3
$\begingroup$

Let $X_K$ be a smooth proper curve over a field $K$, and let $S$ be a Dedekind scheme with function field $K$.

Let $X$ be the proper regular minimal model of $X_K$ over $S$.

Let $Y_K$ be another smooth proper curve over $K$, equipped with a finite map $Y_K\rightarrow X_K$.

Let $Y$ be the normalization of $X$ inside the function field of $Y_K$.

Must $Y$ be the proper regular minimal model of $Y_K$?

If not, what goes wrong, and are there reasonable situations where this is true? (perhaps if we assume $Y_K/X_K$ etale?)

$\endgroup$
5
  • 2
    $\begingroup$ The natural map $f:X_1(11) \rightarrow X_0(11)$ is an isogeny between elliptic curves over $\mathbf{Q}$ (upon using compatible $\mathbf{Q}$-points for the origin of each), and its dual $f^*$ is a counterexample over $R=\mathbf{Z}_{(11)}$. Indeed, these have semistable reduction at 11, so by the link between minimal regular proper models and Neron models for elliptic curves we see via computing $-{\rm{ord}}_{11}(j)$'s that the minimal regular proper model $\mathscr{X}_0(11)$ has irreducible special fiber whereas $\mathscr{X}_1(11)$ has reducible special fiber (so $f^*$ doesn't extend over $R$). $\endgroup$
    – nfdc23
    Oct 1, 2016 at 6:59
  • $\begingroup$ @oxeimon You will find more on your question in arxiv.org/abs/math/0412075 and/or math.u-bordeaux.fr/~qliu/articles/modcove.pdf If I remember correctly, you can even find examples where $Y$ (hence $X$ if $g(X) >0$) has good reduction over $S$ such that $Y\to X$ does not extend to a morphism of minimal regular models. $\endgroup$ Oct 1, 2016 at 7:44
  • $\begingroup$ @AriyanJavanpeykar: Suppose $Y$ and $X$ are smooth proper over local $S$ with geometrically connected $K$-fiber of genus $>0$. Note that $P_X:={\rm{Pic}}_{X/S}^1$ is a torsor for $J_X:={\rm{Pic}}^0_{X/R}$, and likewise for $Y$. For finite $f:Y_K\rightarrow X_K$, the Albanese map $(P_Y)_K=P_{Y_K}\rightarrow P_{Y_K}=(P_X)_K$ of torsors uniquely extends to an $S$-map $A:P_Y\rightarrow P_X$ over $J_Y\rightarrow J_X$ (work etale-locally on $S$ to get $S$-points). But $Y$ is the closure in $P_Y$ of the canonical $Y_K\hookrightarrow P_{Y_K}$, and likewise for $X$, so $A:Y\rightarrow X$ extending $f$. $\endgroup$
    – nfdc23
    Oct 1, 2016 at 20:59
  • $\begingroup$ @nfdc23 Nice argument. Is the obtained morphism $Y\to X$ necessarily finite? (That's what I should have written in the last sentence of my comment, and I think I remember seeing an example where $Y\to X$ is not finite. Am I wrong?) Note that, more generally, if $S$ is a Dedekind scheme with function field $K$, and $X$ and $Y$ are smooth proper curves over $K$ with genus $>0$, then any morphism $Y\to X$ extends to a morphism $\mathcal Y^{sm}\to \mathcal X^{sm}$, where $\mathcal Y$ (resp. $\mathcal X$) is the minimal regular model of $Y$ (resp. $X$) over $S$; see arxiv.org/abs/1312.4822 $\endgroup$ Oct 2, 2016 at 7:48
  • $\begingroup$ @AriyanJavanpeykar: Yes, it must be finite. By Stein factorization considerations, the smooth proper special fiber is geometrically connected and hence irreducible. By properness and flatness over $S$ and surjectivity over $K$, the map must be surjective. Thus, on special fibers it is a surjection between irreducible curves, so quasi-finite, so the map over $S$ is quasi-finite and therefore (by properness) is finite. $\endgroup$
    – nfdc23
    Oct 2, 2016 at 15:48

1 Answer 1

2
$\begingroup$

There is no reason for $Y$ to be minimal, even if $Y_K\rightarrow X_K$ is étale. Take an elliptic curve $Y_K$ over $K=\mathbb{C}(t)$ with a rational point $y$ of order 2. Translation by $y$ gives an involution of $Y_K$, which extends to the minimal model $Y_0$. This involution will have some fixed points; it extends to the surface $Y$ obtained by blowing up the fixed points, and the quotient $X$ is smooth over $\mathbb{C}$, hence regular. $Y$ is the normalization of $X$ in $\mathbb{C}(Y)$ but is not minimal.

$\endgroup$

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.