The Stacks project

Proof. Let $C_ i \subset X$, $i = 1, \ldots , t$ be the irreducible components of dimension $1$ with reduced induced scheme structure and let $m_ i$ be the multiplicity of $C_ i$ in $X$. Then $[X]_1 = \sum m_ i[C_ i]$ and $c_1(\mathcal{E}) \cap [X]_1$ is the sum of the pushforwards of the cycles $m_ i c_1(\mathcal{E}|_{C_ i}) \cap [C_ i]$. Since we have a similar decomposition of the degree of $\mathcal{E}$ by Varieties, Lemma 33.44.6 it suffices to prove the lemma in case $X$ is a proper curve over $k$.

Assume $X$ is a proper curve over $k$. By Divisors, Lemma 31.36.1 there exists a modification $f : X' \to X$ such that $f^*\mathcal{E}$ has a filtration whose successive quotients are invertible $\mathcal{O}_{X'}$-modules. Since $f_*[X']_1 = [X]_1$ we conclude from Lemma 42.38.4 that

Since we have a similar relationship for the degree by Varieties, Lemma 33.44.4 we reduce to the case where $\mathcal{E}$ has a filtration whose successive quotients are invertible $\mathcal{O}_ X$-modules. In this case, we may use additivity of the degree (Varieties, Lemma 33.44.3) and of first Chern classes (Lemma 42.40.3) to reduce to the case discussed in the next paragraph.

Assume $X$ is a proper curve over $k$ and $\mathcal{E}$ is an invertible $\mathcal{O}_ X$-module. By Divisors, Lemma 31.15.12 we see that $\mathcal{E}$ is isomorphic to $\mathcal{O}_ X(D) \otimes \mathcal{O}_ X(D')^{\otimes -1}$ for some effective Cartier divisors $D, D'$ on $X$ (this also uses that $X$ is projective, see Varieties, Lemma 33.43.4 for example). By additivity of degree under tensor product of invertible sheaves (Varieties, Lemma 33.44.7) and additivity of $c_1$ under tensor product of invertible sheaves (Lemma 42.25.2 or 42.39.1) we reduce to the case $\mathcal{E} = \mathcal{O}_ X(D)$. In this case the left hand side gives $\deg (D)$ (Varieties, Lemma 33.44.9) and the right hand side gives $\deg ([D]_0)$ by Lemma 42.25.4. Since

by definition, we see

The penultimate equality by Algebra, Lemma 10.52.12 using that $D$ is affine. $\square$