# Why is flatness needed for the Segre classes of a family of cones to be equal in the Chow ring of the base

Let $X$ be an algebraic scheme and $\mathscr C$ a cone on $X\times\mathbf A^1$ and $C_t$ denote the restriction of $\mathscr C$ to $X\times\{t\}$ ($t=0,1$, or whatever). The claim in Fulton's Intersection Theory is that when $\mathscr C$ is flat over $\mathbf A^1$, then $$s(C_0)=s(C_1)\in A_\ast X$$ where the Segre class $s(C)$ of a cone $C$ on a scheme $X$ is defined as $$s(C)=q_\ast\left(\sum_{i\geq0}c_1(\mathscr O(1))^i\smallfrown [P(C\oplus 1)]\right)$$ where $q:P(C\oplus 1)\to X$ is the projection.

Since if $C_i$ are the irreducible components of a cone $\mathscr C$ on $X$, then $s(C)=\sum_i m_is(C_i)$ and $[P(C\oplus 1)]=\sum_i m_i[P(C_i\oplus1)]$, where $m_i$ are the geometric multiplicities, so I believe we can restrict our attention to the class $[V]$ of an irreducible component of $\mathscr C$. In general a variety is flat over a nonsingular curve iff its generic point maps to the generic point of the curve. But if this is not true, then it means that the image of $V$ under projection to $\mathbf A^1$ is a point, say $t$. But in this case, $i_t^\ast(V)=0\ne V=V_t$, where I take $V_t$ to be the fiber and I take $i_t^\ast$ to be the Gysin morphism $A_\ast(X)\to A_{\ast-1}(D)$ where $C_t\subset\mathscr C$ is the divisor on $\mathscr C$ with local equation $t$ (aka the fiber). Since if $V=V_t$ then $i_t^\ast(V)$ is computed by restricting $\mathscr O_{\mathscr C}(C_t)$ to $V$ and then taking a corresponding Cartier divisor on $V$. But $\mathscr O_{\mathscr C}(C_t)$ restricted to $C_t$ is the normal bundle which is trivial, so $i_t^\ast(V)=0$.

So I see why flatness is needed. My question is, wouldn't the statement $$s(i_0^\ast\mathscr C)=s(i_1^\ast\mathscr C)\in A_\ast X$$ hold without the assumption of flatness? When $\mathscr C$ is flat over $\mathbf A^1$ it restricts to the original statement.

• Your statement is correct. It follows by Example 10.1.7(a) for the Segre class $\alpha=s(\mathcal{C})$ on $X\times \mathbb{A}^1$. The "claim" is Example 10.1.10 in the chapter "Families of Algebraic Cycles" that reconciles the new approach via deformation to the normal cone with the earlier approach via moving families of cycles. So it is reasonable that Fulton would point out that, under a flatness hypothesis, the "naive" construction (without using deformation to the normal cone or refined Gysin pullbacks) already produces rationally equivalent cycles . . . – Jason Starr Apr 26 '17 at 13:17
• . . . Note: this already comes up in Example 4.1.6(c). In some sense, it also comes up in Proposition 5.2, the fundamental result that the specialization to the normal cone respects rational equivalence (and thus is well-defined on cycle classes). For those earlier results, certainly Fulton would not use the properties of specialization to the normal cone. – Jason Starr Apr 26 '17 at 13:20