Why does a flat degeneration induce equality in the K-Theory?

Question

Let $Z_1$ and $Z_2$ be two closed subschemes of a smooth, complex, algebraic variety $X$. We will be interested in the Grothendieck ring of coherent sheaves of $X$, i.e. isomorphism classes of such sheaves modulo exact sequences and a product given by the tensor product of $\mathcal{O}_X$-modules.

Denote by $[\mathcal{O}_{Z_1}]$ the class of $f^\ast\mathcal{O}_{Z_1}$ in the Grothendieck Ring of coherent sheaves on $X$, where $f:Z_1\to X$ is the inclusion.

Assume now that $Z_2$ is a flat degeneration of $Z_1$, i.e. there is a morphism $\pi: Y\to \mathbb{A}^1$ of schemes such that the generic fiber of $\pi$ is isomorphic to $Z_1$ and the fiber of $\pi$ over $0$ is isomorphic to $Z_2$.

I am reading a paper by David Speyer with the title "A matroid invariant via the K-theory of the Grassmannian" and on page 885, right below equation (3) he says that $[\mathcal{O}_{Z_1}]=[\mathcal{O}_{Z_2}]$ holds in this case. I don't see how to prove this implication and I'd be very grateful if someone could explain it to me.

Answer 1

As Allen says, we are talking about flat subfamilies of $X$. I.e. $Y$ is a closed subscheme of $X \times \mathbb{A}^1$, flat over $\mathbb{A}^1$, with $Z_0 \times \{ 0 \}$ and $Z_1 \times \{ 1 \}$ the fibers over $0$ and $1$.

Replacing $Y$ by its closure in $X \times \mathbb{P}^1$, we may assume that we instead have $Y$ closed in $X \times \mathbb{P}^1$, flat over $\mathbb{P}^1$. (Since $\mathbb{P}^1$ is smooth and one dimensional, being flat over it just means all associated primes of $Y$ map to the generic point of $\mathbb{P}^1$.)

On $\mathbb{P}^1$, we have short exact sequences $$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k_0 \to 0$$ and $$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k_1 \to 0$$ where $k_0$ and $k_1$ are the skyscraper sheaves of a point at $0$ and $1$ respectively. So the point $0$ and the point $1$ have the same class in $K_0(\mathbb{P}^1)$. Since $Y \to X$ is flat, we can pull this back to an equality $[\mathcal{O}_{Z_0}] = [ \mathcal{O}_{Z_1}]$ on $Y$.

Since $Y$ is closed in $X \times \mathbb{P}^1$, the map $\pi: Y \to X$ is proper and we can push forward this equality to $X$. We need to check that $\pi_{\ast} \mathcal{O}_{Z_i \times \{ i \}} = \mathcal{O}_{Z_i}$ and $R^j \pi_{\ast} \mathcal{O}_{Z_i \times \{ i \}} =0$ in order to know that the $K$-theoretic push forward is the obvious thing.

I'm not sure where to cite this argument to, it is something that everyone I talked to about this sort of thing seemed to view as obvious. Allen does the analogue for equivariant $K_0$, which is more interesting, in Prop 3.5 of his recent paper.