0
$\begingroup$

In page 55~56 of Qing Liu's Algebraic Geometry and Arithmetic Curves, there is a paragraph as follows:

Definition 3.47. Let $k$ be a field. An affine variety over $k$ is the affine scheme associated to a finitely generated algebra over $k$. An algebraic variety over $k$ is a $k$-scheme $X$ such that there exists a covering by a finite number of affine open subschemes $X_i$ which are affine varieties over $k$.

If $\varphi_i:X_i\to \mathrm{Spec}(k[T_0,T_1,\cdots,T_n]/I)$ is isomorphic, let $\psi:X\to \mathrm{Spec}k$ be the morphism together with $k$-scheme $X$, is $\psi\circ\varphi_i^{-1}$ required to be induced by the canonical morphism $k\to k[T_0,T_1,\cdots,T_n]/I$?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes. When we say an open subscheme $X_i$ is an affine variety, we mean it is isomorphic to an affine variety as a $k$-scheme, so that the isomorphism $\varphi_i$ forms a commutative diagram with the maps to $\operatorname{Spec} k$. This means exactly that $\psi\circ\varphi_i^{-1}$ is required to be induced by the canonical morphism.

$\endgroup$
6
  • $\begingroup$ But I have another comprehension: $\endgroup$ Commented Oct 31, 2017 at 1:02
  • $\begingroup$ An affine variety over $k$ is an affine scheme $\operatorname{Spec} A$ together with a morphism $\chi:\operatorname{Spec} A\to \operatorname{Spec} k$ such that $\chi$ is induced by $f:k\to A$ and $A=f(k)[x_1,x_2,\cdots,x_n]$ where $x_i\in A$. $\endgroup$ Commented Oct 31, 2017 at 1:30
  • $\begingroup$ In the context of the definition of an algebraic variety over $k$, when we say "$X_i$ is an affine variety over $k$", we only mean that up to isomorphism. $\endgroup$ Commented Oct 31, 2017 at 2:03
  • $\begingroup$ Are my two interpretations the same in essence? $\endgroup$ Commented Oct 31, 2017 at 5:27
  • $\begingroup$ I don't know what you mean by your two interpretations. $\endgroup$ Commented Oct 31, 2017 at 5:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .