Example 50.19.1. Here is an example where we do not have a trace map on de Rham complexes. For example, consider the $\mathbf{C}$-algebra $B = \mathbf{C}[x, y]$ with action of $G = \{ \pm 1\} $ given by $x \mapsto -x$ and $y \mapsto -y$. The invariants $A = B^ G$ form a normal domain of finite type over $\mathbf{C}$ generated by $x^2, xy, y^2$. We claim that for the inclusion $A \subset B$ there is no reasonable trace map $\Omega _{B/\mathbf{C}} \to \Omega _{A/\mathbf{C}}$ on $1$-forms. Namely, consider the element $\omega = x \text{d} y \in \Omega _{B/\mathbf{C}}$. Since $\omega $ is invariant under the action of $G$ if a “reasonable” trace map exists, then $2\omega $ should be in the image of $\Omega _{A/\mathbf{C}} \to \Omega _{B/\mathbf{C}}$. This is not the case: there is no way to write $2\omega $ as a linear combination of $\text{d}(x^2)$, $\text{d}(xy)$, and $\text{d}(y^2)$ even with coefficients in $B$. This example contradicts the main theorem in [Zannier].

## 50.19 Trace maps on de Rham complexes

A reference for some of the material in this section is [Garel]. Let $S$ be a scheme. Let $f : Y \to X$ be a finite locally free morphism of schemes over $S$. Then there is a trace map $\text{Trace}_ f : f_*\mathcal{O}_ Y \to \mathcal{O}_ X$, see Discriminants, Section 49.3. In this situation a trace map on de Rham complexes is a map of complexes

such that $\Theta _{Y/X}$ is equal to $\text{Trace}_ f$ in degree $0$ and satisfies

for local sections $\omega $ of $\Omega ^\bullet _{X/S}$ and $\eta $ of $f_*\Omega ^\bullet _{Y/S}$. It is not clear to us whether such a trace map $\Theta _{Y/X}$ exists for every finite locally free morphism $Y \to X$; please email stacks.project@gmail.com if you have a counterexample or a proof.

Lemma 50.19.2. There exists a unique rule that to every finite syntomic morphism of schemes $f : Y \to X$ assigns $\mathcal{O}_ X$-module maps

satisfying the following properties

the composition with $\Omega ^ p_{X/\mathbf{Z}} \otimes _{\mathcal{O}_ X} f_*\mathcal{O}_ Y \to f_*\Omega ^ p_{Y/\mathbf{Z}}$ is equal to $\text{id} \otimes \text{Trace}_ f$ where $\text{Trace}_ f : f_*\mathcal{O}_ Y \to \mathcal{O}_ X$ is the map from Discriminants, Section 49.3,

the rule is compatible with base change.

**Proof.**
First, assume that $X$ is locally Noetherian. By Lemma 50.18.3 we have a canonical map

By Discriminants, Proposition 49.13.2 we have a canonical isomorphism

mapping $\delta (\mathop{N\! L}\nolimits _{Y/X})$ to $\tau _{Y/X}$. Combined these maps give

By Discriminants, Section 49.5 this is the same thing as a map

Recall that the relationship between $c^ p_{Y/X} \otimes c_{Y/X}$ and $\Theta _{Y/X}^ p$ uses the evaluation map $f_*\omega _{Y/X} \to \mathcal{O}_ X$ which sends $\tau _{Y/X}$ to $\text{Trace}_ f(1)$, see Discriminants, Section 49.5. Hence property (1) holds. Property (2) holds for base changes by $X' \to X$ with $X'$ locally Noetherian because both $c^ p_{Y/X}$ and $c_{Y/X}$ are compatible with such base changes. For $f : Y \to X$ finite syntomic and $X$ locally Noetherian, we will continue to denote $\Theta ^ p_{Y/X}$ the solution we've just found.

Uniqueness. Suppose that we have a finite syntomic morphism $f: Y \to X$ such that $X$ is smooth over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ and $f$ is étale over a dense open of $X$. We claim that in this case $\Theta ^ p_{Y/X}$ is uniquely determined by property (1). Namely, consider the maps

The sheaf $\Omega ^ p_{X/\mathbf{Z}}$ is torsion free (by the assumed smoothness), hence it suffices to check that the restriction of $\Theta ^ p_{Y/X}$ is uniquely determined over the dense open over which $f$ is étale, i.e., we may assume $f$ is étale. However, if $f$ is étale, then $f^*\Omega _{X/\mathbf{Z}} = \Omega _{Y/\mathbf{Z}}$ hence the first arrow in the displayed equation is an isomorphism. Since we've pinned down the composition, this guarantees uniqueness.

