# Is a toric variety over a field of positive characteristic complete if and only if the support is all of $N_{\mathbb{R}}$?

In Cox, Little and Schenck's book Toric Varieties they show that a toric variety $$X_{\Sigma}$$ over a field of characteristic zero is complete if and only if the support is all of $$N_{\mathbb{R}}$$. This proof was very specific to varieties over fields of characteristic zero. When I looked at Oda and Fulton's books they seemed to work over characteristic zero as well.

Is a toric variety $$X_{\Sigma}$$ over an algebraically closed field $$k$$ of characteristic $$p>0$$ complete if and only if the support of $$\Sigma$$ is all of $$N_{\mathbb{R}}$$? I believe the answer should be yes for the following reason, but if anyone could give me a reference, then I would appreciate it.

Let $$R$$ be a DVR of mixed characteristic with special fibre $$k$$ and generic fibre $$L$$. Also, let us define a scheme $$\mathcal{X}_{\Sigma}$$ over $$\operatorname{Spec}(R)$$ to be $$\begin{equation*} \mathcal{X}_{\Sigma} =\left(\coprod_{\sigma \in \Sigma} \operatorname{Spec}(R[\sigma^{\vee} \cap M])\right)/\sim. \end{equation*}$$ We claim that the scheme $$\mathcal{X}_{\Sigma}$$ is faithfully flat over $$\operatorname{Spec}(R)$$. The claim is local, so it suffices to prove that $$\operatorname{Spec}(R[\sigma^{\vee} \cap M])$$ is faithfully flat over $$\operatorname{Spec}(R)$$. Since $$R[\sigma^{\vee} \cap M]$$ is an $$R$$-algebra, the claim follows.

If $$A$$ is a DVR over $$k$$, then let $$\operatorname{Spec}(B)$$ be a faithfully flat, family of DVRs over $$\operatorname{Spec}(R)$$, such that the special fibre is $$\operatorname{Spec}(A)$$.

A normal, toric variety $$X_{\Sigma}$$ over a field $$L$$ of characteristic zero is complete if and only if the support of $$\Sigma$$ is all of $$N_{\mathbb{R}}$$. So since $$\operatorname{Spec}(B)$$ is faithfully flat over $$\operatorname{Spec}(R)$$, the valuative criterion of properness will hold for the morphism $$\mathcal{X}_{\Sigma} \times_{\operatorname{Spec}(R)} \operatorname{Spec}(L) \to \operatorname{Spec}(L)$$ if and only if it holds for the morphism $$X_{\Sigma} \to \operatorname{Spec}(k)$$.

• Often it is good to read the masters. The treatise of Kempf-Mumford-Knudsen-Saint-Donat already treats toric varieties over algebraically closed fields of arbitrary characteristic. Commented May 2, 2023 at 5:34

The answer is yes. Fulton's proof is in fact valid positive characteristic as well. In the proof of "full support $$\Rightarrow$$ completeness" he uses the valuative criterion which, as you noticed, is true for all characteristic. For the opposite direction you only need to replace the zero characteristic language from Fulton's proof: if there is $$v \in N$$ which is not in any cone of $$\Sigma$$, then consider the corresponding one parameter subgroup $$\lambda_v$$. If you choose a system of coordinates on $$N$$ so that $$v = (v_1, \ldots, v_n)$$, then $$\lambda_v$$ is simply the map $$k^* \to X$$ given by $$t \mapsto (t^{v_1}, \ldots, t^{v_n})$$. If $$X$$ were complete, then $$\lambda_v$$ would extend to a map from $$k \to X$$. However, for every cone $$\sigma \in \Sigma$$, since $$v \not\in \sigma$$, there is $$u\in \check{\sigma}$$ such that $$\langle u, v \rangle < 0$$. Then $$\chi^u$$ is not defined at $$0$$, so that $$0$$ can not be mapped to anywhere on the open affine subset of $$X$$ corresponding to $$\sigma$$. Since $$X$$ is a union of these subsets, it follows that $$X$$ is not complete, as required.