔ Scroll wide equations horizontally.

A stalk records behavior after shrinking neighborhoods

The stalk at a point consists of germs of sections defined near that point. Two representatives give the same germ when they agree on some smaller common neighborhood; equality on their original entire domains is unnecessary. Algebraically this is a filtered colimit over neighborhoods. Stalks are effective for checking equality and exactness of sheaf maps, but one cannot replace a sheaf by a bare list of stalks without the information explaining how germs vary and extend locally.

A locally ringed space is a ringed space whose stalk at every point is a local ring. Its unique maximal ideal consists of the germs that are nonunits, and the quotient is the residue field at that point. Residue fields need not equal a fixed ground field: the generic point of an integral affine scheme has its function field. This is why evaluating only at familiar rational points does not recover the full geometric object.

Pullbacks must be local on every stalk

A ringed-space morphism includes a continuous map f and a sheaf map from the target's functions to the direct image of the source's functions. It induces a ring homomorphism from the target stalk at f(x) to the source stalk at x. For a morphism of locally ringed spaces this homomorphism must be local: the inverse image of the source maximal ideal equals the target maximal ideal. This ensures that a germ vanishing at the image pulls back to a germ vanishing at the source point.

A unital homomorphism between local rings need not be local. For example, the inclusion of the local ring of integers at the prime three into the rationals sends the nonunit three to a unit. It cannot be a stalk map at the closed point with the proposed image. The associated scheme morphism from the spectrum of the rationals instead lands at the generic point, where the appropriate target stalk is already a field. Keeping the point and stalk map together resolves the apparent discrepancy.

fx♯:OY,f(x)⟶OX,x,(fx♯)−1(mx)=mf(x),κ(f(x))⟶κ(x)f_x^\sharp:\mathcal O_{Y,f(x)}\longrightarrow\mathcal O_{X,x},\qquad (f_x^\sharp)^{-1}(\mathfrak m_x)=\mathfrak m_{f(x)},\qquad \kappa(f(x))\longrightarrow\kappa(x)
The maximal-ideal condition makes the induced map on residue fields well-defined. For unital maps of local rings it is equivalent to sending the target maximal ideal into the source maximal ideal.↔ Scroll wide equations horizontally.

Check a ramified local pullback

Let k be a field. Show that t↦u² defines a local homomorphism k[t]_(t)→k[u]_(u), and compute the induced residue-field map.

  1. An element inverted in k[t]_(t) is a polynomial q(t) with q(0) nonzero. Its image q(u²) has the same nonzero constant term, so it is invertible in k[u]_(u). The polynomial homomorphism therefore extends to the stated localization.

  2. The maximal ideal (t) maps into (u), since t maps to u². Conversely, a fraction with nonzero value at t=0 maps to a fraction with nonzero value at u=0, so no source unit enters the maximal ideal. The contraction is exactly (t).

  3. Reducing both local rings modulo their maximal ideals gives k. Constants are unchanged, so the residue-field homomorphism is the identity. The induced map on cotangent spaces sends the class of t to the class of u², which is zero modulo u²; locality does not require an invertible tangent map.

The result

This is a legitimate local geometric pullback even though its first-order cotangent map vanishes.

Explore the model

Interactive lab

Interactive lab · modular

Inspect units before making a local-ring claim

Pause and Reset; set Modulus m=8,Additive step a=3. Predict period 8, reflecting that 3 is a unit modulo 8. For a fresh run Reset and set m=8,a=2; predict period 4, reflecting the nonunit 2. In this finite ring the nonunits form the unique maximal ideal of even residues. The instrument computes additive orbits only; the equivalence between full period and a unit follows from gcd arithmetic, not a displayed multiplication table or a sheaf computation.

Inspect units before making a local-ring claimAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA local-ring morphism must respect which germs belong to the maximal ideal; preserving addition alone is insufficient.

Check your understanding

What extra condition distinguishes a locally ringed-space morphism from a ringed-space morphism?

Try it yourself

Does t↦1+u extend to a homomorphism k[t]_(t)→k[u]_(u)? If not, identify the target-point stalk on the t-line that is appropriate for the source point u=0.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Test the polynomial t−1, which is inverted in k[t]_(t), and evaluate the proposed image of t at u=0.

Compare with the solution

The element t−1 has nonzero constant term and is a unit in k[t]_(t), but it would map to the nonunit u. A unital ring homomorphism preserves units, so no such extension exists. The polynomial map sends u=0 to t=1. The appropriate stalk is k[t]_(t−1), and t↦1+u defines a local homomorphism from that stalk to k[u]_(u).

Take it with you

  • The sheaf axiom glues precisely the compatible local sections.
  • A point includes its local ring and residue field, not just its position in a drawing.
  • Geometric pullbacks reverse arrows and must be local on stalks.

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