This is a soft question and really a request for pointers towards a certain rigorous formulation of geometric intuition I've had for some "arithmetic schemes". I'm looking for ideas and key references of areas to look into.

Anyway, I've been thinking about the "arithmetic surface" $X = \mathbb{A}^1_{\mathbb{Z}} = \text{Spec}\mathbb{Z}[x]$ and the closed subscheme $Y = \text{Spec}\mathbb{Z}[\sqrt{3}]\subseteq X$ along with the natural morphism $\pi: Y\rightarrow \text{Spec}\mathbb{Z}$. Following the treatment in Eisenbud and Harris' The Geometry of Schemes (p.83) we can draw $Y$ as a scheme fibred over $\text{Spec}\mathbb{Z}$ that looks something like this:

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The fibre above a prime $p\in\mathbb{Z}$ has one of three types:

  1. Type 1, like the points above $p=11,13$, consists of a fibre with two distinct points. This occurs in general when the Legendre symbol $(3/p)=1$ and $p$ does not divide the discriminant $12$ of $K=\mathbb{Q}(\sqrt{3})$, and these are reduced points with residue field $\mathbb{F}_p$.This occurs when the ideal $p\mathcal{O}_K$ factors into two distinct prime ideals in the ring of integers $\mathcal{O}_K$ of $K$. I'm not going to talk more about these points - it's the distinction between the following two that counts;
  2. Type 2, like the points above $p=5,7$ and more generally when $(3/p)=-1$ and $p$ does not divide $\text{disc}(K)$, is a fibre consisting of one reduced point when the ideal $p\mathcal{O}_K$ remains prime in $\mathcal{O}_K$. Here the residue field of this point is a quadratic extension $\mathbb{F}_{p^2}$. In my picture - and in Eisenbud/Harris - these are drawn as single points where the generic point of $Y$ ramifies and becomes tangent to a horizonal line;
  3. Type 3 like the fibres above primes $p=2,3$ that divide $\text{disc}(K)$; these are the ramified primes, for obvious geometrical reasons. Here the single point in the fibre is nonreduced but has residue field $\mathbb{F}_p$. For these primes the ideal $p\mathcal{O}_K$ is the square of a prime ideal in $\mathcal{O}_K$. Here in my picture the generic point ramifies and becomes tangent to a vertical line at these points.

Now I was wondering why, in algebraic number theory, Type 3 primes are called ramified whilst Type 2 ones aren't, given that the fibres "ramify" as point sets above these primes and become singleton sets rather than pairs of distinct points. One intuitive reason to me is that the non-reducedness of the truly ramified points underlies that two distinct fibres are joining together at this point. However, in the Type 2 case these points are reduced, so although these look "ramified" in the picture there is a different phenomenon occuring here.

The way I tried to understand it is this: the generic point of $Y$ consists of two "strands" which truly ramify only at the primes $p=2,3$. When they pass through a Type 2 prime like 5 they each intersect at a pair of distinct Galois-conjugate points $\pm \alpha\in \mathbb{F}_{25}$ each of which is a root of $x^2-3\in\mathbb{F}_5 [x]$. Now in the fibre above $5$, which is isomorphic to $\mathbb{A}^1_5 := \text{Spec}\mathbb{F}_5 [x]$, this pair of elements of $\mathbb{F}_{25}$ constitutes a single point (corresponding to the prime ideal $(x^2-3)$), which is why the fibre is a singleton. However, where we able to "lift" this fibre to $\text{Spec}\mathbb{F}_{25}[x]$ we would see that really there are two branches of this generic point passing through the pair of distinct points $\pm\alpha$, and these only "arithmetically ramify" because they map to the same point under the map $\text{Spec}\mathbb{F}_{25}[x]\rightarrow\text{Spec}\mathbb{F}_5 [x]$. So we could somehow "resolve" this ramification over (the affine line over) a quadratic extension field of $\mathbb{F}_5$ to see that "geometrically" this isn't true ramification.

In contrast, the point in the fibre above a Type 3 prime like $p=2,3$ corresponds to a Galois orbit with one element since it has residue field $\mathbb{F}_p$, and represents the "geometric" branches of the generic point of $Y$ genuinely coming together at this point. To me it seems that these points are where real ramification occurs, because no "lift" to an extension field can separate the branches of the generic point.

So my questions are:

  1. Is this geometric picture the reason ramified (Type 3) primes are so named?
  2. For Type 2 primes, is there a formal construction for this "lifting" process that I described, whereby we "resolve" the ramification at Type 2 primes by lifting the whole of Y to another scheme on which the fibres above Type 2 primes are isomorphic to/contain $\text{Spec}\mathbb{F}_{p^2}[x]$ and hence distinguish these Galois-conjugate points?
  • $\begingroup$ PS, for future reference, use larger tags. Schemes, and arithmetic geometry, while not terrible descriptors of this post, are not tags that people commonly check. In particular, next time put algebraic geometry. :P $\endgroup$ Feb 26, 2016 at 9:39

1 Answer 1

  1. Yes, this is correct. Whenever we define a property $P$ of a morphism $f:X\to Y$, one of the FIRST things one verifies, to make sure it's a reasonable class, is that

a) Property $P$ is closed under composition.

b) Property $P$ is closed under base change.

