The Stacks project

Proof. First assume $X$ is nonempty. Observe that $[X]$ is the pullback of $[\mathop{\mathrm{Spec}}(k)]$ by the structure morphism $p : X \to \mathop{\mathrm{Spec}}(k)$. Hence we get $\gamma ([X]) = 1$ by axiom (C)(a) and Lemma 45.9.4. Let $X' \subset X$ be an irreducible component. By functoriality it suffices to show $1 \not= 0$ in $H^0(X')$. Thus we may and do assume $X$ is irreducible, and in particular nonempty and equidimensional, say of dimension $d$. To see that $1 \not= 0$ it suffices to show that $H^*(X)$ is nonzero.

Let $x \in X$ be a closed point whose residue field $k'$ is separable over $k$, see Varieties, Lemma 33.25.6. Let $i : \mathop{\mathrm{Spec}}(k') \to X$ be the inclusion morphism. Denote $p : X \to \mathop{\mathrm{Spec}}(k)$ is the structure morphism. Observe that $p_*i_*[\mathop{\mathrm{Spec}}(k')] = [k' : k][\mathop{\mathrm{Spec}}(k)]$ in $\mathop{\mathrm{CH}}\nolimits _0(\mathop{\mathrm{Spec}}(k))$. Using axiom (C)(b) twice and Lemma 45.9.4 we conclude that

is nonzero. Thus $i_*\gamma ([\mathop{\mathrm{Spec}}(k)]) \in H^{2d}(X)(d)$ is nonzero (because it maps to something nonzero via $p_*$). This concludes the proof in case $X$ is nonempty.

Finally, we consider the case of the empty scheme. Axiom (B)(a) gives $H^*(\emptyset ) \otimes H^*(\emptyset ) = H^*(\emptyset )$ and we get that $H^*(\emptyset )$ is either zero or $1$-dimensional in degree $0$. Then axiom (B)(a) again shows that $H^*(\emptyset ) \otimes H^*(X) = H^*(\emptyset )$ for all smooth projective schemes $X$ over $k$. Using axiom (A)(b) and the nonvanishing of $H^0(X)$ we've seen above we find that $H^*(X)$ is nonzero in at least two degrees if $\dim (X) > 0$. This then forces $H^*(\emptyset )$ to be zero. $\square$