# Unramified Morphisms

In this post we will develop the notion of unramified morphisms, in their various guises.

In this post we will be assuming the reader is familiar with the (relative) cotangent sheaf. If not, I suggest they first read chapter 21 of Vakil’s Foundations of Algebraic Geometry or II.8 of Hartshorne. I’ll try to be explicit when I can though, and reference the needed theorems, and sometimes be overly hand-holdy with computations.

# Motivation

Our current goal is to construct the right analogue of “local isomorphisms” that were hinted at in our motivation for etale cohomology. Unramified morphisms, along with flatness, will be the properties that end up defining this correct idea of local isomorphisms.

Just like flat morphisms, unramified morphisms are a type of map that plays correctly into our intuition of a “nice morphisms”. They also have the pleasant property of not only being intuitive to understand, but relatively painless to work with technically, thanks to the formalism of the cotangent sheaf.

To understand the idea behind unramified morphisms, let us look at a picture (taken from Wikipedia):

In this picture, what we say is that generically $f$ is a $3$-to-$1$ function, and that $X$ is split into $3$ sheets. Of course, the $3$-to-$1$ness is only generic since over two points of $Y$, the ones marked with dots, we see that $f$ does stop being $3$-to-$1$. At these points, we see that the three sheets that generically make up $X$ come together, or “ramify”.

If we think about this picture in the analytic topology (the “usual one” from complex analysis) these points of ramification are precisely the points of $X$ for which no neighborhood is injective. Indeed, the other points of $x$ have neighborhoods where $f$ is injective by just taking an open segment (the intersection of a small Euclidean ball in the ambient space with $X$) along the sheet they reside. But, at the point of ramification, every Euclidean ball intersects all three sheets, and so no neighborhood of these points verifies $f$‘s injectivity.

Now, we see that if we want to create a type of morphisms modeling “local isomorphisms” (something like a local biholomorphism in complex geometry), we clearly want to exclude ramified morphisms. Indeed, since “local isomorphisms” should, in particular, be locally injective, we see that ramification points are a clear and present obstruction to “local isomorphism”ness.

So, how do we formalize unramifiedness? The problem with the above intuition is that it is just that–intuition. General schemes don’t usually have the luxury of some kind of analytic topology from which to work off of (the whole point of etale cohomology!) and so we must use some other properties which are intrinsic to the schemes themselves. But, what are possible properties of a general scheme for which the above picture violates at the point of ramification?

It turns out, fortunately, that there are several ways to proceed:

## Open Diagonal

Perhaps the most easily explained is the openess of the diagonal map. Let us suppose that we have a morphism $f:X\to Y$. Very informally, we can think about the fiber product $X\times_Y X$ as being the pairs of points $(x,x')\in X\times X$ such that $f(x)=f(x')$. The diagonal $\Delta_X\subseteq X\times_Y X$ can then be thought of merely as the set of points $(x,x)$ or, said differently, points of the fiber product whose coordinates are equal.

Now, notice that in the picture above each point $y\in Y$, which is not the image of a ramification point, there are points in the set $X\times_Y X-\Delta_X$ where each coordinate is in $f^{-1}(y)$. Indeed, just pick a tuple $(x,x')$ where $x$ and $x'$ both lie above $y$, but are in different sheets. Now, a point of ramification, call it $c$, is one where two sheets come together. This intuitively allows us to take a sequence of points $(x_n,x'_n)$ in $X\times_Y X-\Delta_X$, where $x_n$ is in one of the sheets coming together and $x'_n$ is in the other, whose limit is precisely the point $(c,c)$. And, intuitively, this is how all ramification points should arise, as points of the diagonal which are limits of points not in the diagonal.

This gives us one way to define unramifiedness. In particular, being ramified is the ability to find points of the diagonal which are limits of points in the fiber product, but not on the diagonal. This is precisely the statement that $X\times_Y X-\Delta_X$ is not closed. Thus, if we want to force this to not happen, we merely want to specify that $X\times_Y X-\Delta_X$ is closed or, in other words, $\Delta_X$ is open.

## Unicity of Tangent Vector Lifts

Another point of attack is through the notion of “unicity of tangent vector lifts”. This point of view falls under the banner of “formally unramified”, but, it is also a fairly intuitive idea. It shows that unramified morphisms are the algebraic geometry analogue of the notion of immersion from differential topology.

Let’s consider, once again, a morphism $f:X\to Y$. Let’s fix a point $y\in Y$, and a point $x\in X$ which is in the fiber $f^{-1}(y)$. Given a tangent vector $p$ of $X$ based at $x$ we can push it forward, through $f$, to a tangent vector of $Y$ based at $y$. Or, in other terms, we have the derivative map $df_x:T_x X\to T_yY$.

Let’s consider this derivative map in the case of the following picture.

Clearly here this is only one ramification point (the intersection of the two sheets). Notice that if $x$ is not a ramification point, then the derivative map is injective. Indeed, fix a point $y\in Y$, and a tangent vector at that point. For example, the following picture:

Then, there is clearly only one choice of tangent vector at the point above $y\in Y$ in the top sheet, whose image is the specified tangent vector. Namely:

But, consider the following tangent vector lying below the ramification point

Then, there are two tangent vectors at the ramification point that could map to this tangent vector. Namely:

Thus, we might hope to to define unramified via some type of “injectivity on tangent vectors”, or “unicity of tangent vector lifts” criterion. Of course, it’s not quite that simple. To correctly define it, we must, in fact, specify injectivity on all “differential like data”.

What this means rigorously is the following. Let $f:X\to Y$ be a morphism. For any ring $A$, square-zero ideal $I$ (i.e. $I^2=0$), and morphism $\text{Spec}(A)\to Y$ the map

$\text{Hom}_Y(\text{Spec}(A),X)\to\text{Hom}_Y(\text{Spec}(A/I),X)$

is injective, where $\text{Spec}(A/I)$ is considered a $Y$-scheme via the composition

$\text{Spec}(A/I)\to\text{Spec}(A)\to Y$

To see how this relates to our picture above, let’s consider the case where $X$ and $Y$ are $k$-varieties, $A=k[\varepsilon]$ (where $\varepsilon^2=0$), and $I=(\varepsilon)$. Then, intuitively, giving a morphism $\text{Spec}(k[\varepsilon])\to Y$ amounts to choosing a tangent vector $p$ on $Y$. The induced map $\text{Spec}(k)\to Y$ (note that $k[\varepsilon]/I=k$) is merely picking out the base $y$ of this tangent vector. Note that for the rest of this discussion, this choice of tangent vector (and induced choice of base point) on $Y$ is fixed.

An element of $\text{Hom}_Y(\text{Spec}(k),X)$ is then choosing a point of $X$ which maps, under $f$, to the chosen point $y$ of $Y$. Or, in other words, $\text{Hom}_Y(\text{Spec}(k),X)$ is just $f^{-1}(y)$. An element $\text{Hom}_Y(\text{Spec}(k[\varepsilon]),X)$ is just a choice of tangent vector on $X$ which, under $f$, maps to the chosen tangent vector on $Y$. Finally, the map

$\text{Hom}_Y(\text{Spec}(k[\varepsilon]),X)\to\text{Hom}_Y(\text{Spec}(k),X)=f^{-1}(y)$

just takes a tangent vector in the domain, and picks out its base point (an element of the codomain).

