43.17 Intersection product using Tor formula
Let $X$ be a nonsingular variety. Let $\alpha = \sum n_ i [W_ i]$ be an $r$-cycle and $\beta = \sum _ j m_ j [V_ j]$ be an $s$-cycle on $X$. Assume that $\alpha $ and $\beta $ intersect properly, see Definition 43.13.5. In this case we define
\[ \alpha \cdot \beta = \sum \nolimits _{i,j} n_ i m_ j W_ i \cdot V_ j. \]
where $W_ i \cdot V_ j$ is as defined in Section 43.14. If $\beta = [V]$ where $V$ is a closed subvariety of dimension $s$, then we sometimes write $\alpha \cdot \beta = \alpha \cdot V$.
Lemma 43.17.1. Let $X$ be a nonsingular variety. Let $a, b \in \mathbf{P}^1$ be distinct closed points. Let $k \geq 0$.
If $W \subset X \times \mathbf{P}^1$ is a closed subvariety of dimension $k + 1$ which intersects $X \times a$ properly, then
$[W_ a]_ k = W \cdot X \times a$ as cycles on $X \times \mathbf{P}^1$, and
$[W_ a]_ k = \text{pr}_{X, *}(W \cdot X \times a)$ as cycles on $X$.
Let $\alpha $ be a $(k + 1)$-cycle on $X \times \mathbf{P}^1$ which intersects $X \times a$ and $X \times b$ properly. Then $pr_{X,*}( \alpha \cdot X \times a - \alpha \cdot X \times b)$ is rationally equivalent to zero.
Conversely, any $k$-cycle which is rationally equivalent to $0$ is of this form.
Proof.
First we observe that $X \times a$ is an effective Cartier divisor in $X \times \mathbf{P}^1$ and that $W_ a$ is the scheme theoretic intersection of $W$ with $X \times a$. Hence the equality in (1)(a) is immediate from the definitions and the calculation of intersection multiplicity in case of a Cartier divisor given in Lemma 43.16.4. Part (1)(b) holds because $W_ a \to X \times \mathbf{P}^1 \to X$ maps isomorphically onto its image which is how we viewed $W_ a$ as a closed subscheme of $X$ in Section 43.8. Parts (2) and (3) are formal consequences of part (1) and the definitions.
$\square$
For transversal intersections of closed subschemes the intersection multiplicity is $1$.
Lemma 43.17.2. Let $X$ be a nonsingular variety. Let $r, s \geq 0$ and let $Y, Z \subset X$ be closed subschemes with $\dim (Y) \leq r$ and $\dim (Z) \leq s$. Assume $[Y]_ r = \sum n_ i[Y_ i]$ and $[Z]_ s = \sum m_ j[Z_ j]$ intersect properly. Let $T$ be an irreducible component of $Y_{i_0} \cap Z_{j_0}$ for some $i_0$ and $j_0$ and assume that the multiplicity (in the sense of Section 43.4) of $T$ in the closed subscheme $Y \cap Z$ is $1$. Then
the coefficient of $T$ in $[Y]_ r \cdot [Z]_ s$ is $1$,
$Y$ and $Z$ are nonsingular at the generic point of $T$,
$n_{i_0} = 1$, $m_{j_0} = 1$, and
$T$ is not contained in $Y_ i$ or $Z_ j$ for $i \not= i_0$ and $j \not= j_0$.
Proof.
Set $n = \dim (X)$, $a = n - r$, $b = n - s$. Observe that $\dim (T) = r + s - n = n - a - b$ by the assumption that the intersections are proper. Let $(A, \mathfrak m, \kappa ) = (\mathcal{O}_{X, \xi }, \mathfrak m_\xi , \kappa (\xi ))$ where $\xi \in T$ is the generic point. Then $\dim (A) = a + b$, see Varieties, Lemma 33.20.3. Let $I_0, I, J_0, J \subset A$ cut out the trace of $Y_{i_0}$, $Y$, $Z_{j_0}$, $Z$ in $\mathop{\mathrm{Spec}}(A)$. Then $\dim (A/I) = \dim (A/I_0) = b$ and $\dim (A/J) = \dim (A/J_0) = a$ by the same reference. Set $\overline{I} = I + \mathfrak m^2/\mathfrak m^2$. Then $I \subset I_0 \subset \mathfrak m$ and $J \subset J_0 \subset \mathfrak m$ and $I + J = \mathfrak m$. By Lemma 43.14.3 and its proof we see that $I_0 = (f_1, \ldots , f_ a)$ and $J_0 = (g_1, \ldots , g_ b)$ where $f_1, \ldots , g_ b$ is a regular system of parameters for the regular local ring $A$. Since $I + J = \mathfrak m$, the map
\[ I \oplus J \to \mathfrak m/\mathfrak m^2 = \kappa f_1 \oplus \ldots \oplus \kappa f_ a \oplus \kappa g_1 \oplus \ldots \oplus \kappa g_ b \]
is surjective. We conclude that we can find $f_1', \ldots , f_ a' \in I$ and $g'_1, \ldots , g_ b' \in J$ whose residue classes in $\mathfrak m/\mathfrak m^2$ are equal to the residue classes of $f_1, \ldots , f_ a$ and $g_1, \ldots , g_ b$. Then $f'_1, \ldots , g'_ b$ is a regular system of parameters of $A$. By Algebra, Lemma 10.106.3 we find that $A/(f'_1, \ldots , f'_ a)$ is a regular local ring of dimension $b$. Thus any nontrivial quotient of $A/(f'_1, \ldots , f'_ a)$ has strictly smaller dimension (Algebra, Lemmas 10.106.2 and 10.60.13). Hence $I = (f'_1, \ldots , f'_ a) = I_0$. By symmetry $J = J_0$. This proves (2), (3), and (4). Finally, the coefficient of $T$ in $[Y]_ r \cdot [Z]_ s$ is the coefficient of $T$ in $Y_{i_0} \cdot Z_{j_0}$ which is $1$ by Lemma 43.14.3.
$\square$
Comments (2)
Comment #7518 by Hao Peng on
Comment #7650 by Stacks Project on