Let $f : Y \to X$ be a finite syntomic morphism of locally Noetherian schemes. Let $x \in X$. By Discriminants, Lemma 49.11.7 we can find $d \geq 1$ and a commutative diagram

such that $x \in U \subset X$ is open, $V = f^{-1}(U)$ and $V = U \times _{U_ d} V_ d$. Thus $\Theta ^ p_{Y/X}|_ V$ is the pullback of the map $\Theta ^ p_{V_ d/U_ d}$. However, by the discussion on uniqueness above and Discriminants, Lemmas 49.11.4 and 49.11.5 the map $\Theta ^ p_{V_ d/U_ d}$ is uniquely determined by the requirement (1). Hence uniqueness holds.

At this point we know that we have existence and uniqueness for all finite syntomic morphisms $Y \to X$ with $X$ locally Noetherian. We could now give an argument similar to the proof of Lemma 50.18.3 to extend to general $X$. However, instead it possible to directly use absolute Noetherian approximation to finish the proof. Namely, to construct $\Theta ^ p_{Y/X}$ it suffices to do so Zariski locally on $X$ (provided we also show the uniqueness). Hence we may assume $X$ is affine (small detail omitted). Then we can write $X = \mathop{\mathrm{lim}}\nolimits _{i \in I} X_ i$ as the limit over a directed set $I$ of Noetherian affine schemes. By Algebra, Lemma 10.127.8 we can find $0 \in I$ and a finitely presented morphism of affines $f_0 : Y_0 \to X_0$ whose base change to $X$ is $Y \to X$. After increasing $0$ we may assume $Y_0 \to X_0$ is finite and syntomic, see Algebra, Lemma 10.168.9 and 10.168.3. For $i \geq 0$ also the base change $f_ i : Y_ i = Y_0 \times _{X_0} X_ i \to X_ i$ is finite syntomic. Then

Hence we can (and are forced to) define $\Theta ^ p_{Y/X}$ as the colimit of the maps $\Theta ^ p_{Y_ i/X_ i}$. This map is compatible with any cartesian diagram

with $X'$ affine as we know this for the case of Noetherian affine schemes by the arguments given above (small detail omitted; hint: if we also write $X' = \mathop{\mathrm{lim}}\nolimits _{j \in J} X'_ j$ then for every $i \in I$ there is a $j \in J$ and a morphism $X'_ j \to X_ i$ compatible with the morphism $X' \to X$). This finishes the proof. $\square$

Proposition 50.19.3. Let $f : Y \to X$ be a finite syntomic morphism of schemes. The maps $\Theta ^ p_{Y/X}$ of Lemma 50.19.2 define a map of complexes

with the following properties

in degree $0$ we get $\text{Trace}_ f : f_*\mathcal{O}_ Y \to \mathcal{O}_ X$, see Discriminants, Section 49.3,

we have $\Theta _{Y/X}(\omega \wedge \eta ) = \omega \wedge \Theta _{Y/X}(\eta )$ for $\omega $ in $\Omega ^\bullet _{X/\mathbf{Z}}$ and $\eta $ in $f_*\Omega ^\bullet _{Y/\mathbf{Z}}$,

if $f$ is a morphism over a base scheme $S$, then $\Theta _{Y/X}$ induces a map of complexes $f_*\Omega ^\bullet _{Y/S} \to \Omega ^\bullet _{X/S}$.

**Proof.**
By Discriminants, Lemma 49.11.7 for every $x \in X$ we can find $d \geq 1$ and a commutative diagram

such that $x \in U \subset X$ is affine open, $V = f^{-1}(U)$ and $V = U \times _{U_ d} V_ d$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$ and observe that $B = A \otimes _{A_ d} B_ d$ and recall that $B_ d = A_ d e_1 \oplus \ldots \oplus A_ d e_ d$. Suppose we have $a_1, \ldots , a_ r \in A$ and $b_1, \ldots , b_ s \in B$. We may write $b_ j = \sum a_{j, l} e_ d$ with $a_{j, l} \in A$. Set $N = r + sd$ and consider the factorizations

Here the horizontal lower right arrow is given by the morphism $U \to U_ d$ (from the earlier diagram) and the morphism $U \to \mathbf{A}^ N$ given by $a_1, \ldots , a_ r, a_{1, 1}, \ldots , a_{s, d}$. Then we see that the functions $a_1, \ldots , a_ r$ are in the image of $\Gamma (U', \mathcal{O}_{U'}) \to \Gamma (U, \mathcal{O}_ U)$ and the functions $b_1, \ldots , b_ s$ are in the image of $\Gamma (V', \mathcal{O}_{V'}) \to \Gamma (V, \mathcal{O}_ V)$. In this way we see that for any finite collection of elements^{1} of the groups

we can find a factorizations $V \to V' \to V_ d$ and $U \to U' \to U_ d$ with $V' = \mathbf{A}^ N \times V_ d$ and $U' = \mathbf{A}^ N \times U_ d$ as above such that these sections are the pullbacks of sections from

The upshot of this is that to check $\text{d} \circ \Theta _{Y/X} = \Theta _{Y/X} \circ \text{d}$ it suffices to check this is true for $\Theta _{V'/U'}$. Similarly, for property (2) of the lemma.