Thus, the injectivity of map

$\text{Hom}_Y(\text{Spec}(k[\varepsilon]),X)\to\text{Hom}_Y(\text{Spec}(k),X)=f^{-1}(y)$

amounts to the assertion that given a point of the fiber $f^{-1}(y)$, there is at most one tangent vector at that point which maps to the prechosen tangent vector at $y$. This is precisely the property we were pictorially describing above. One can even make the connection to the tangent space rigorous by taking all of the above discussion relative $k$.

Thus, we see via an example, of how rings and square-zero ideals come into play. The full definition is then just an expansion to all “differential like data”. In other words, each ring $A$ and square-zero ideal $I$, is some analogue of $k[\varepsilon]$ and $(\varepsilon)$. In other words, it’s an affine scheme $\text{Spec}(A)$ with some “fuzz” (representing differential data) and the underlying space $\text{Spec}(A/I)$ with the fuzz removed. It’s not shocking that in the total generality of schemes (to which the notion of unramified applies), we need much stronger conditions than just the injectivity when $A=k[\varepsilon]$ and $I=(\varepsilon)$.

Note that these other “differential like data” don’t just consist of ‘strange’ affine schemes with fuzz, but also consist of examples in the same vein of $k[\varepsilon]$ but with higher order data–think about $A=k[\varepsilon^n]$ and $I=(\varepsilon^{n-1})$. Thus, if this makes sense to you, this injectivity condition is not only keeping track of injectivity of tangent spaces, but injectivity of jet spaces.

## Vanishing Relative Cotangent Sheaf

Another avenue of attack, is through the relative cotangent sheaf $\Omega_{X/Y}:=\Omega^1_{X/Y}$.

Recall that, intuitively, the relevant cotangent sheaf is the analogy of the “relative cotangent bundle”, or the cotangent space of $X$ “modulo” that of $Y$. So, at a point $p\in X$, we can imagine $(\Omega_{X/Y})_p$ being non-zero as saying, roughly, that the space of “tangent vectors” or “directions” at $p$ is more than the corresponding set of “tangent vectors” or “directions” at $f(p)$ since, again, $(\Omega_{X/Y})_p$ is the “quotient” of these.

Of course, this then allows us to use our intuition from the “unicity of tangent vector lifts” to tell that non-vanishing of $\Omega_{X/Y}$ at a point should indicate ramification. So, to guarantee unramifiedness we should specify that $(\Omega_{X/Y})_p=0$ for all $p\in X$ or, equivalently, that $\Omega_{X/Y}=0$.

## Ideals Lifting Correctly

The last intuitive condition for unramifiedness, takes a heavy hint from its complex geometric cousin. Namely, let’s recall that if $f:X\to Y$ is a non-constant map of compact Riemann surfaces, then we define the ramification index of a point $p\in X$ as follows. There exists a unique integer $e_p\geqslant 1$ such that, up to a change of coordinates, $f$ looks like $z\mapsto z^{e_p}$ near $p$. Intuitively, this says that near $p$ (but not at $p$), $f$ always looks like a $e_p$-to-$1$ cover of its image. In particular, $f$ is never a local biholomorphism at $p$ unless $e_p=1$. It was then the maps of Riemann surfaces $f:X\to Y$ for which $e_p=1$ for all $p\in X$ which we called unramified. They were, as alluded to above, precisely the local biholomorphisms (holomorphic covering maps, in fact).

Now, it seems impossible to rephrase this notion of unramifiedness, from the geometry of compact Riemann surfaces, in the language of schemes–the above is very reliant upon analysis to make precise what “locally looks like” means. But, this is just because we’ve cast it this way. There is another, equivalent, way to phrase the notion of ramification index of a map of Riemann surfaces which is wholly amenable to algebraic recasting.

Namely, let us denote the meromorphic function fields of $X$ and $Y$ by $M(X)$ and $M(Y)$ respectively. We know that our map $f:X\to Y$ induces a morphism $f^\ast:M(Y)\to M(X)$. For our point $p\in X$ we have the subring $\mathcal{O}_p\subseteq M(X)$ consisting of those meromorphic functions which are holomorphic at $p$. If $q:=f(p)$, then we similarly have the subring $\mathcal{O}_q\subseteq M(Y)$. Note then that the mapping $M(Y)\to M(X)$ induces a mapping $\mathcal{O}_q\to\mathcal{O}_p$. Moreover, a little thought shows that $\mathfrak{m}_q\mathcal{O}_p=\mathfrak{m}_p^{e_p}$. Thus, we see that the non-local injectivity of a map of Riemann surfaces can be measured by the inability of $\mathfrak{m}_q$ to generate all of $\mathfrak{m}_p$. In particular, we would get that $f:X\to Y$ is unramified at $p$ if and only if $\mathfrak{m}_q\mathcal{O}_p=\mathfrak{m}_p$.

This is precisely one of the ways to define unramifiedness in the context of schemes. Namely, given a morphism $X\to Y$, if $x$ is to be unramified, we want $\mathfrak{m}_{f(x)}$ to be able to take up all of $\mathcal{O}_{X,x}$ just as in the above. The only difference is one of geometric vs. arithmetic. Namely, in the case of Riemann surfaces, all of our points are visible (i.e. all our points are closed points). In the case of schemes, this is not the case. There may exist “hidden” points, that come from things like base changing our scheme to extensions of the base field, if we were dealing with varieties. These hidden points always manifest themselves in the residue field. And, since unramifiedness is a geometric notion, and “sees all the points” (e.g. if we were on a variety it can’t distinguish between the variety and its base change to the algebraic closure of the base field) we need to take these “hidden” points into account. For more intuition about “hidden” points, see this math.se answer of mine.

So, how do we measure “unramifiedness” for the extension of residue fields? Namely, if $f:X\to Y$ is a morphism of schemes, with $y=f(x)$, then how does one attempt to measure ramification for the extension $k(x)/k(y)$? For clarity, let’s assume that the extension is finite (we’ll see below that finiteness is, in fact, forced). Now, if $[k(x):k(y)]=n$ then this should mean that, in some sense, there are $n$ “points” of $k(x)$ lying above the one point of $k(y)$. But, how can we make rigorous sense of this? Taking a cue from the previous paragraph, to make these hidden points become seen, we need to move to the geometric setting. Or, in other words, we need to base change our situation to $\overline{k(y)}$ where all points become visible.

But, the base change of $k(x)$ to $\overline{k(y)}$ is merely $k(x)\otimes_{k(y)}\overline{k(y)}$. So, we would want to say that $k(x)/k(y)$ is unramified if and only if $k(x)\otimes_{k(y)}\overline{k(y)}$ has $n$ points. But, this happens precisely when $k(x)/k(y)$ is separable. Thus, we see that the right notion of unramifiedness for extensions of fields is separability.

