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I have been reading the article "A conjecture in the arithmetic theory of differential equations" of Katz and I have a doubt regarding the "spreading out". My question is about the following paragraph, in section VI, which says

Let $X$ be a connected, smooth $\mathbb{C}$-scheme of finite type. It is standard that we can find a subring $R\subseteq\mathbb{C}$ which is finitely generated as a $\mathbb{Z}$-algebra, and a connected smooth $R$-scheme $\mathbb{X}/R$ of finite type, with geometrically connected fibres, from which we recover $X/\mathbb{C}$ by making the extension of scalars $R\hookrightarrow\mathbb{C}$.

I don't understand why it holds and wasn't able to find any concrete reference.

My attempt was to first try to understand this locally, that is, for the coordinate ring $\mathbb{C}[x_1,\ldots,x_n]/(f_1,\ldots,f_r)$ of an affine variety over $\mathbb{C}$ and I thought that one may take $R$ as the ring obtained by adjoining the coefficients of the $f_i$ to $\mathbb{Z}$. However in another article, "Nilpotent connections and the monodromy theorem: applications of a result of Turritin", Katz considers the following example

For instance, the Legendre family of elliptic curves, given in homogeneous coordinates by $$Y^2 Z - X(X-Z)(X-\lambda Z) \;\;\text{ in }\mathrm{Spec}\left(\mathbb{C}\left[\lambda,\dfrac{1}{\lambda(1-\lambda)}\right]\right)\times\mathbb{P}^2$$ is (projective and) smooth over $\mathrm{Spec}\left(\mathbb{C}\left[\lambda,\dfrac{1}{\lambda(1-\lambda)}\right]\right)=\mathbb{A}^1\smallsetminus\{0,1\}$. A natural thickening is just to keep the previous equation, but replace $\mathbb{C}\left[\lambda,\dfrac{1}{\lambda(1-\lambda)}\right]$ by $\mathbb{Z}\left[\lambda,\dfrac{1}{2\lambda(1-\lambda)}\right]$, and replace $\mathbb{C}$ by $\mathbb{Z}[1/2]$.

and in that example he adds a more than just the coefficients to $\mathbb{Z}$.

Edit: The main part that bugs me in this construction is that I don't see why the fibers are geometrically connected. When I saw the Legendre family example I thought that maybe one needed to add more things to the ring. Any help for how to do this?

Thanks in advance for any hints, reference or anything that will help me to better understand this.

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    $\begingroup$ You're correct, you just adjoin the coefficients of the $f_i$. The extra adjoining of $\frac{1}{2}$ in this example is to accomplish something else, I think. $\endgroup$ – Qiaochu Yuan Aug 17 '16 at 23:32
  • $\begingroup$ Thanks for the comment. However I don't see how the part of "geometrically connected fibres" works in what I'm saying and that's where I thought the $\frac{1}{2}$ came into play. Could you give me a hint about how to prove it? (I should edit this in the question to make it more clear!) $\endgroup$ – Nacho Darago Aug 18 '16 at 13:33
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    $\begingroup$ @NachoDarago Katz inverts $2$ so that the resulting scheme is smooth over the base. Indeed, the Legendre family $y^2=x(x-1)(x-\lambda)$ does not define an elliptic curve in characteristic two. $\endgroup$ – Ariyan Javanpeykar Aug 18 '16 at 20:36
  • $\begingroup$ See also mathoverflow.net/questions/312108 $\endgroup$ – Watson Nov 7 '18 at 9:37
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For the general case, you need to know some of constructibility and openness results in EGA. But let me explain two "baby" cases which I hope will make clear what is going on.

${\color{red}{\text{Example 1:}}}$ We are given a monic polynomial $f(X)$ in one variable $X$, with coefficients in $\mathbb{C}$, and with all distinct roots. We want to find a finitely generated subring $R_0$ of $\mathbb{C}$ and a monic polynomial $f_0(X)$ with coefficieints in $R_0$ such that after the extension of scalars from $R_0$ back to $\mathbb{C}$, we get our original polynomial $f(X)$, and for every homomorphism from $R_0$ to a field, call it $k$, the polynomial over $k$ we get by applying the homomorphism to the coefficients of $f_0$ has all distinct roots (in an algebraic closure of $k$).

${\color{blue}{\text{First method:}}}$ Start with$$R_1 := \mathbb{Z}[\text{all the coefficients of }f].$$The discriminant $\Delta_1$ of $f_1$ is an element of $R_1$ which is nonzero in $\mathbb{C}$, so certainly nonzero in $R_1$. Take $R_0$ to $R_1[1/\Delta_1]$. What we have achieved is that the discriminant is now an invertible element of $R_0$.

${\color{blue}{\text{Second method:}}}$ Again start with $R_1$ as above. We know that over $\mathbb{C}$, $f$ and its derivative $f'$ generate the unit ideal in $\mathbb{C}[X]$. Explicitly, there exist complex polynomials $A$ and $B$ such that $Af + Bf' = 1$. Now adjoin to $R_1$ all the coefficients of both $A$ and $B$. Over this $R_0$, $f$ and $f'$ generate the unit ideal in $R_0[X]$.