Namely, whatever it should mean for a morphism $f:X\to Y$ to be 'unramified' it certainly should be the case that $f_S:X_S\to S$ is also 'unramified' for any $Y$-scheme $S\to Y$.

In particular, for a morphism $f:X\to\text{Spec}(k)$, it should be true that if $f$ satisfies $P$ then $f_{\overline{k}}:X_{\overline{k}}\to \text{Spec}(\overline{k})$ satisfies $P$.

In particular, for ramifiedness, one can truly only get a GEOMETRIC picture (i.e. ramified morphisms are those where strands intersection) only in a GEOMETRIC setting. Namely, working rationally (i.e. over $\text{Spec}(k)$) and expecting to visually be able to 'see' a property is a little misguided—all the points of the situation aren't 'visible' yet, you have to first move to $\text{Spec}(\overline{k})$.

That said, there are many, many beautiful scheme-theoretic definitions of unramifiedness that one can easily verify agree with our intuitive definition when $f:C\to C'$ is a map of curves over an algebraically closed field (the situation that you are most thinking about).

For example, let us make the definition that a map $f:C\to C'$ of curves over $\text{Spec}(\overline{k})$ is unramfied if for all $y\in C'$ a closed point the fiber $f^{-1}(y)\to \text{Spec}(k(y))$ is reduced—that it's just a disjoint union of $\text{Spec}(\overline{k})$ if $y$ is a closed point. This is intuitively what we want since, we think that a point $x\in f^{-1}(y)$ of ramifification should 'count' for more than just a point, since it's picking up the data of multiple strands, and so we expect that point in the fiber to have global sections greater than dimension $1$, which, since we've over $\overline{k}$ can only happen if that point is non-reduced.

If we're NOT working over $\overline{k}$, but just $k$, then it's obvious how to fix this. Namely, $f:C\to C'$ non-constant should be unramified when $f^{-1}(y)\to\text{Spec}(k(y))$ is geometrically reduced. Thus, perhaps a point has global sections of dimension greater than $1$, but not because it has multiple strands meeting at that point, but that it's a Galois orbit of points glued together, that would separate into distrinct strands geometrically.

So, let us make the preliminary definition that a finite map $f:X\to Y$ of Dedekind schemes (i.e. integral normal schemes of dimension $1$) of finite type over $\text{Spec}(\mathbb{Z})$ is unramified if for all $y\in Y$ the scheme $f^{-1}(y)\to\text{Spec}(k(y))$ is geometrically reduced carrying the same intuition we had for curves over a field.

Let us then verify that this definition agrees with the one we learn in number theory. Namely, let's assume, for the time being, that $X$ and $Y$ are affine (we can reduce to this case since $f$ is affine). Namely, let's assume that $Y=\text{Spec}(A)$ and $X=\text{Spec}(B)$. Then, for a non-zero prime $\mathfrak{p}\in Y$ we know that $f^{-1}(\mathfrak{p})\subseteq X$ is given by


Let us then use that $B$ is Dedekind to factor $\mathfrak{p}$ as follows


with $\mathfrak{P}_i$ distinct primes of $B$.

We claim that the $k(\mathfrak{p})=A/\mathfrak{p}$-algebra $B/\mathfrak{p}B$ is geometrically reduced if and only if $e_i=1$ for all $i$. Indeed,

$$B/\mathfrak{p}B\cong \prod_i B/\mathfrak{P}_i^{e_i}$$

and so

$$(B/\mathfrak{p}B)\otimes_{k(\mathfrak{p})}\overline{k(\mathfrak{p})}=\prod_i B_{\overline{k(\mathfrak{p})}}/\mathfrak{P}^{e_i}B_{\overline{k(\mathfrak{p})}}$$

Thus, if $e_i>1$ for some $i$, this is evidently still nonreduced.

Suppose now that all the $e_i=1$, so that $B/\mathfrak{p}B$ above is isomorphic to a product with terms $B/\mathfrak{P}B$. Note that this is a finite separable extension of $k(\mathfrak{p})$ (here I am using that we're finite type over $\text{Spec}(\mathbb{Z})$, so these are both finite fields!). Then, by the primitive element theorem


for some separable polynomial $f(x)\in k(\mathfrak{p})[x]$. Then, $(B/\mathfrak{P}B)\otimes_{k(\mathfrak{p})}\overline{k(\mathfrak{p})}$ is just $\overline{k(\mathfrak{p})}[x]/(f(x))$. Since $f(x)$ was separable it becomes a product of distinct linear factors in $\overline{k(\mathfrak{p})}$ which implies that $\overline{k(\mathfrak{p})}[x]/(f(x))$ is isomorphic to a product of $\overline{k(\mathfrak{p})}$ and so separable.

Thus, our geometric definition, that we can geometrically see ramification (the coming together of strands) by reducedness of fibers agrees with the arithmetic definition!