So, finally, using our intuition from Riemann surfaces, appropriately adapted to the general, possibly arithmetic, world of schemes, we see that we should define $f:X\to Y$ to be unramified at $x$ if $\mathfrak{m}_{f(x)}\mathcal{O}_{X,x}=\mathfrak{m}_x$ (it is unramified at the visible points) and $k(x)/k(y)$ is separable (it is unramified at the “hidden” points).

# Unramified Morphisms

## Definition ands Basic Theorems

Let us now show that, when defined correctly, the three notions above, which are conducive to definitions pointwise (all but unicity of tangent vector lifts), coincide:

Theorem 1: Let $f:X\to Y$ be locally of finite type and let $x\in X$. Then, the following are equivalent:

1. If $y=f(x)$, then $\mathfrak{m}_y\mathcal{O}_{X,x}=\mathfrak{m}_x$, and $k(x)/k(y)$ is a finite separable extension.
2. The equality $(\Omega_{X/Y})_x=0$ holds.
3. There is a neighborhood $U$ of $x$ for which the restriction of the diagonal map $\Delta:X\to X\times_Y X$ is an open immersion.

Proof: Let’s prove that 1. implies 2. Since this is clearly a local condition, let us assume that $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$, with $x=\mathfrak{q}$ and $y=\mathfrak{p}$. By assumption, we then have that $\mathfrak{p}B_\mathfrak{q}=\mathfrak{q}B_\mathfrak{q}$. Now, let us then note that

$B_\mathfrak{q}\otimes_{A_\mathfrak{p}} k(\mathfrak{p})=B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q}=B_\mathfrak{q}/\mathfrak{q}B_\mathfrak{q}=k(\mathfrak{q})$

Thus, we see that the diagram

$\begin{matrix}\text{Spec}(k(\mathfrak{q})) & \rightarrow & \text{Spec}(B_\mathfrak{q})\\ \downarrow & & \downarrow\\ \text{Spec}(k(\mathfrak{p})) & \rightarrow & \text{Spec}(A_\mathfrak{p})\end{matrix}$

is fibered. Thus, we know (Hartshorne Proposition II.8.2.A) that

$\Omega_{B_\mathfrak{q}/A_\mathfrak{p}}\otimes_{B_\mathfrak{q}}k(\mathfrak{q})=\Omega_{k(\mathfrak{q})/k(\mathfrak{p})}$

Now, since $k(\mathfrak{q})/k(\mathfrak{p})$ is finite separable, we know that $\Omega_{k(\mathfrak{q})/k(\mathfrak{p})}=0$ (see discussion in “Geometric Examples” for proof). Since $\Omega_{B_\mathfrak{q}/A\mathfrak{p}}$ is a finitely generated $B_\mathfrak{q}$-module (because we assumed locally of finite type) we can conclude from Nakayama’s lemma that $\Omega_{B_\mathfrak{q}/A_\mathfrak{p}}=0$. Next, we note that $\Omega_{B_\mathfrak{q}/A_\mathfrak{p}}=\Omega_{B_\mathfrak{q}/A}$. This is because we have the set of ring maps $A\to A_\mathfrak{p}\to B_\mathfrak{q}$ which (by Hartshorne II.8.3.A) gives us a short exact sequence

$\Omega_{A_\mathfrak{p}/A}\otimes_{A_\mathfrak{p}}B_\mathfrak{q}\to \Omega_{B_\mathfrak{q}/A}\to \Omega_{B\mathfrak{q}/A_\mathfrak{p}}\to 0$

which implies our desired inequality since

$\Omega_{A_\mathfrak{p}/A}=(\Omega_{A/A})_\mathfrak{p}=0$

Finally, we see that $\Omega_{B_\mathfrak{q}/A}=(\Omega_{B/A})_\mathfrak{q}$. Thus, $(\Omega_{B/A})_\mathfrak{q}=0$ as desired.

To prove that 2. implies 3., we may, once again, assume that we’re affine, and that $f$ is of finite type. Say, $X=\text{Spec}(B)$ and $Y=\text{Spec}(A)$. Now, since we’ve restricted to the affine case, we know that the diagonal map $\Delta:X\to X\times_Y X$ is a closed embedding, corresponding to the usual (surjective!) multiplication map $\mu:B\otimes_A B\to B$. Note then that the ideal sheaf corresponding to the closed subset $\Delta(X)$ (with the reduced structure) is $I:=\ker\mu$. Now, since $\Omega_{B/A}=\widetilde{I/I^2}$ (Hartshorne 8.1.A), we know, by assumption, that

$0=(\Omega_{B/A})_\mathfrak{p}=(I/I^2)_{\Delta(\mathfrak{p})}=I_{\Delta(\mathfrak{p})}/I^2_{\Delta(\mathfrak{p})}$

Now, since $f$ is of finite type, it’s easy to see that $I_{\Delta(\mathfrak{p})}$ is a finitely generated $(B\otimes_A B)_{\Delta(\mathfrak{p})}$-module. Thus, by Nakayama’s lemma we may conclude that $I_{\Delta(\mathfrak{p})}=0$. But, then, by geometric Nakayama’s lemma (see Vakil exercise 13.7.E) we may conclude that $I$ vanishes in a neighborhood $U$ of $\Delta(x)$. Then, by definition, the map $\Delta\mid_{\Delta^{-1}(U)}:\Delta^{-1}(U)\to U$ is an isomorphism. Thus, $\Delta$ is an open immersion in a neighborhood of $x$ as desired.

Finally, to prove that 3. implies 1., we make a set of reductions. As always, we may assume that $X=\text{Spec}(B)$, $Y=\text{Spec}(A)$, $x=\mathfrak{q}$, and $y=\mathfrak{p}$, and by shrinking small enough we may assume that $\Delta_{X/Y}$ is an open immersion. Let us first claim that we can base change to the fiber over $y$. Namely, consider the following fibered diagram

$\begin{matrix}\text{Spec}(B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p}) & \to & \text{Spec}(B)\\ \downarrow & & \downarrow\\ \text{Spec}(k(\mathfrak{p})) & \to & \text{Spec}(A)\end{matrix}$

For notational convenience, let us denote ideals of $B_{\mathfrak{p}}/\mathfrak{p}B_\mathfrak{p}$ by $\overline{I}$, where $I$ is an ideal of $B_\mathfrak{p}$. Note then that if $\overline{\mathfrak{q}B_\mathfrak{p}}$ is the point corresponding to $x$, then the localization of $B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p}$ at this prime is $B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q}$, and the residue field at this point is $(B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q})/(\mathfrak{q}B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q})\cong k(\mathfrak{q})$. Thus, the property of 1. holding for $\text{Spec}(B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p})$ for the point $\overline{\mathfrak{q}B_\mathfrak{p}}$ (in the domain) and $(0)$ (in the codomain) are equivalent to

$0=(0)B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q}=\mathfrak{q}B_\mathfrak{q}/\mathfrak{p}B_\mathfrak{q}$

and that $k(\overline{\mathfrak{q}B_\mathfrak{p}})=k(\mathfrak{q})$ is a finite separable extension of $k(\mathfrak{p})$. Since the centered equation is equivalent to $\mathfrak{p}B_\mathfrak{q}=\mathfrak{q}B_\mathfrak{q}$, we may thus conclude that the property of 1. holding for $\text{Spec}(B_\mathfrak{p}/\mathfrak{p}B_\mathfrak{p})$ for the point $\overline{\mathfrak{q}B_\mathfrak{p}}$ and $(0)$ is equivalent to 1. holding for $X\to Y$ with $x$ and $y$.