${\color{red}{\text{Example 2:}}}$ We start with a nonsingular hypersurface in affine $n$-space, defined by one equation $f(X_1, \ldots, X_n) = 0$. The nonsingularity means that in $\mathbb{C}[X_1, \ldots, X_n]$, $f$ and its partial derivatives $df/dX_i$ generate the unit ideal. So there exist polynomials $A$, $B_1$, $B_2$, $\ldots$ , $B_n$ in $\mathbb{C}[X_1, \ldots, X_n]$ such that$$Af + \sum_i B_i {{df}\over{dX_i}} = 1.$$ Okay, start with$$R_1 := \mathbb{Z}[\text{coefficients of }f],$$then pass to$$R_0 := R_1[\text{all coefficients of }A\text{ and all the }B_i].$$Once you have a spreading out, say $X_1/R_1$ with structure map $\pi$ which is now smooth and whose $\mathbb{C}$-fiber is geometrically connected (which for a smooth scheme is the same as geometrically irreducible), you can use the fact for $n := \text{the relative dimension of }X_1/R_1$, and $\ell$ a prime invertible in $R_1$ (if there isn't one pass to $R_1[1/\ell]$ for your favorite $\ell$), $R^{2n}\pi_!(\mathbb{Z}/\ell\mathbb{Z})$ is constructible, so "locally constant" or lisse, on a dense open set of $\text{Spec}\,R_1$; Its rank at each point is the number of geometrically irreducible components of the fiber: as this rank is one at the point corresponding to $\mathbb{C}$ (a point lying over the generic point), this rank is one on an open dense set, so certainly over a set of the form $R_1[1/g]$ for some nonzero element $g$ in $R_1$.

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  • $\begingroup$ I appreciate these examples, they've been helpful for understanding this! $\endgroup$ – Nacho Darago Aug 19 '16 at 14:42
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I think that I have answered my question.

The Nagata compactification theorem shows that if $X$ is a connected, smooth $\mathbb C$-scheme of finite type, there exists a proper $\mathbb C$-scheme $\overline{X}$ and an open inmersion $j:X\to\overline{X}$ over $\mathbb C$. Therefore, $\overline{X}$ is a finite union of coordinate rings of affine varieties and we may consider $R$ the $\mathbb Z$-algebra generated by the coefficients of all the polynomials defining these varieties. If we spread-out each variety over $R$ as in the question, then a glueing procedure gives us a connected, smooth $R$-scheme $\overline{\mathbb{X}}$ of finite type such that extension of scalars $R\hookrightarrow\mathbb C$ gives back $\overline{X}$.

edit: this was wrong, as compactifying may produce singularities. The $R$-scheme obtained $\mathbb{X}$ is only proper. The argument will work if $R$ can be chosen such that $\overline{\mathbb{X}}$ it is also flat.

Now, define $\mathbb{X}$ as the image of $X$ via the composition map $X\stackrel{j}{\to}\overline{X}\to\overline{\mathbb{X}}$, which is a connected, smooth $R$-scheme of finite type. The assertion regarding the geometrically connectedness of the fibers follows from this post https://mathoverflow.net/questions/201016/connectedness-of-fibers-for-flat-proper-morphism since $\overline{\mathbb{X}}$ is a proper $R$-scheme and the generic fiber is geometrically connected (since it is $\overline{X}$). Because every fiber in $\mathbb{X}$ is simply the pullback along $j$ of the fibers of $\overline{\mathbb{X}}$ the connectedness follows.

Is this alright? Thanks once again.

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  • $\begingroup$ The proper scheme you obtain over $R$ is not necessarily flat. You need to use that flatness also spreads out. Thus, you should "shrink" Spec $R$ a bit. Then you can use the result from EGA to conclude that $\mathbb X$ has geometrically connected fibres. $\endgroup$ – Ariyan Javanpeykar Aug 18 '16 at 20:44
  • $\begingroup$ Great, thanks! I didn't notice this. Could you please elaborate on the shrinking part? I don't know how should I do it. $\endgroup$ – Nacho Darago Aug 18 '16 at 22:02
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    $\begingroup$ Let $X\to S$ be a morphism of schemes of finite presentation. Suppose that $S$ is integral. Note that $f$ is generically flat, i.e., $f_K : X_K\to Spec K$ is flat (with $K =K(S)$ the function field of $S$). Indeed, "everything" is flat over a field. In particular, the morphism $f$ is flat over some dense open $U$ of $S$. This is called spreading out of flatness; see Poonen's notes on rational points. Thus, in your case: There exists an open (hence open affine) of Spec $R$ over which your morphism is flat. $\endgroup$ – Ariyan Javanpeykar Aug 19 '16 at 7:05
  • $\begingroup$ Everything is clear now, thanks a lot for the help! (and also for that reference, those notes seem really good) $\endgroup$ – Nacho Darago Aug 19 '16 at 14:41

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