I am trying to get comfortable with flat limits. This question is motivated by Section 3.5.3 of Eisenbud and Harris's '3264 And All That' and Exercises 3.35 and 3.36. This section and the surrounding ones are dedicated to working with flat limits, but unfortunately, a lot of the arguments seem "handwavey" to me. I feel like this is a general issue for arguments like this: they all either seem not entirely convincing to me, or rely on some software to compute a saturation of an ideal.
Let $C$ be a curve of $\mathbb{P}^3$ over an algebraically closed field. Our goal is to compute the rational equivalence class of $\Gamma_C\subset\mathbb{G}(1,3)$ of lines meeting $C$.
Choose a plane $H$ intersecting $C$ transversely in points $p_1,\ldots,p_d$, and a point $q$ not lying on $C$. Choose coordinates $z_0,\ldots,z_3$ such that $q=(1:0:0:0)$ and $H=\{z_0=0\}$. Consider the subgroup of $PGL_4$ consisting of the automorphisms with matrices
$$A_t=\left(
\begin{matrix}
1 & 0&0&0\\
0&t&0&0\\
0&0&t&0\\
0&0&0&t
\end{matrix}\right),\ t\neq0.$$
To do this, we consider the closure $\overline{\Psi}$ of the locus
$$\Psi=\{(t,L)\subset \mathbb{A}^1\times\mathbb{G}(1,3)\ |\ t\neq0,\ L\cap A_t(C)\neq\varnothing\}.$$
I would like to identify the fiber $\overline\Psi_0$ as the union of the Schubert cycles $\Sigma_1(\overline{p_i,q})$ of lines meeting the lines $\overline{p_i,q}$ with multiplicity 1 along each. The support of $\overline\Psi_0$ must be contained in the locus of lines meeting the flat limit $C_0=\lim_{t\to0} A_t(C)$, which is supported on the union of lines $\overline{p_i,q}$, and it certainly contains the lines $\overline{p_i,q}$ themselves, since they meet each of the $A_t(C)$, so it is equal to the union of the Schubert cycles $\Sigma_1(\overline{p_i,q})$.
The above is basically the argument written for a different problem in the book, but I do not quite get it. How can I rigorously show that the support of $\overline\Psi_0$ must be contained in the locus of lines meeting the flat limit $C_0=\lim_{t\to0} A_t(C)$
? It is certainly belieavable from a sort of 'continuity', but I have no idea how to prove something like this without digging into the ideals of the various subvarieties involved, which does not seem tractable by hand.
The authors also write that the flat limit $C_0$ may have an embedded point at $q$, but this does not happen for $\Psi_0$, because it is a divisor in $\mathbb{G}(1,3)$. Is it true that a flat limit of divisors is a divisor? I can see this from a Hilbert polynomial argument for divisors of $\mathbb{P}^n$.
Finally, it remains to compute the multiplicity of the limit along each $\Sigma_1(\overline{p_i,q})$. I guess it is 1, since we have stable lines in $H$ that meet $A_t(C)$ for each $t\neq 0$ only at one of the points $p_i$ and these lines are certainly not the limit of any other lines. However, this is the point I consider the least convincing. I would like to avoid saying the word "limit" here, since most lines in $\Sigma_1(\overline{p_i,q})$, the ones not passing through either $p_i$ or $q$ are not limits of anything. What would be the best way to compute the multiplicity of each component here?
Best Answer
Really reasonable questions.
The flat limit is by definition the fiber of the closure of $\Psi$ (note that this is true scheme theoretically not just set theoretically).
So to show that the flat limit is contained in something, one way is to exhibit an appropriate closed set containing $\Psi$, more precisely to exhibit the fiber of such a set. In this case take the set
$$\Psi’ = \{(t,L) \in \mathbb{A}^1 \times \mathbb{G}(1,3) : L \cap C_t \ne \emptyset \}$$
where $C_t$ is $A_t(C)$ for $t\ne0$ and is the flat limit of that curve at $t=0$. Note that this definition also dropped the $t\ne 0$ requirement, so it’s almost like a “guess” as to what the closure of $\Psi$ is.
This gives a closed set (since the total space of the flat family $C_t$ is closed in $\mathbb{A}^1 \times \mathbb{P}^3$), so it contains the closure of $\Psi$. And the fiber of it is set-theoretically the set of lines meeting $C_0$.
Next: is it true that a flat limit of divisors is a divisor?
Yes. First of all fiber dimension is constant in flat families. But more concretely, the closure of $\Psi$ is a divisor on $\mathbb{A}^1 \times \mathbb{G}(1,3)$, so restricting it to $t=0$ gives a divisor on $\mathbb{G}(1,3)$. (In general the cycle class is also preserved under flat limits over $\mathbb{P^1}$.)
As for multiplicity. One nice thing here is that divisors on smooth varieties are Cohen-Macaulay, so they are reduced iff generically reduced. So, no embedded points. In this case the key fact is that $C_0$ is reduced away from $q$. This is actually a consequence of the fact that $C$ is transverse to $H$ at each of the $p_i$’s. As $t\to 0$, the curve “stretches” away from $H$ and flattens into a line through $p_i$. (If $C$ were tangent to $H$ at $p_i$ it would become a double line, etc.)
The local model of this would be to take $C$ itself to be a line passing through $H$ at just one point, in which case the calculation is (I think) easy to do directly.