So, since base change of open immersions are open immersions (and the base change of the diagonal is the diagonal of the base change), we can assume, without loss of generality, that $X=\text{Spec}(B)$, $Y=\text{Spec}(k)$ (for some field $k$).

To prove that 3. implies 1. in this case, let’s first prove it in the even more specific case when $k=\overline{k}$. Let us first show that $X$ must be finite. Since $X$ is quasi-compact, it suffices to show that it’s closed points are discrete. Indeed, this will then imply (from basic algebra) that $B$ is Artinian, and so has finite spectrum. So, take any closed point $p\in X$, and consider the associated morphism $s:\text{Spec}(k)\to X$ (here we are using that $k=\overline{k}$ so that all closed points are $k$-points). Now, we have the following fibered diagram

$\begin{matrix}\text{Spec}(k) & \to & \text{Spec}(k)\times_{\text{Spec}(k)} X\\ \downarrow & & \downarrow\\ X & \to & X\times_{\text{Spec}(k)} X\end{matrix}$

where $\text{Spec}(k)\to\text{Spec}(k)\times_{\text{Spec}(k)} X$ is the identity on the first entry, and $s$ on the second (this makes clear why we needed that all closed points are $k$-rational). But, there is a natural isomorphism $\text{Spec}(k)\times_{\text{Spec}(k)}X\to X$, and so the composition

$\text{Spec}(k)\to\text{Spec}(k)\times_{\text{Spec}(k)}X\to X$

is open. But, the composition is just $s$. Thus, $p$ is an open point. Thus, from the above (and the fact that $B/k$ is finite type) we may conclude that $B$ is Artinian, and so $X$ is finite. But, we’ve also shown that for each point $p$ we have an isomorphism $k \to \mathcal{O}_{X,p}$ which implies, in particular that 1. holds in this setup.

Finally, let us deduce the result for a general $k$. Now, we have a surjection $X_{\overline{k}}\to X$ (where $X_{\overline{k}}$ is base change to $\text{Spec}(\overline{k})$) and since base change of open immersions are open immersions (and base change of the diagonal is the diagonal of the base change) we know from the previous paragraph that $X_{\overline{k}}$ is finite. Thus, so is $X$, and so $B$ is Artinian. Since this problem is purely local, we may as well now work with $\text{Spec}(\mathcal{O}_{X,x})$ which is an open subscheme of $X$. Now, we know from the previous paragraph that $\mathcal{O}_{X,x}\otimes_k \overline{k}$ is $\overline{k}^m$, for some $m$, and thus $\mathcal{O}_{X,x}$ is reduced. We claim that this implies that $\mathcal{O}_{X,x}$ is a finite separable field extension of $k$. It’s finite since

$\dim_k\mathcal{O}_{X,x}=\dim_{\overline{k}}(\mathcal{O}_{X,x}\otimes_k \overline{k})=m$

It’s reduced since $\mathcal{O}_{X,x}$ embeds into $\mathcal{O}_{X,x}\otimes_k\overline{k}$ which is reduced. This implies it’s a field since it’s then a reduced local Artinian ring. Finally, it’s separable since it’s base change to $\overline{k}$ is reduced which is a common equivalent definition of separability. $\blacksquare$

Thus, we may unabashedly define a morphism $f:X\to Y$, locally of finite presentation, to be unramified at $x$ if any of the equivalent properties in Theorem 1 hold. We then define $f$ to be unramified if it is locally of finite presentation and it is unramified at all $x\in X$. If $f$ is not unramified at a point $x\in X$, we say that $f$ is ramified there.

One important thing to note, is that implicit in our proof of Theorem 1 (namely in 3. implies 1.) we’ve shown that

Corollary 2: If $f:X\to Y$ is unramified, then it is locally quasifinite.

This intuitively makes sense, insomuch as the obvious case of non-quasifiniteness is ramified. In particular, when one things of a point for which has no neighborhood on which $f$ is quasifinite, one probably thinks of a point where infinitely many sheets (maybe so many that they can’t be ‘discretely’ distinguished!) come together. This is clearly not a picture of an unramified morphism.

Just as in the case of flatness, unramifiedness is not necessarily a black-and-white notion. Some points of $f$ can be unramified, and others can be ramified. In particular, we can define the unramified locus of a, locally of finite presentation, morphism $f:X\to Y$ to be the set of $x\in X$ for which $f$ is unramified at $x$. Similarly, we define the ramification locus of $f$ to be the points of $X$ for which $f$ is ramified there.

Note that the unramified locus is simply the complement of the support of $\Omega_{X/Y}$ (as a sheaf). Said differently, the ramification locus of $f$ is the support of $\Omega_{X/Y}$. In particular, we see that $f$ is unramified if and only if the unramified locus is all of $X$, if and only if $\text{Supp}(\Omega_{X/Y})=0$. But, this is equivalent to $\Omega_{X/Y}=0$. Thus, we see that unramifiedness, for a morphism locally of finite presentation, is equivalent to $\Omega_{X/Y}$ being the zero sheaf.

Also, this point of view allows us to see that if $X$ is locally Noetherian, then $\text{Supp}(\Omega_{X/Y})$, being the support of a coherent sheaf on a locally Noetherian scheme, is closed. Thus, we have the following:

Theorem 3: If $f:X\to Y$ is unramified, and $X$ is locally Noetherian, then the unramified locus of $f$ is open.

## Geometric Examples

Let us now do some examples, to show that our intuition about unramified morphisms developed in the motivation, does translate well into this formal setting.

Let us first consider the natural example of a map $\mathbb{A}^n_k\to\mathbb{A}^m_k$, where $k$ is a field, and $n> m$. Now, intuitively we would expect this to be unramified. For example. it’s somewhat obvious that there is not unicity of tangent vector lifts–this isn’t a “differential immersion” (it is a differential submersion, which will be validated when we discuss smooth morphisms). To prove this though, we need to compute $\Omega_{\mathbb{A}^n/\mathbb{A}^m}=\widetilde{\Omega_{A/B}}$ where $B=k[x_1,\ldots,x_n]$ and $A=k[x_1,\ldots,x_m]$. Since $B$ is a free $A$-algebra of rank $n-m$, it follows from the basic theory of differentials that $\Omega_{B/A}$ is a free $B$-module of rank $n-m$. In particular, we see that not only is $\mathbb{A}^n_k\to\mathbb{A}^m_k$ not unramified (i.e. $\Omega_{X/Y}=0$), but that it’s ramification locus is empty–it is everywhere ramified. This matches our intuitive picture.

Let’s now consider another classic geometric example. In particular, let’s consider the normalization $\mathbb{A}^1_k\to C$, where $C$ is the cuspidal cubic curve $C=\text{Spec}(k[x,y]/(y^2-x^3))=\text{Spec}(k[t^2,t^3])$. For the sake of convenience, we assume that $\text{char}(k)\ne 2$.

