The Stacks project

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

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). $\square$