Lemma 45.14.10. Assume given data (D0), (D1), and (D2') satisfying axioms (A1) – (A7). If $k'/k$ is a Galois extension, then we have $\int _{\mathop{\mathrm{Spec}}(k')} 1 = [k' : k]$.
Proof. We observe that
\[ \mathop{\mathrm{Spec}}(k') \times \mathop{\mathrm{Spec}}(k') = \coprod \nolimits _{\sigma \in \text{Gal}(k'/k)} (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id})^{-1} \Delta \]
as cycles on $\mathop{\mathrm{Spec}}(k') \times \mathop{\mathrm{Spec}}(k')$. Our construction of $\gamma $ always sends $[X]$ to $1$ in $H^0(X)$. Thus $1 \otimes 1 = 1 = \sum (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id})^*\gamma ([\Delta ])$. Denote $\lambda : H^0(\mathop{\mathrm{Spec}}(k')) \to F$ the map from axiom (A6), in other words $(\text{id} \otimes \lambda )(\gamma (\Delta )) = 1$ in $H^0(\mathop{\mathrm{Spec}}(k'))$. We obtain
\begin{align*} \lambda (1) 1 & = (\text{id} \otimes \lambda )(1 \otimes 1) \\ & = (\text{id} \otimes \lambda )( \sum (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id})^*\gamma ([\Delta ])) \\ & = \sum (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id})^*( (\text{id} \otimes \lambda )(\gamma ([\Delta ])) \\ & = \sum (\mathop{\mathrm{Spec}}(\sigma ) \times \text{id})^*(1) \\ & = [k' : k] \end{align*}
Since $\lambda $ is another name for $\int _{\mathop{\mathrm{Spec}}(k')}$ (Remark 45.14.6) the proof is complete. $\square$
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