Lemma 45.9.8. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Then $H^*(\mathbf{P}^1_ k)$ is $1$-dimensional in dimensions $0$ and $2$ and zero in other degrees.

Proof. Let $x \in \mathbf{P}^1_ k$ be a $k$-rational point. Observe that $\Delta = \text{pr}_1^*x + \text{pr}_2^*x$ as divisors on $\mathbf{P}^1_ k \times \mathbf{P}^1_ k$. Using axiom (C)(a) and additivity of $\gamma$ we see that

$\gamma ([\Delta ]) = \text{pr}_1^*\gamma ([x]) + \text{pr}_2^*\gamma ([x]) = \gamma ([x]) \otimes 1 + 1 \otimes \gamma ([x])$

in $H^*(\mathbf{P}^1_ k \times \mathbf{P}^1_ k) = H^*(\mathbf{P}^1_ k) \otimes _ F H^*(\mathbf{P}^1_ k)$. However, by Lemma 45.9.7 we know that $\gamma ([\Delta ])$ cannot be written as a sum of fewer than $\sum \beta _ i$ pure tensors where $\beta _ i = \dim _ F H^ i(\mathbf{P}^1_ k)$. Thus we see that $\sum \beta _ i \leq 2$. By Lemma 45.9.5 we have $H^0(\mathbf{P}^1_ k) \not= 0$. By PoincarĂ© duality, more precisely axiom (A)(b), we have $\beta _0 = \beta _2$. Therefore the lemma holds. $\square$

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