In his Algebraic surfaces book, Beauville gives a result allowing one to "absorb ramification" for certain maps (see below). There are also something similar one can do with number fields. I would really like to know how I should think about these results, (that is, are they specific to these two situations or are they vastly more general.) I'm really hoping for a reference (to EGA?) where such a version is stated, assuming it is true.

This question naturally lends itself to questions 2,3; which are hopefully covered in the same place as #1 (assuming again that they are true).

$\textbf{1.}\textrm{ Absorbing ramification}$

If $K|k$ is a finite extension of number fields, then there are infinitely many finite extensions $E|k$, so that $K\cap E = k$ and $EK|E$ is unramified. (You can get such extensions from the approximation lemma.) (Does this fact have a name?)

$\textbf{Setup:}$ Let $X \xrightarrow{\phi} Y$ be a finite, faithfully flat map of noetherian schemes. Feel free to assume conditions on $X, Y, \phi$, as you need.

Can we have something similar for faithfully flat maps $\phi$? More specifically, let $\phi, X,Y$ be as in the setup. Does there exist $Z \xrightarrow{\tau} Y$ a finite, faithfully flat map so that the base change along $\tau$ is etale. Is there a lemma (like the approximation lemma) showing that there are "many" such maps in good conditions, and allowing us to control the ramification of $\tau$.

-In certain cases, we can absorb ramification for a fibration of a smooth surface (Beauville, p73 Lemma VI.7):


$\textbf{2.}\textrm{ Maximal etale subextensions}$

Let $\phi, X,Y$ be as in the setup.

Does there exist a maximal etale subextension? That is, can we factor $\phi$ as $X \rightarrow Y_{et} \xrightarrow{\psi} Y$ with $\psi$ etale, and so that $\psi$ is maximal amongst all such factorizations. That is, if $\phi$ also factors as $$X \rightarrow Y' \xrightarrow{\psi'} Y$$ with $\psi'$ etale, then the map $Y_{et} \xrightarrow{\psi} Y$ factors as $Y_{et} \rightarrow Y' \xrightarrow{\psi'} Y$.

$\textbf{3.}\textrm{ Galois closure}$

The natural question is take $X, Y, \phi$ be as our initial setup, (assume that $\phi$ is separable, $X,Y$ are projective over a field, integral) and ask if there a galois closure $Z \rightarrow X$, so that $Z/Aut(Z/Y) \cong Y$ naturally. However, it's not even clear to me that the automorphism group would be finite (or more generally that the quotient exists as a scheme) - or there exists such $Z$ which is "minimal".

These questions bear a strong relationship to theorems true for number fields. One issue of course is that we use composition of number fields in number theory and in algebraic geometry we have base change along maps which are a little different operations.

  • $\begingroup$ Can someone tell me if the link to google books is visible to them? I could write up the theorem, but since mathoverflow doesn't really support commutative diagrams it won't look nearly as good as the version in Beauville's book. $\endgroup$
    – LMN
    Jan 25, 2013 at 22:08
  • $\begingroup$ You might want to look at Abhyankar's Lemma and its variations. See SGA 1, Exp. XIII, Appendice I, which says that if $Y$ is regular, and $\phi$ is tamely ramified, then you can remove the ramification via an appropriate base change of $Y$. There's also the Nagata-Zariski purity theorem (SGA 1, Exp. X, Theoreme 3.1), which says that, when $Y$ is regular and $X$ normal, then the locus in $Y$ where $\phi$ is ramified has to be of pure co-dimension $1$. $\endgroup$ Jan 25, 2013 at 22:53
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    $\begingroup$ The link was fine. As I recall Abhyankar's Lemma applies to this situation but only applies to dvr's. It allows you to shift tame ramification from phi to the base extension. A good reference is an early LNM volume by Murre and Grothendieck. This will not answer your general question without some additional condition on Y. $\endgroup$ Jan 25, 2013 at 22:58
  • $\begingroup$ Isn't relative dimension preserved by sufficiently nice base change? It would be hard to change the relative dimension of a map. Do you want to base change a flat morphism to a smooth morphism, maybe $\endgroup$
    – Will Sawin
    Jan 25, 2013 at 23:21
  • $\begingroup$ @Keerthi, Ray, Thanks! I'll look into it. Will, I don't understand, could you clarify please? The picture I have in mind is what you say - I'm taking finite flat maps and absorbing their ramification to become etale. $\endgroup$
    – LMN
    Jan 26, 2013 at 0:10

1 Answer 1


If $X$ and $Y$ are normal and separated then they each have a unique generic point, so the map betwen generic points gives a field extension, and $X$ is determined over $Y$ by its field extension. Indeed, if $U$ is an affine open set of $Y$, and $U'$ is its inverse image in $X$, then $\Gamma(U',\mathcal O_X)$ is the normalization of $\Gamma(U,\mathcal O_Y)$ in the residue field of the generic point of $X$. Moreover, this is an equivalence of categories - the morphisms are the same.

This, for instance, immediately tells us how to construct the Galois closure - just take the Galois closure of the corresponding field extension.

One could study maximal etale subextensions this way, but I think it's better to argue like this: If $X \to Y_1 \to Y$ and $X \to Y_2 \to Y$ are both etale, then $Y_1 \times_Y Y_2$ is etale and $X$ maps to it. If you multiply all the finite etale things that $X$ factors through like this, and take the image of $X$, you should get a maximal etale cover.

  • $\begingroup$ The fiber product $Y_1\times_Y Y_2$ may be not connected. So you should add taking the connected component containing $X$ after taking the fiber product of all etale subextensions. $\endgroup$
    – Sasha
    Jan 26, 2013 at 3:38
  • $\begingroup$ The image of $X$ is connnected. Saying "the connected component containing $X$" may make it slightly easier to prove, but unless I'm missing some weird degeneracy $X$ should always equal its connected component. $\endgroup$
    – Will Sawin
    Jan 26, 2013 at 4:08
  • $\begingroup$ You wind up, of course, with a pro-object, not a scheme. $\endgroup$ Jan 26, 2013 at 18:39
  • $\begingroup$ Not if there are only finitely many things that factor through, which you can get if you assume enough finiteness conditions. I haven't worked out exactly which. $\endgroup$
    – Will Sawin
    Jan 26, 2013 at 20:20
  • $\begingroup$ Each $Y_i$ is given by the subextension $K(Y_i)$ of $K(X)/K(Y)$. As $K(X)$ is finite separable over $K(Y)$, there are only finitely many subextensions. $\endgroup$
    – Qing Liu
    Feb 26, 2013 at 15:19

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