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

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).