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 $p_i \in C_i$:

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:

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.