User:TakuyaMurata/sandbox2

• Let X be an integral scheme of finite type over $\mathbb {Z}$ and $K$ the function field of X (= the residue field of the generic point) and $F$ the prime field of $K$. If the characteristic of $K$ is positive, say, p, then $F=\mathbb {Z} /p\mathbb {Z}$ and $X$ is also an integral scheme of finite type over F (i.e., an algebraic pre-variety) and so $\operatorname {tr.deg} _{F}K=\dim X$ where tr.deg means transcendence degree. If the characteristic is zero, then we consider the structure map $f:X\to \operatorname {Spec} (\mathbb {Z} )$ is flat and thus

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Let L be a line bundle on an algebraic variety and $V\subset \Gamma (X,L)$ a finite-dimensional vector subspace. For the sake of clarity, we first consider the case when V is base-point-free; in other words, the natural map $V\otimes _{k}{\mathcal {O}}_{X}\to L$ is surjective (here, k = the base field). Or equivalently, $\operatorname {Sym} ((V\otimes _{k}{\mathcal {O}}_{X})\otimes _{{\mathcal {O}}_{X}}L^{-1})\to \bigoplus _{n=0}^{\infty }{\mathcal {O}}_{X}$ is surjective. Hence, writing $V_{X}=V\times X$ for the trivial vector bundle and passing the surjection to the relative Proj, there is a closed immersion:

$i:X\hookrightarrow \mathbb {P} (V_{X}^{*}\otimes L)\simeq \mathbb {P} (V_{X}^{*})=\mathbb {P} (V^{*})\times X$

where $\simeq$ on the right is the invariance of the projective bundle under a twist by a line bundle. Following i by a projection, there results in the map:

$f:X\to \mathbb {P} (V^{*}).$

When the base locus of V is not empty, the above discussion still goes through with ${\mathcal {O}}_{X}$ replaced by an ideal sheaf defining the base locus and X replaced by the blow-up ${\widetilde {X}}$ of it along the (scheme-theoretic) base locus B. Precisely, as above, there is a surjection $\operatorname {Sym} ((V\otimes _{k}{\mathcal {O}}_{X})\otimes _{{\mathcal {O}}_{X}}L^{-1})\to \bigoplus _{n=0}^{\infty }{\mathcal {I}}^{n}$ where ${\mathcal {I}}$ is the ideal sheaf of B and that gives rise to

$i:{\widetilde {X}}\hookrightarrow \mathbb {P} (V^{*})\times X.$

Since $X-B\simeq$ an open subset of ${\widetilde {X}}$, there results in the map:

$f:X-B\to \mathbb {P} (V^{*}).$

Finally, when a basis of V is chosen, the above discussion becomes more down-to-earth (and that is the style used in Hartshorne, Algebraic Geometry).