Home » Mathematics » Algebraic geometry » A small gluing construction

A small gluing construction

This came up in my research recently.

In geometry, “simple nodes” are the simplest, nicest kinds of curve singularities — they have length 1, they’re resolved after a single blowup, they are the only singularities that occur “generically”, and their scheme-theoretic properties seem generally well-understood and well-behaved. (For example, the moduli space \mathcal{M}_{g,n} of marked points on stable curves of genus g exists because of, among other things, the restriction to only nodal curves.)

Intuitively, to obtain a nodal curve N, one takes two points p_1 \in C_1 and p_2 \in C_2 (on smooth curves, say) and glues the curves together at those two points. This is a minimal gluing — the pushout of the diagram showing the inclusion of a point at each p_i \in C_i:

diagram1

In particular, the glued curve N has a universal property: if f : C_1 \to X and g : C_2 \to X are two morphisms, such that f(p_1) = g(p_2) (considering only the closed points), then f and g glue to give a unique morphism N \to X.

But does the same construction work for, say, gluing a complicated singularity of C_1 to another complicated singularity on C_2 with ‘minimal contact’?

Turns out, the answer is yes, and in fact more is true:

Proposition. Let X, Y be integral schemes over an algebraically closed field k, and pick x \in X and y \in Y. Then there exists a scheme X \sqcup_{x \sim y} Y, the gluing of X to Y at the points x \in X, \ y\in Y, with the following properties:

  • It has two irreducible components, isomorphic to X and Y, which intersect scheme-theoretically in one reduced point;
  • It satisfies the universal property stated above — namely, a map X \sqcup_{x \sim y} Y \to Z is the same as a pair of maps X \to Z,\ Y \to Z that agree set-theoretically at x and y.

Moreover, the first property implies the second, and so uniquely characterizes the gluing (up to unique isomorphism).

Corollary. Let f : Z \to X \sqcup_{x \sim y} Y be any morphism whose restrictions to X and to Y are isomorphisms. Then f is an isomorphism.

(This follows by using the universal property to construct the inverse map back to Z.)

The first property ends up not being so hard to prove: we look inside X \times Y, and take the union of the fibers X \times \{y\} and \{x\} \times Y. This obviously satisfies the first bullet point. (If this is hard to picture when X and Y are highly singular, consider the case where they are projective, and consider them inside \mathbb{P}^n \times \mathbb{P}^m. Clearly, the “horizontal fiber” intersects the “vertical fiber” in one reduced point.)

To show that the first property implies the second, here’s what’s going on algebraically. We have a reduced local ring (R,\mathfrak{m}) with exactly two minimal prime ideals P,Q. (In particular P \cap Q = 0.) The fact that the irreducible components meet scheme-theoretically in one point translates to P + Q = \mathfrak{m}.

The statement we want is the following: let f : S \to R/P and g : S \to R/Q be two maps such that \pi \circ f = \pi \circ g, where \pi is the projection map R/P \to R/\mathfrak{m} or R/Q \to R/\mathfrak{m}.

Then there exists a unique homomorphism S \to R such that the following diagram commutes:

diagram

Constructing this map basically comes down to the standard but slightly-tricky-to-prove short exact sequence

0 \to R/(P \cap Q) \to R/P \oplus R/Q \to R/(P+Q) \to 0,

where the first map is restriction and the second is subtraction. In our case, this sequence becomes

0 \to R \to R/P \oplus R/Q \to R/\mathfrak{m} \to 0.

In particular, let s \in S with images a \in R/P and b \in R/Q. Since a,b have the same image in the residue field, by the sequence above, they lift to a unique element r \in R. Then the assignment s \mapsto r is the desired ring homomorphism.

(There are many other nice properties of R. For example, the ‘dual’ sequence to the one above is the more well-known

0 \to P \cap Q \to P \oplus Q \to P+Q \to 0,

which in this case gives a direct sum decomposition \mathfrak{m} \cong P \oplus Q.)

So it really does make sense to glue arbitrary varieties together “transversely”, even at highly singular points! I haven’t thought about what happens if you try to include nonreducedness, though.

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