# Wednesday, November 10


:::{.remark}
We defined projective space $\PP^n\slice Y \da \PP^n\slice \ZZ \fiberprod{\spec \ZZ} Y$, and a projective morphism as one that factors as a closed immersion into $\PP^n\slice Y$ for some $Y$ followed by projection onto $Y$.
Continuing the proof from last time: we reduced to $\PP^n\slice \ZZ$ and produced a map $\spec R\to \PP^n\slice \ZZ$.
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:::{.proof title="continued"}
We noted that $\phi_{ij}\in k\units$ since $\phi_{ij}\phi{ji} = 1$, and more generally $\phi_{ij}\phi_{jk} = \phi_{ik}$.
We chose $v_i = \val(\phi_{i, 0}) \in \ZZ$ minimally, so assume without loss of generality by relabeling that it is $v_1$, so $v_i\geq v_1$ for all $i$.
Then use that $\phi_{ij} = \phi_{i1}/\phi_{j1} \implies v(\phi_{i1}) = v(\phi_{i0}) - v(\phi_{10}) = v_2 -v_1 \geq 0$.
Writing $R = \ts{\phi \in k\ \st v(\phi) \geq 0}$, we have $\phi_{i1} \in R \subseteq k$.
Consider the ring map
\[
\zadjoin{{x_0 \over x_1}, \cdots, {x_n \over x_1}} &\to R \\
{x_i \over x_1} &\mapsto \phi_{i1}
.\]
This yields a map $\spec R\to \AA^n\slice \ZZ \cong D(x_1)$, which restricts to a map $\spec K\to D(x_0, \cdots, x_n)$, a smaller open set.


> Fire alarm! Class canceled.

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