Lemma 33.44.6. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $\mathcal{E}$ be a locally free $\mathcal{O}_ X$-module of rank $n$. Then

$\deg (\mathcal{E}) = \sum m_ i \deg (\mathcal{E}|_{C_ i})$

where $C_ i \subset X$, $i = 1, \ldots , t$ are the irreducible components of dimension $1$ with reduced induced scheme structure and $m_ i$ is the multiplicity of $C_ i$ in $X$.

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

$0 \to \mathcal{F} \to \mathcal{O}_ X \to \mathcal{O}_{X'} \to 0$

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

$m = \text{length}_{\mathcal{O}_{X, \xi }} \mathcal{O}_{X, \xi } = \dim _{\kappa (\xi )} \mathcal{F}_\xi + \text{length}_{\mathcal{O}_{X', \xi }} \mathcal{O}_{X', \xi } = r + m'$

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

$0 \to \mathcal{E}|_ C \otimes \mathcal{F} \to \mathcal{E} \to \mathcal{E} \otimes \mathcal{O}_{X'} \to 0$

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. $\square$

Comment #7495 by Hao Peng on

i noticed that proof of tag01YV works for a natural extension of statements to proper schemes of dimensione$\le2$, and then this proposition will follow as a direct consequence. The proof will be much shorter, because the current proof is essentially doing devissage for coherent sheaves again.

There are also:

• 4 comment(s) on Section 33.44: Degrees on curves

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).