This normalization map corresponds to the $k$-algebra map coming from the inclusion mapping $A=[t^2,t^3]\to k[t]=B$. To find the ramification locus we need to compute $\Omega_{\mathbb{A}^1/C}=\widetilde{\Omega_{B/A}}$. To do this, we note that $B=A[x]/(x^2-t^2)$. Thus, from the basic theory of differentials, we know that $\Omega_{B/A}$ is the $B$-module generated by the symbol $dx$, and subject to the relation

$0=d(x^2-t^2)=d(x^2)-d(t^2)=2dx-0=2dx$

Thus, $\Omega_{B/A}$ is isomorphic, as a $B$-module, to $Bdx/B(2dx)$ which, is isomorphic to

$A[x]/(2x,x^2-t^2)=k[t]/(2t)=k[t]/(t)$

So, we see that the ramification locus of the normalization, $\text{Supp}(\Omega_{X/Y})$, which is just the support of $k[t]/(t)$ as a $k[t]$-module, is just $(t)$. Thus, we see that the unramified locus of the normalization is $\mathbb{A}^1_k-\{(t)\}$. So, the only point of ramification is the origin. This is not unexpected, since we know that the normalization is an isomorphism off of the cusp, and there the picture “looks ramified”, since the cusp has “several directions” in which to lift tangent vectors.

Let us compute another geometric example. Consider the squaring map $\mathbb{A}^1_k\to\mathbb{A}^1_k$, where $\text{char}(k)\ne 2$. This corresponds to the inclusion $A=k[t^2]\hookrightarrow k[t]=B$. To find there this squaring map is ramified, we should, as always, compute the differentials. Namely, we want to compute $\Omega_{\mathbb{A}^1_k/\mathbb{A}^1_k}$, which is just $\widetilde{\Omega_{B/A}}$. But, just as in the last computation, we find that

$\Omega_{B/A}=\Omega_{B[x]/(x^2-t^2)/B}=Bdx/B(2dx)=k[t]/(t)$

and so the squaring map is ramified only at the origin. This is not shocking. In fact, if one thinks about it, the squaring map was the map depicted in the “Unicity of Tangent Vectors” section.

Intuitively, for every $p\in k^\times$, we have that there are two preimages of the squaring map. But, at the point $0$, there is only one preimage. Of course, this isn’t totally correct since there is no reason that $k$ must be closed under squareroots, but since ramification is an intuitively geometric notion, and shouldn’t depend on base change to field extensions, we may as well assume $k=\overline{k}$.

A good example to talk about, even though we implicitly used it in the proof of Theorem 1, is the case of $f:\text{Spec}(L)\to\text{Spec}(K)$ where $L/K$ is a finite field extension. In particular, if $L/K$ is finite and separable, then $f$ is unramified. To prove this, we must merely show that $\Omega_{L/K}=0$. But, since $L/K$ is finite separable, the primitive element theorem tells us that $L=K[x]/(p(x))$ for some irreducible, separable $p(x)\in K[x]$. So,

$\Omega_{L/K}=L dx/ L d(p(x))=Ldx/L p'(x)dx$

which, if we write $L=K[x]/(p(x))$ is the same as $K[x]/(p(x),p'(x))$. But, this is equal to zero since $p$ is separable so that $(p(x),p'(x))=K[x]$.

Of course, while all of the above examples have been affine, since ramification is a local (stalk-local!) condition, we can always reduce to the affine case. For example, we know that the structure map $\mathbb{P}^1_k\to\text{Spec}(k)$ to be ramified everywhere. Indeed, locally, this is just the $n=1$ and $m=0$ case of our first example. That being said, one can compute though that $\Omega_{\mathbb{P}^1_k/k}=\mathcal{O}(-2)$. This is just because a rational section of $\Omega_{\mathbb{P}^1_k/k}$, where $\mathbb{P}^1_k$ has coordinates $[x:y]$, is $dx$. But, at infinity this section transforms to $d(x^{-1})=x^{-2}dx$, which gives the desired equality. Then the everywhere ramifiedness is clear since $\Omega_{\mathbb{P}^1_k/k}$ is vector bundle of positive rank, and so has nowhere vanishing stalks.

Just to mention, there is not a general formula for $\Omega_{\mathbb{P}^n_k/k}$. There is just the Euler sequence which says that $\Omega_{\mathbb{P}^n_k/k}$ sits in the exact sequence of sheaves

$0\to\Omega_{\mathbb{P}^n_k/k}\to \mathcal{O}_{\mathbb{P}^n_k}^{\oplus (n+1)}\to \mathcal{O}_{\mathbb{P}^n_k}\to 0$

This tells us that $\Omega_{\mathbb{P}^n_k/k}$ is locally free of rank $n$ (which, once we talk about smooth morphisms, will be obvious). Also, taking determinants, this shows that the canonical sheaf of $\mathbb{P}^n$ is just $\mathcal{O}_{\mathbb{P}^n_k}(-n-1)$. Our computation for $n=1$ is a coincidence, since in dimension one the differentials and the cotangent sheaf coincide.

## Relation to Number Theory

Let us now address which, to the arithmetically minded, may be the elephant in the room. While we have just defined the notion of ramification for schemes, there is already a well-developed theory of ramification for Dedekind domains. In particular, for an extension of number fields $L/K$, there is the well-known notion of ramification for the associated extension of number rings $\mathcal{O}_L/\mathcal{O}_K$.

Just to recall, suppose that $R\subseteq S$ is an integral extension of Dedekind domains. Then, for any non-zero prime $\mathfrak{p}\in\text{Spec}(R)$, we can look at its extension to $S$, defined to be $\mathfrak{p}S$. Note that $\mathfrak{p}S$ is a proper ideal of $S$. Indeed, since $R\subseteq S$ is an integral extension, the Lying Over Theorem implies that there is some prime $\mathfrak{P}\in\text{Spec}(S)$ restricting to $\mathfrak{p}$. Thus, $\mathfrak{p}S\subseteq\mathfrak{P}$. Thus, since $S$ is a Dedekind domain, we can factor $\mathfrak{p}S$ as a finite product of primes

$\mathfrak{p}S=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_n^{e_n}$

One then says that $\mathfrak{p}$ is ramified if one of the $e_i$ is greater than one. Of course, we say that a prime $\mathfrak{p}$ is unramified, if it is not ramified.

The obvious question is how ramification of the ring extensions $R\subseteq S$, a number theoretic consideration, corresponds to ramification of the associated map $\text{Spec}(S)\to\text{Spec}(R)$, a geometric consideration. The first step towards understanding the connection, is the realization that the two different notions of ramification are focused on different parts of the map $\text{Spec}(S)\to\text{Spec}(R)$. Namely, ramification in the sense of algebraic geometry is a $\text{Spec}(S)$-centric notion, whereas the number theoretic idea of ramification is $\text{Spec}(R)$-centric.