NB: There are two things that I glossed over in the above. First, I didn't talk about the fiber over the generic point of $X$ and $Y$. What do you think happens there? Secondly, I made the artificial assumption that we were finite type over $\text{Spec}(\mathbb{Z})$ so that I didn't have to talk about assuming that the residue field extensions were separable. So, you can now guess the correct definition for an arbitrary map of Dedekind schemes—that the fibers are a product of separable extensions of $k(y)$ (or $k(\mathfrak{p})$.

Let me give one more possible nice interpretation of unramifiedness, and let you work out the details yourself. So, let's suppose that we have a surjective finite map $X\to Y$ of Dedekind schemes. Then, we imagine that $f$ should be unramied at $x\in X$ if 'locally around $x$' the map $f$ is an isomorphism.

Here is one way of interpretting this. Since $X$ is a Dedekind scheme, the ring $\mathcal{O}_{X,x}$ is a DVR and so it's maximal ideal $\mathfrak{m}_x\mathcal{O}_{X,x}$ is principal, say with uniformizer $t_x$. We then think about $t_x$ as being a COORDINATE of $X$ at $x$—a chart. Then, one way to think about $f$ being a local isomorphism at $x$ is that if we take the chart $t_y$ at $y=f(x)$ (defined in the same way) then '$t_y\circ f$' should be a chart at $x$—that it should generate the same ideal as $t_x$.

Thus, another intuitive definition of unramifiedness is that for all $x\in X$ the map $\mathcal{O}_{Y,f(x)}\to \mathcal{O}_{X,x}$ has the property that


that the chart at $y$ is still, after being pulled back by $f$, a chart.

I leave it to you to check that this is also equivalent, in the arithmetic situation, to the usual definition of being unramified, that there are no higher powers of primes showing up in the factorization.

Let me end with two more remarks:

a) All of the above, in some sense, has been a ruse. Namely, by using the crutch that we were dealing with Dedekind schemes, we were able to ignore an important property—flatness. Namely, what you're picturing in your head is not really unramifiedness, it's étaleness (google this!). Although, frankly, this doesn't really factor into your picture because every morphism we picture naturally is flat!

So, if you want to learn more about this topic and the beautiful, beautiful details it entails, I HIGHLY recommend picking up the Book Galois Groups and Fundamental Groups by Szamuely.

b) I can't resist saying this. Another interpretation of unramifiedness (again, I'm lying! I really mean étaleness, but this is the same for surjective maps of Dedekind schemes) one might take is that the morphism if `smooth of relative dimension $0$'. If one understands enough algebraic geometry, one can see that this really means that $f$ is flat and that $\Omega_{X/Y}$, the relative cotangent sheaf, is zero.

So, where is this cotangent sheaf showing up arithmetically? Well, if $\Omega_{X/Y}$ being zero is measuring unramifiedness, then ramifiedness has to do with the points $x\in X$ where $(\Omega_{X/Y})_x\ne 0$. Or, thought about differently, we want to know that the annihilator of $\Omega_{X/Y}$ is the whole ring (at least in the case that $X$ and $Y$ are affine).

But, what is the annihilator of $\Omega_{X/Y}$? It's the different ideal! The mysterious ideal which dictates the ramification properties of our map/extension.

  1. I think this is answered in my above post. Basically, one just wants to move everything to the geometric situation over the algebraic closure of the residue field of the point in the target.
  • $\begingroup$ This is probably the most comprehensive answer I've ever received on MSE. Thank you! In particular the discussion of two different interpretations/definitions of ramification was really helpful (I guess the second is essentially the same definition as for morphisms of varieties). I have to say I'm not entirely sure what happens for the generic fibre but I will think about this today - unless you have enough energy left after this to give a brief explanation! Also, +1 for mentioning étale morphisms. I have been reading Szamuely's book for my undergrad thesis and part of your (continued) ... $\endgroup$
    – Alex Saad
    Feb 26, 2016 at 13:42
  • $\begingroup$ first discussion on ramification seems very familiar to his discussion precluding étale morphisms of curves. Since I'm only focusing on elliptic curves in my project from a fairly classical viewpoint I chickened out at this point and instead use a more ad hoc definition involving tangent spaces (which is all I really required). However it is helpful to see the same phenomenon occurring here with schemes in a more abstract sense. After reading your fantastic answer and seeing how naturally étaleness appears with schemes I'll go back to Szamuely and read this section fully. $\endgroup$
    – Alex Saad
    Feb 26, 2016 at 13:49
  • $\begingroup$ @AlexSaad Glad it was a help to you. :) If you'd like to read more about this, I just remembered that I actually wrote a post about étale morphisms! And, since you don't seem to mind schemes, I think it'll be of use to you: ayoucis.wordpress.com/2014/04/06/unramified-morphisms $\endgroup$ Feb 26, 2016 at 13:52
  • $\begingroup$ Great, I look forward to reading it :) And thanks for the blog link too - it looks really helpful for lots of other algebraic geometry (which reminds me I will reply to your comments on mine in due course!). Finally, I'm going to make a couple of edits to the above answer where I think there have been small typos - if any of them are deliberate and I'm just being stupid, please just change them back. $\endgroup$
    – Alex Saad
    Feb 26, 2016 at 14:01

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