# index of smooth varieties

What are the simplest examples of smooth, projective varieties defined over the fraction field of an Henselian DVR of characteristic $0$ which have index $>1$?

EDIT: Also assume that the residue field of the DVR is algebraically closed.

• What do you mean by index here? May 6, 2018 at 18:39
• @R.vanDobbendeBruyn Index is the gcd of the set of degrees of zero dimensional cycles on the variety May 6, 2018 at 18:40
• Conics without rational points are usually the easiest examples. For example, the conic over $\mathbb C((a,b))$ given by $\{ax^2 + by^2 = z^2\} \subseteq \mathbb P^2_{\mathbb C((a,b))}$ should be an example. May 6, 2018 at 18:48
• @R.vanDobbendeBruyn What is the index in this case? Could you please elaborate a little on the computation or cite a reference. I also wanted to mention that absence of rational point does not imply index greater than $1$. May 6, 2018 at 18:53
• @R.vanDobbendeBruyn. "Note that now complete DVRs are also out" I do not understand this comment. If the OP were asking for a smooth, projective scheme over $R$ whose generic fiber has index $>1$, then I would completely agree. However, the OP only asks that the generic fiber is smooth, not that the entire $R$-scheme is $R$-smooth. May 7, 2018 at 15:44

Let $d\geq 3$ be any integer that is relatively prime to the residue characteristic of $R$. Let $s$ be any generator of the maximal ideal $\mathfrak{m}$ of $R$. Let $\widetilde{R}$ be the finite, flat extension $R[\sigma]/\langle \sigma^d -s \rangle$ with its natural action of the group scheme $\mu_d=\text{Spec} \ \mathbb{Z}[\zeta]/\langle \zeta^d-1\rangle$, $$\text{Spec}\ \mathbb{Z}[\zeta]\langle \zeta^d-1 \rangle \times \text{Spec}\ R[\sigma]/\langle \sigma^d-s\rangle \to \text{Spec}\ R[\sigma]/\langle \sigma^d - s\rangle, \ \ \sigma \mapsto \zeta\otimes \sigma.$$ Denote the induced morphism of schemes by $$q:\text{Spec}\ \widetilde{R} \to \text{Spec}\ R.$$ The generic fiber of $q$ is a $\mu_d$-torsor.
Let $X_R$ be the smooth, projective $R$-scheme, $$X_R = \text{Proj}\ R[t_0,t_1,\dots,t_{d-1}]/\langle t_0^d + t_1^d + \dots + t_{d-1}^d \rangle.$$ This has a natural action of $\mu_d$ by $$\zeta\bullet [t_0,t_1,\dots,t_{d-1}] = [\zeta^0t_0,\zeta^1 t_1, \dots, \zeta^{d-1}t_{d-1}].$$ Denote by $X_{\widetilde{R}}$ the base change, $$X_{\widetilde{R}} = \text{Spec}\ \widetilde{R} \times_{\text{Spec}\ R} X_R.$$ This has a diagonal action of $\mu_d$ that is a free action. The projection to the first factor is equivariant for this action, $$\text{pr}_1:X_{\widetilde{R}} \to \text{Spec}\ \widetilde{R}.$$ Thus, for the geometric quotient by this free $\mu_d$-action, $$q_X:X_{\widetilde{R}} \to \mathcal{X}_R^s,$$ there is a unique morphism, $$\pi:\widetilde{X}_R\to \text{Spec}\ R,$$ such that $\pi\circ q_X$ equals $q\circ \text{pr}_1$. Since the base change of $\pi$ by the fppf morphism $q$ equals the smooth morphism $\text{pr}_1$, also the proper morphism $\pi$ is smooth. Since the geometric generic fiber of $\text{pr}_1$ is a smooth hypersurface of dimension $d-2$, it is integral. Thus, the geometric generic fiber of $\pi$ is also integral.
If $d$ is a prime, then the index of the generic fiber of $\pi$ equals $d$. Indeed, every zero-dimensional, reduced, closed subscheme of the generic fiber pulls back to a zero-dimensional, reduced, closed subscheme of the generic fiber of $\text{pr}_1$ that is $\mu_d$-invariant. Thus, for every irreducible component of this closed subscheme, the closure in $X_{\widetilde{R}}$ is an integral, closed subscheme that is $\mu_d$-invariant and finite over $\text{Spec}\ \widetilde{R}$. By the valuative criterion of properness for $\text{pr}_1$, the intersection of this closed subscheme with the closed fiber of $\text{pr}_1$ is a closed subscheme that is $\mu_d$-invariant. Since the action of $\mu_d$ on the closed fiber is free, the length is divisible by $d$. Since this holds for every irreducible component, the total length is divisible by $d$.
• Is it obvious that the generic fiber of $\pi$ is smooth? May 7, 2018 at 0:10
• @user43198. "Is it obvious that the generic fiber of $\pi$ is smooth?" I do not know what is obvious to you. However, that follows from the Jacobian criterion, since the Jacobian ideal of the Fermat polynomial equals $\langle d t_0^{d-1},\dots, dt_{d-1}^{d-1} \rangle$. May 7, 2018 at 0:16
• One further question. Redoing your arguments for a finite, unramified extension $L$ of $K$ (with residue degree $0$) and $R_L$ the integral closure of $R$ in $L$, we will be able to conclude that the generic fiber of the corresponding $\pi_L:\mathcal{X}_{R_L}^s \to \mbox{Spec}(R_L)$ is also of index $d$. The generic fiber of $\pi_L$ should be the base change to $\mbox{Spec}(L)$ of the generic fiber of $\pi$ (unramified extension will ensure that the uniformizer of $R$ and $R_L$ can be the same). Is this correct? May 7, 2018 at 8:23
• Thank you. One last question. Your construction does not seem to require the residue field is algebraically closed. Is this correct i.e., will your construction also work if the residue field of $R$ is not algebraically closed? May 7, 2018 at 9:04