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 Poincaré duality, more precisely axiom (A)(b), we have $\beta _0 = \beta _2$. Therefore the lemma holds. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)