With this disparity in mind, we can make clear the connection between the two uses of the word “ramified”. Namely, we claim that $R\subseteq S$ is unramified at a prime $\mathfrak{p}\in\text{Spec}(R)$ if and only if $\text{Spec}(S)\to\text{Spec}(R)$ is unramified at all primes $\mathfrak{P}\in\text{Spec}(S)$ which lie above $\mathfrak{p}$. Indeed, note that for any prime $\mathfrak{P}_i$ lying above $\mathfrak{p}$, we have that

\begin{aligned}\mathfrak{p} S_{\mathfrak{P}_i} &=(\mathfrak{p} S)_{\mathfrak{P}_i}\\ &=(\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_n^{e_n})_{\mathfrak{P}_i}\\ &= (\mathfrak{P}_1 S_{\mathfrak{P}_i})^{e_1}\cdots (\mathfrak{P}_n S_{\mathfrak{P}_i})^{e_n}\\ &= (\mathfrak{P}_i S_{\mathfrak{P}_i})^{e_i}\end{aligned}

But, $\text{Spec}(S)\to\text{Spec}(R)$ being unramified (in the algebraic geometry sense) at $\mathfrak{P}_i$, is equivalent to $\mathfrak{p}S_{\mathfrak{P}_i}=\mathfrak{P}_i S_{\mathfrak{P}_i}$. So, we see that being unramified at $\mathfrak{P}_i$ is equivalent to $e_i=1$. From this, our claim follows.

In fact, there is an even stronger connection of the algebro-geometric theory of ramification to the ramification theory in number theory. Let us assume that $L/K$ is an extension of number fields, and $\mathcal{O}_L/\mathcal{O}_K$ the associated extension of number rings. Then, we know that cotangent sheaf associated to the map $\text{Spec}(\mathcal{O}_L)\to\text{Spec}(\mathcal{O}_K)$ is just $\widetilde{\Omega_{\mathcal{O}_L/\mathcal{O}_K}}$. Now, let us consider the points of ramification, the support $\text{Supp}(\Omega_{\mathcal{O}_L/\mathcal{O}_K})$. Then, from basic commutative algebra, since $\Omega_{\mathcal{O}_L/\mathcal{O}_K}$ is a finite $\mathcal{O}_L$-module, we have the following equality

$\text{Supp}(\Omega_{\mathcal{O}_L/\mathcal{O}_K})=V(\mathfrak{d})$

where $\mathfrak{d}=\text{Ann}_{\mathcal{O}_L}(\Omega_{\mathcal{O}_L/\mathcal{O}_K})$. Now, in general, if $R$ is any Dedekind domain, and $I$ an ideal of $R$, then $V(I)$ is exactly the same as the set of primes which divide $I$. Thus, we see that $\mathfrak{d}$ is an ideal of $\mathcal{O}_L$ with the property that a prime $\mathfrak{P}\in\text{Spec}(\mathcal{O}_L)$ divides $\mathfrak{d}$ if and only if $\mathfrak{P}\in\text{Supp}(\Omega_{\mathcal{O}_L/\mathcal{O}_K})$ which is true if and only if $\mathfrak{P}$ is a ramified prime.

But, there is another ideal of $\mathcal{O}_L$, familiar to any number theory fans, which satisfies similar properties. Namely, the different (if you are unfamiliar, I suggest you read these fantastic notes of Keith Conrad) of $L/K$, denoted $\mathfrak{d}_{L/K}$, is an ideal of $\mathcal{O}_L$ which satisfies the same properties as our above $\mathfrak{d}$. Namely, that $\mathfrak{P}\mid\mathfrak{d}_{L/K}$ if and only if $\mathfrak{P}$ is a ramified prime (with respect to $L/K$). It is not as well-known as it’s brother, the discriminant, the ideal $\Delta_{L/K}\subseteq\mathcal{O}_K$ which satisfies similar properties (a prime divides it if and only if it ramifies in the sense of number theory). They are related by the ideal norm $N_{L/K}$–in particular, $\Delta_{L/K}=N_{L/K}(\mathfrak{d}_{L/K})$.

The similarities between $\mathfrak{d}$ and $\mathfrak{d}_{L/K}$ aren’t coincidental. As a matter of fact, $\mathfrak{d}=\mathfrak{d}_{L/K}$. The proof of this is not too difficult, and can be found as Proposition 14, III.7 of Serre’s Local Fields. This is nice, because it tells us that not only is our notion of ramification in algebraic geometry consistent with that of number theory, but the objects dictating ramification (cotangent sheaf vs. different/discriminant) are actually intimately related.

Conversely, this connection is also nice since it validates a common, but slightly confusing, practice in number theory. In particular, when one is usually introduced to the notion of ramification, they are often give some sort of geometric motivation. They are drawn a pretty picture like those in the motivation section of this post, and told “look, ramification!” Usually one is told that they should analogize to ramification in, say, the theory or Riemann surfaces, and that these pictures are “actually accurate”. It is never made clear, or was never made clear to me at least, in what way this random geometric picture was “accurate” (beyond the obvious similarities). The above tells us how. The pictures we drew do have logical and easy-to-see connections to our geometric definition of ramification, and the above discussion tells us that this really is what is going on in the arithmetic case. Those pictures aren’t made up, they really are describing the geometry of the situation!

## Arithmetic Examples

With the above theory, we see that giving geometrically minded, but arithmetically motivated, examples is actually entirely arithmetic–this is just plain old number theory. Regardless, we shall give two examples to illustrate the concepts.

Consider first the map $\text{Spec}(\mathbb{Z}[i])\to\text{Spec}(\mathbb{Z})$. It is a classic theorem of number theory, that the only ramified primes of $\mathbb{Z}$ (i.e. ramified in the sense of number theory) in this case are $(2)$. And, since the only prime of $\mathbb{Z}[i]$ which lies above $(2)$ is $(1+i)$, we know (from our above analysis) that the only ramified prime in this situation (in the sense of algebraic geometry) is $(1+i)$. Let us verify this by hand though, just to see that it all works out. Namely, let us being by computing $\Omega_{\mathbb{Z}[i]/\mathbb{Z}}$. To do this, we first express $\mathbb{Z}[i]$ as $\mathbb{Z}[T]/(T^2+1)$. Then, we can compute $\Omega_{\mathbb{Z}[i]/\mathbb{Z}}$ in the usual way. It is the $\mathbb{Z}[i]$-module generated by the symbol $dT$, subject to the relation that $0=d(T^2+1)=2dT$. In particular, we see that

$\Omega_{\mathbb{Z}[i]/\mathbb{Z}}\cong \mathbb{Z}[i]dT/\mathbb{Z}[i]2dT\cong \mathbb{Z}[i]/2\mathbb{Z}[i]$

From this, it’s easy to see that if $2\notin\mathfrak{P}$, for $\mathfrak{P}\in\text{Spec}(\mathbb{Z}[i])$, then $(\Omega_{\mathbb{Z}[i]/\mathbb{Z}})_\mathfrak{P}$, being just $(\mathbb{Z}[i]/2\mathbb{Z}[i])_\mathfrak{P}$, is zero. And, if $2\in\mathfrak{P}$, then $(\Omega_{\mathbb{Z}[i]/\mathbb{Z}})_\mathfrak{P}$, being just $(\mathbb{Z}[i]/2\mathbb{Z}[i])_\mathfrak{P}$, is non-zero since $2$ is not inverted. Thus, we see that the only ramified primes are those containing $2$. Or, in other words, the only ramified prime is $(1+i)$. Thus, we arrive at the same conclusion as in number theory (like we knew we must).

