Proposition 52.25.4. Let (\mathcal{F}, \mathcal{F}_0, \alpha ) be a coherent triple. Let (\mathcal{L}, \mathcal{L}_0, \lambda ) be an invertible coherent triple. Then the function
is a polynomial of degree \leq \dim (\text{Supp}(\mathcal{F})).
Proposition 52.25.4. Let (\mathcal{F}, \mathcal{F}_0, \alpha ) be a coherent triple. Let (\mathcal{L}, \mathcal{L}_0, \lambda ) be an invertible coherent triple. Then the function
is a polynomial of degree \leq \dim (\text{Supp}(\mathcal{F})).
Proof. We will prove this by induction on the dimension of the support of \mathcal{F}.
The base case is when \mathcal{F} = 0. Then either \mathcal{F}_0 is zero or its support is \{ \mathfrak m\} . In this case we have
Thus the function of the lemma is constant with value equal to the length of \mathcal{F}_0.
Induction step. Assume the support of \mathcal{F} is nonempty. Let \mathcal{G}_0 \subset \mathcal{F}_0 denote the submodule of sections supported on \{ \mathfrak m\} . Then we get a short exact sequence
This sequence remains exact if we tensor by the invertible coherent triple (\mathcal{L}, \mathcal{L}_0, \lambda ), see discussion above. Thus by additivity of \chi (Lemma 52.25.3) and the base case explained above, it suffices to prove the induction step for (\mathcal{F}, \mathcal{F}_0/\mathcal{G}_0, \alpha ). In this way we see that we may assume \mathfrak m is not an associated point of \mathcal{F}_0.
Let T = \text{Ass}(\mathcal{F}) \cup \text{Ass}(\mathcal{F}/f\mathcal{F}). Since U is quasi-affine, we can find s \in \Gamma (U, \mathcal{L}) which does not vanish at any u \in T, see Properties, Lemma 28.29.7. After multiplying s by a suitable element of \mathfrak m we may assume \lambda (s \bmod f) = s_0|_{U_0} for some s_0 \in \Gamma (X_0, \mathcal{L}_0); details omitted. We obtain a morphism
in the category of coherent triples. Let \mathcal{G} = \mathop{\mathrm{Coker}}(s : \mathcal{F} \to \mathcal{F} \otimes \mathcal{L}) and \mathcal{G}_0 = \mathop{\mathrm{Coker}}(s_0 : \mathcal{F}_0 \to \mathcal{F}_0 \otimes \mathcal{L}_0). Observe that s_0 : \mathcal{F}_0 \to \mathcal{F}_0 \otimes \mathcal{L}_0 is injective as it is injective on U_0 by our choice of s and as \mathfrak m isn't an associated point of \mathcal{F}_0. It follows that there exists an isomorphism \beta : \mathcal{G}/f\mathcal{G} \to \mathcal{G}_0|_{U_0} such that we obtain a short exact sequence
By induction on the dimension of the support we know the proposition holds for the coherent triple (\mathcal{G}, \mathcal{G}_0, \beta ). Using the additivity of Lemma 52.25.3 we see that
is a polynomial. We conclude by a variant of Algebra, Lemma 10.58.5 for functions defined for all integers (details omitted). \square
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