The Stacks project

Proof. Observe that the statement makes sense because $C_ i \to \mathop{\mathrm{Spec}}(k)$ is proper of dimension $1$ (Morphisms, Lemmas 29.41.6 and 29.41.4). Consider the open subscheme $U_ i = X \setminus (\bigcup _{j \not= i} C_ j)$ and let $X_ i \subset X$ be the scheme theoretic closure of $U_ i$. Note that $X_ i \cap U_ i = U_ i$ (scheme theoretically) and that $X_ i \cap U_ j = \emptyset $ (set theoretically) for $i \not= j$; this follows from the description of scheme theoretic closure in Morphisms, Lemma 29.7.7. Thus we may apply Lemma 33.44.4 to the morphism $X' = \bigcup X_ i \to X$. Since it is clear that $C_ i \subset X_ i$ (scheme theoretically) and that the multiplicity of $C_ i$ in $X_ i$ is equal to the multiplicity of $C_ i$ in $X$, we see that we reduce to the case discussed in the following paragraph.

Assume $X$ is irreducible with generic point $\xi $. Let $C = X_{red}$ have multiplicity $m$. We have to show that $\deg (\mathcal{E}) = m \deg (\mathcal{E}|_ C)$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal defining the closed subscheme $C$. Let $e \geq 0$ be minimal such that $\mathcal{I}^{e + 1} = 0$ (Cohomology of Schemes, Lemma 30.10.2). We argue by induction on $e$. If $e = 0$, then $X = C$ and the result is immediate. Otherwise we set $\mathcal{F} = \mathcal{I}^ e$ viewed as a coherent $\mathcal{O}_ C$-module (Cohomology of Schemes, Lemma 30.9.8). Let $X' \subset X$ be the closed subscheme cut out by the coherent ideal $\mathcal{I}^ e$ and let $m'$ be the multiplicity of $C$ in $X'$. Taking stalks at $\xi $ of the short exact sequence

we find (use Algebra, Lemmas 10.52.3, 10.52.6, and 10.52.5) that

where $r$ is the rank of $\mathcal{F}$ as a coherent sheaf on $C$. Tensoring with $\mathcal{E}$ we obtain a short exact sequence

By induction we have $\chi (\mathcal{E} \otimes \mathcal{O}_{X'}) = m' \deg (\mathcal{E}|_ C)$. By Lemma 33.44.5 we have $\chi (\mathcal{E}|_ C \otimes \mathcal{F}) = r \deg (\mathcal{E}|_ C) + n \chi (\mathcal{F})$. Putting everything together we obtain the result. $\square$