Consider now the extension of number field $L/K$, where $L=\mathbb{Q}(\sqrt{-5},\sqrt{3})$ and $K=\mathbb{Q}(\sqrt{-15})$. The number theoretic amongst us might recognize $L$ as the Hilbert class field of $K$. Thus, in particular, we know that $\mathcal{O}_L/\mathcal{O}_K$ is unramified. So, we know that

$X=\text{Spec}(\mathcal{O}_L)\to\text{Spec}(\mathcal{O}_K)=Y$

must be unramified, and so

$0=\Omega_{X/Y}=\widetilde{\Omega_{\mathcal{O}_L/\mathcal{O}_K}}$

from which we may conclude that $\Omega_{\mathcal{O}_L/\mathcal{O}_K}=0$. Of course, if we really wanted to get our hands really dirty, we could work the other way. Namely, to show that $\mathcal{O}_L/\mathcal{O}_K$ is unramified, we could show that $\Omega_{\mathcal{O}_L/\mathcal{O}_K}=0$. This would be computationally intensive though. Indeed, we’d have to find a presentation for $\mathcal{O}_L$ as a $\mathcal{O}_K$-algebra, and then proceed our normal way. This isn’t totally infeasible, with computer assistance from something like SAGE, but is not particularly enlightening in this instance.

## Properties of Unramified Morphisms

Let us now discuss some of the properties that hold true for unramified morphisms.

First and foremost the permanence properties of unramified morphisms

Theorem 4: The class of unramified morphisms is closed under composition and base-change.

Proof: Suppose that $f:X\to Y$ and $g:Y\to Z$ is unramified. Let us show that $g\circ f:X\to Z$ is unramified. To see this, let $x\in X$ be arbitrary. We then want to show that $\mathfrak{m}_z\mathcal{O}_{X,x}=\mathfrak{m}_x$, and $k(x)/k(z)$ is finite separable. But,

$\mathfrak{m}_z\mathcal{O}_{X,x}=(\mathfrak{m}_z\mathcal{O}_{Y,y})\mathcal{O}_{X,x}=\mathfrak{m}_y\mathcal{O}_{X,x}=\mathfrak{m}_x$

and since $k(x)/k(z)$ has the subextension $k(x)/k(y)$, and both $k(x)/k(y)$ and $k(y)/k(z)$ are finite separable, so is $k(x)/k(z)$.

Suppose now that we have a fibered diagram of the following form

$\begin{matrix}P & \xrightarrow{p} & X\\ \downarrow & & \downarrow\\ Y & \to & S\end{matrix}$

where $X\to S$ is unramified. We want to show that $P\to Y$ is unramified. But, from basic theory of differentials we know that $\Omega_{P/Y}=p^\ast(\Omega_{X/S})$. In particular, since $p^\ast$ is additive, and $\Omega_{X/S}=0$, we know that $\Omega_{P/Y}=0$ from which the conclusion follows. $\blacksquare$

Now, while we gave many examples of ramified and unramified morphisms above, we have yet to discuss an entire class of morphisms which belong to either camp. To this end, we have the following:

Theorem 5: Every monomorphism, locally of finite presentation, is unramified.

Proof: Let $f:X\to Y$ be a monomorphism locally of finite presentation. Then, it suffices to show that the diagonal morphism of $f$ is an open embedding. But, since $f$ is a monomorphism, we know that the diagonal map is actually an isomorphism. $\blacksquare$

Corollary 6: Every locally closed embedding is unramified.

This allows us to prove that the class of unramified morphisms has the so-called “cancellation property”. Namely:

Theorem 7: If $f:X\to Y$ and $g:Y\to Z$ be morphisms, and $g\circ f$ is unramified, then $f$ is unramified.

Proof: We may apply the Cancellation Theorem (cf. Vakil’s Fundamentals of Algebraic Geometry 10.1.19), since $\Delta_g$, being locally closed, is unramified. $\blacksquare$

We also have somewhat of a converse to “base change of unramified morphisms are unramified”, if we assume the extra condition of faithful flatness:

Theorem 8: Suppose that the following diagram is fibered:

$\begin{matrix} P & \xrightarrow{p} & X\\ ^{g}\downarrow & & \downarrow ^{f}\\ Y & \xrightarrow{q} & S\end{matrix}$

$f$ is locally of finite presentation, and $p$ faithfully flat. Then, if $g$ is unramified, then so is $f$.

Proof: Since $f$ is locally of finite presentation, it suffices to check that $\Omega_{X/S}=0$. But, s

$0=\Omega_{P/Y}=p^\ast(\Omega_{X/S})$

Since $p$ is faithfully flat, this implies that $\Omega_{X/S}=0$ as desired. $\blacksquare$

In particular, this tells us that unramifiedness really is a “geometric” notion, as hinted at in the motivation. Namely, let’s suppose that $X$ and $Y$ are $k$-schemes. Then, the following diagram is fibered

$\begin{matrix}X_{\overline{k}} & \to & X\\ \downarrow & & \downarrow\\ Y_{\overline{k}} & \to & Y\end{matrix}$

and, since $Y_{\overline{k}}\to Y$ is faithfully flat (being the base change of the faithfully flat map $\text{Spec}(\overline{k})\to\text{Spec}(k)$) Theorem 8 tells us that $X_{\overline{k}}\to Y_{\overline{k}}$ is unramified if and only if $X\to Y$ is.

Finally, we prove a theorem which says that it’s hard to be a section of an unramified morphism. Indeed, intuitively, from our pictures, if you’re a section, then by continuity you should stay within one of the sheets. So, intuitively, sections of “nice” unramified morphisms should be just picking out one of (or a subset of) the sheets. This is what the following proposition says:

Theorem 9:  Let $f:X\to Y$ be unramified and separated. Then, the sections of $f$ are open embeddings.

We can’t prove this quite yet–we need first to talk about etale morphisms, and in particular characterize open embeddings as radiciel etale morphisms.

## Lifting Lemma: Unicity of Tangent Vectors

Let us now discuss the final rigorous definition of unramifiedness, and relate it back to the above definitions. As the header suggests, this is the unicity of tangent vector lifts. More rigorously, let $f:X\to Y$ be any morphism. We say that $f$ is formally unramified if for all pairs $(A,I)$, where $A$ is a ring, and $I$ is a square-zero ideal (i.e. $I^2=0$), and for any choice of a $Y$-structure $\text{Spec}(A)\to Y$ for $\text{Spec}(A)$, the following map is injective

$\text{Hom}_Y(\text{Spec}(A),X)\to\text{Hom}_Y(\text{Spec}(A/I),X)$

where, of course, $\text{Spec}(A/I)$ is given the structure of a $Y$-scheme via the natural composition $\text{Spec}(A/I)\to\text{Spec}(A)\to Y$.

