0
$\begingroup$

I am currently working through a paper (related to tropical geometry, but this is not important in the following context) which utilizes the concept of very affine varieties in the following way (I'll give a quick summary of the concept):

Let $X$ be a variety over a field $K$, i.e. an irreducible, reduced, separated $K$-scheme of finite type. Then an open subset $U$ of $X$ is called very affine if $U$ has a closed immersion to some multiplicative split torus $\mathbb G^r_m := Spec(K[T_1^{\pm 1}, \dots, T_r^{\pm 1}])$.

The author states: "[...] The very affine open subsets of $X$ form a basis for the Zariski topology."

I am wondering why this is the case. Here is what I know so far, if any of it helps: First, let $U$ be any open affine subset of $X$. Then \begin{equation} \text{Hom}_{K-\text{Sch.}}(U, \mathbb G^r_m) \cong \text{Hom}_{K-\text{alg.}}(K[T_1^{\pm 1}, \dots, T_r^{\pm 1}], \mathcal O_X(U)). \end{equation}

Any $K$-algebra morphism $f \colon K[T_1^{\pm 1}, \dots, T_r^{\pm 1}] \to \mathcal O_X(U)$ is uniquely given by the units $f(T_i) \in \mathcal O_X(U)^*$.
Furthermore it is known that the abelian group $\mathcal O_X(U)^* / K^*$ is free of finite rank, say $r$. Choose representatives $\varphi_1, \dots, \varphi_r$ in $\mathcal O_X(U)^*$ of a basis, we get a (more or less canonical) morphism $$\varphi \colon K[T_1^{\pm 1}, \dots, T_r^{\pm 1}] \to \mathcal O_X(U), T_i \mapsto \varphi_i.$$

So far so good. Using these identities it is easy to show that the following statements are equivalent:

a.) $U$ is very affine;
b.) $\mathcal O_X(U)$ is generated as a $K$-algebra by $\mathcal O_X(U)^*$;
c.) the canonical map $\varphi$ above is surjective.

Also I have shown that intersections of very affine opens are very affine again.

My approach now would be to take any point $x \in X$, choose any affine open $U$ around it and then try to 'make it very affine' by successively localizing until I get an open subscheme $V$ of $U$ that satisfies statement b.) above. However I can't really get it to work.

Long explanation, but if anyone could help or has any comments, that would be super awesome!

Thanks a lot in advance!

$\endgroup$
0
$\begingroup$

I thought about it once more and I think I have a solution which might work:

Claim: Let $X$ be variety over field $K$. Then the very affine open subsets of $X$ form a basis for the Zariski topology.

Proof: Let $x \in X$ and $U \subseteq X$ be an open neighborhood of $x$. It suffices to show that there is very affine $V$ around $x$ with $V \subseteq U$.
As open subschemes of varieties are varieties again, and by possibly passing to a smaller open neighborhood, we can assume that $U$ is affine with $U = \text{Spec}(A)$, where $A$ is a $K$-algebra of finite type, i.e. it is of the form $A = K[T_1, \dots, K_n]/I$ for some ideal $I$. Let $\mathfrak p$ denote the prime ideal of $A$ corresponding to $x$. Let $\overline{T_i}$ denote the class in $A$ of $T_i \in K[T_1, \dots, K_n]$.

We first suppose that there is no $i \in \{1, \dots, n\}$ with $\overline{T_i} \in \mathfrak p$. Then $V := D(\overline{T_1}) \cap \dots \cap D(\overline{T_n}) = D(\overline{T_1 \cdot \cdots \cdot T_n}) \subseteq U = \text{Spec}(A)$ is open around $x$ and corresponds to localization $A[\frac{1}{\overline{T_1 \cdot \cdots \cdot T_n}}] = A[\frac{1}{\overline{T_1}}]\cdots[\frac{1}{\overline{T_n}}] =: B$.
As $\overline{T_i} \in B^* \ \forall i \in \{1, \dots, n\}$ we obtain a surjective $K$-algebra morphism $$ K[T^{\pm 1}_1, \dots, T^{\pm 1}_n] \to B, T_i \mapsto \overline{T_i}, $$ which shows that $V$ is very affine.

On the other hand, suppose $\exists i$ with $\overline{T_i} \in \mathfrak p$. Denote $J \subset \{1, \dots, n\}$ the set of such indices. For any $j \in J$ we then must have $\overline{T_j} + 1 \notin \mathfrak p$ and we can proceed as above by successively localizing to obtain $V$ (however instead of localizing at $\overline{T_j}$ we use $\overline{T_j} + 1$ for all $j \in J$). Then again obtain surjective $K$-algebra morphism $$ K[T^{\pm 1}_1, \dots, T^{\pm 1}_n] \to B, T_i \mapsto \begin{cases}\overline{T_i}\ \ \text{for } i \notin J \\ \overline{T_i} + 1 \ \ \text{for } i \in J \\ \end{cases}, $$ which shows the claim.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.