Lemma 45.9.4. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Then $H^ i(\mathop{\mathrm{Spec}}(k)) = 0$ for $i \not= 0$ and there is a unique $F$-algebra isomorphism $F = H^0(\mathop{\mathrm{Spec}}(k))$. We have $\gamma ([\mathop{\mathrm{Spec}}(k)]) = 1$ and $\int _{\mathop{\mathrm{Spec}}(k)} 1 = 1$.
Proof. By axiom (C)(d) we see that $H^0(\mathop{\mathrm{Spec}}(k))$ is nonzero and even $\gamma ([\mathop{\mathrm{Spec}}(k)])$ is nonzero. Since $\mathop{\mathrm{Spec}}(k) \times \mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(k)$ we get
\[ H^*(\mathop{\mathrm{Spec}}(k)) \otimes _ F H^*(\mathop{\mathrm{Spec}}(k)) = H^*(\mathop{\mathrm{Spec}}(k)) \]
by axiom (B)(a) which implies (look at dimensions) that only $H^0$ is nonzero and moreover has dimension $1$. Thus $F = H^0(\mathop{\mathrm{Spec}}(k))$ via the unique $F$-algebra isomorphism given by mapping $1 \in F$ to $1 \in H^0(\mathop{\mathrm{Spec}}(k))$. Since $[\mathop{\mathrm{Spec}}(k)] \cdot [\mathop{\mathrm{Spec}}(k)] = [\mathop{\mathrm{Spec}}(k)]$ in the Chow ring of $\mathop{\mathrm{Spec}}(k)$ we conclude that $\gamma ([\mathop{\mathrm{Spec}}(k)) \cup \gamma ([\mathop{\mathrm{Spec}}(k)]) = \gamma ([\mathop{\mathrm{Spec}}(k)])$ by axiom (C)(c). Since we already know that $\gamma ([\mathop{\mathrm{Spec}}(k)])$ is nonzero we conclude that it has to be equal to $1$. Finally, we have $\int _{\mathop{\mathrm{Spec}}(k)} 1 = 1$ since $\int _{\mathop{\mathrm{Spec}}(k)} \gamma ([\mathop{\mathrm{Spec}}(k)]) = 1$ by axiom (C)(d). $\square$
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