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

\[ \alpha \cdot \beta = \sum \nolimits _{i,j} n_ i m_ j W_ i \cdot V_ j. \]

For transversal intersections of closed subschemes the intersection multiplicity is $1$.

**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 transversal. 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