Our formal claim then becomes the following:

Theorem 10: Let $f:X\to Y$ be locally of finite presentation. Then, $f$ is unramified if and only if it is formally unramified.

Before we prove this though, we need one quick, but important lemma:

Lemma 11: Let $B$ be a ring, and $A$ and $C$ two $B$-algebras. Let $I$ be a square-zero ideal of $A$ and $p$ the projection map $A\to A/I$. Suppose that $\phi:C\to A$ is a $B$-algebra map. Then there exists a bijective correspondence between the set

$\{\psi:C\to A\mid \psi\text{ is a }B\text{-morphism satisfying }p\circ\phi=p\circ\psi\}$

and $\text{Hom}_{A/I}(\Omega_{C/B}\otimes_C (A/I),I)$, where $A/I$ is regarded as a $C$-algebra via the composition $p\circ\phi$.

This is a somewhat nebulous lemma, but is quite familiar to those who have studied some amount of deformation theory. Good intuition for this lemma, and a proof, can be found in Hartshorne’s Deformation Theory–the actual theorem is Lemma 4.5.

Proof: Let’s first suppose that $f$ is unramified. Let $A$ be a ring, and $I$ a square-zero ideal. Suppose further that we have given $\text{Spec}(A)$ the structure of a $Y$-scheme. We want to show that the natural map

$\text{Hom}_Y(\text{Spec}(A),X)\to\text{Hom}_Y(\text{Spec}(A/I),X)$

is injective. But,

$\text{Hom}_Y(\text{Spec}(A),X)=\text{Hom}_A (\text{Spec}(A),X\times_Y \text{Spec}(A))$

and

$\text{Hom}_A(\text{Spec}(A),X\times_Y \text{Spec}(A)$

Thus, by base-change, we may assume that $Y=\text{Spec}(A)$. Or, in other words, we need to show that if $A$ is a ring and $I$ is any square-zero ideal, then

$\text{Hom}_A(\text{Spec}(A),X)\to\text{Hom}_A(\text{Spec}(A/I),X)$

is injective. But, now any element $\varphi$ of $\text{Hom}_A(\text{Spec}(A),X)$ is, by definition, a section of the unramified map $f:X\to\text{Spec}(A)$. But, by Theorem 9, we know then that $\varphi$ is an open embedding. In particular, $\varphi$ is determined by its image. But, if $\varphi,\psi\in\text{Hom}_A(\text{Spec}(A),X)$ are mapped to the same element of $\text{Hom}_A(\text{Spec}(A/I),X)$ then they have the same images, since $\text{Spec}(A/I)\to\text{Spec}(A)$ is surjective. From previous comment, it follows that $\varphi=\psi$.

Conversely, let’s suppose that we have unicity of tangent vector lifts. Since unramifiedness is a local condition, we may as well assume that $X$ and $Y$ are affine, say $X=\text{Spec}(C)$ and $Y=\text{Spec}(B)$. We must show that $\Omega_{C/B}=0$. Now, note that under the assumptions of the unicity of tangent vector lifts, we have that $C$ and $A$ are both $B$-algebras. Now, if there is a $B$-algebra homomorphism $\phi:C\to A$ then, note that

$\{\psi:C\to A\mid \psi\text{ is a }B\text{-morphism satisfying }p\circ\phi=p\circ\psi\}$

is nothing more than the preimage of $p\circ\phi$ under

$\text{Hom}_\text{Spec}(B)(\text{Spec}(A),\text{Spec}(C))\to\text{Hom}_\text{Spec}(B)(\text{Spec}(A/I),\text{Spec}(C))$

Indeed, “un-affinizing” this, we see that this is merely the (natural) map of sets

$\text{Hom}_B(C,A)\to\text{Hom}_B(C,A/I)$

Then, $p\circ\phi\in \text{Hom}_B(C,A/I)$, and $\psi\in\text{Hom}_B(C,A)$ is in the preimage of of $p\circ\phi$ if and only if $\psi$ is a $B$-algebra homomorphism $C\to A$ such that $p\circ\psi=p\circ\phi$.

Thus, we see that for any $B$-algebra map $\phi:C\to A$, we have, by the injectivity of

$\text{Hom}_\text{Spec}(B)(\text{Spec}(A),\text{Spec}(C))\to\text{Hom}_\text{Spec}(B)(\text{Spec}(A/I),\text{Spec}(C))$

that

$\{\psi:C\to A\mid \psi\text{ is a }B\text{-morphism satisfying }p\circ\phi=p\circ\psi\}$

is a singleton. But, by the lemma, this implies that

$\text{Hom}_{A/I}(\Omega_{C/B}\otimes_C (A/I),I)$

is a singleton, and so must be the zero module.

Now, the goal is to cleverly choose $A$ such that the zeroness of this Hom-module implies $\Omega_{C/B}$. But, we know that $\Omega_{C/B}=0$ if and only if $\text{Hom}_C(\Omega_{C/B},\Omega_{C/B})=0$ (where this is Hom of $C$-modules). But, $\Omega_{C/B}$ is merely $J/J^2$, where $J=\ker(\mu)$ where $\mu$ is the multiplication map $C\otimes_B C\to C$. Thus, we see that if we let $A=C\otimes_B C/J^2$, and $I=J/J^2$, then by the tensor-hom adjunction:

$\text{Hom}_C(\Omega_{C/B},\Omega_{C/B})=\text{Hom}_C(\Omega_{C/B},I)=\text{Hom}_{A/I}(\Omega_{C/B}\otimes_C (A/I),I)$

But, note that $C$ and $A$ are both $B$-algebras, $I^2=0$, and $\phi:C\to A$ defined by the natural composition $C\to C\otimes_B C\to C\otimes_B C/J^2$ (where we can map into either side of the tensor in the first map) is a $B$-algebra map. Thus, from the above discussion, we may conclude that $\Omega_{C/B}=0$ as desired. $\blacksquare$

1. It is easy to figure out examples of schemes X,S and an algebra A over S with a square zero ideal I such that X/S is ramified and the lifting lemma fails for this particular choice. For instance take the squaring map on $\mathbb A^1$ and A = k[e] with $e^2 = 0$.

There is a similar lifting characterization of (formal) smoothness where we want the induced map on Hom spaces to be surjective instead of injective. Do you know an example of a non smooth scheme X/S such that one can explicitly demonstrate the failure of the lifting lemma?

I tried the nodal/cuspidal curve over k and over $\mathbb A^1$ but I wasn’t able to find an A that worked…

1. Perhaps I should add that I do know an example which works: the squaring map. It looks like the tangent spaces at 0 gets mapped all to 0 (which one should expect from non injectivity anyway…). I would still like to understand the case of a cusp or a node over the field…

1. Hey Asvin,

I haven’t verified it, but I have a hunch that $\text{Spec}(k[s,t]/(s^2 t^2))$ might work.

2. Thanks for the suggestion Alex. The algebra seems quite difficult but I am really after intuition anyway, could you explain what intuition guided you towards that example?

2. Hi Alex, I found out that simply taking $A = k[\epsilon]/(\epsilon^3)$ and $A/I = k[\epsilon]/(\epsilon^2)$ works in the cuspidal case, thanks for the help anyway!