By Discriminants, Lemmas 49.11.4 and 49.11.5 the scheme $U_ d$ is smooth and the morphism $V_ d \to U_ d$ is étale over a dense open of $U_ d$. Hence the same is true for the morphism $V' \to U'$. Since $\Omega _{U'/\mathbf{Z}}$ is locally free and hence $\Omega ^ p_{U'/\mathbf{Z}}$ is torsion free, it suffices to check the desired relations after restricting to the open over which $V'$ is finite étale. Then we may check the relations after a surjective étale base change. Hence we may split the finite étale cover and assume we are looking at a morphism of the form

with $W$ smooth over $\mathbf{Z}$. In this case any local properties of our construction are trivial to check (provided they are true). This finishes the proof of (1) and (2).

Finally, we observe that (3) follows from (2) because $\Omega _{Y/S}$ is the quotient of $\Omega _{Y/\mathbf{Z}}$ by the submodule generated by pullbacks of local sections of $\Omega _{S/\mathbf{Z}}$. $\square$

Example 50.19.4. Let $A$ be a ring. Let $f = x^ d + \sum _{0 \leq i < d} a_{d - i} x^ i \in A[x]$. Let $B = A[x]/(f)$. By Proposition 50.19.3 we have a morphism of complexes

In particular, if $t \in B$ denotes the image of $x \in A[x]$ we can consider the elements

What are these elements? By the same principle as used in the proof of Proposition 50.19.3 it suffices to compute this in the universal case, i.e., when $A = \mathbf{Z}[a_1, \ldots , a_ d]$ or even when $A$ is replaced by the fraction field $\mathbf{Q}(a_1, \ldots , a_ d)$. Writing symbolically

we see that over $\mathbf{Q}(\alpha _1, \ldots , \alpha _ d)$ the algebra $B$ becomes split:

Thus for example

Next, we have

Next, we have

(modulo calculation error), and so on. This suggests that if $f(x) = x^ d - a$ then

in $\Omega _ A$. This is true for in this particular case one can do the calculation for the extension $\mathbf{Q}(a)[x]/(x^ d - a)$ to verify this directly.

Lemma 50.19.5. Let $p$ be a prime number. Let $X \to S$ be a smooth morphism of relative dimension $d$ of schemes in characteristic $p$. The relative Frobenius $F_{X/S} : X \to X^{(p)}$ of $X/S$ (Varieties, Definition 33.36.4) is finite syntomic and the corresponding map

is zero in all degrees except in degree $d$ where it defines a surjection.

**Proof.**
Observe that $F_{X/S}$ is a finite morphism by Varieties, Lemma 33.36.8. To prove that $F_{X/S}$ is flat, it suffices to show that the morphism $F_{X/S, s} : X_ s \to X^{(p)}_ s$ between fibres is flat for all $s \in S$, see More on Morphisms, Theorem 37.16.2. Flatness of $X_ s \to X^{(p)}_ s$ follows from Algebra, Lemma 10.128.1 (and the finiteness already shown). By More on Morphisms, Lemma 37.59.10 the morphism $F_{X/S}$ is a local complete intersection morphism. Hence $F_{X/S}$ is finite syntomic (see More on Morphisms, Lemma 37.59.8).

For every point $x \in X$ we may choose a commutative diagram

where $\pi $ is étale and $x \in U$ is open in $X$, see Morphisms, Lemma 29.36.20. Observe that $\mathbf{A}^ d_ S \to \mathbf{A}^ d_ S$, $(x_1, \ldots , x_ d) \mapsto (x_1^ p, \ldots , x_ d^ p)$ is the relative Frobenius for $\mathcal{A}^ d_ S$ over $S$. The commutative diagram

of Varieties, Lemma 33.36.5 for $\pi : U \to \mathbf{A}^ d_ S$ is cartesian by Étale Morphisms, Lemma 41.14.3. Since the construction of $\Theta $ is compatible with base change and since $\Omega _{U/S} = \pi ^*\Omega _{\mathbf{A}^ d_ S/S}$ (Lemma 50.2.2) we conclude that it suffices to show the lemma for $\mathbf{A}^ d_ S$.

Let $A$ be a ring of characteristic $p$. Consider the unique $A$-algebra homomorphism $A[y_1, \ldots , y_ d] \to A[x_1, \ldots , x_ d]$ sending $y_ i$ to $x_ i^ p$. The arguments above reduce us to computing the map

We urge the reader to do the computation in this case for themselves. As in Example 50.19.4 we may reduce this to computing a formula for $\Theta ^ i$ in the universal case

In turn, we can find the formula for $\Theta ^ i$ by computing in the complex case, i.e., for the $\mathbf{C}$-algebra map

We may even invert $x_1, \ldots , x_ d$ and $y_1, \ldots , y_ d$. In this case, we have $\text{d}x_ i = p^{-1} x_ i^{- p + 1}\text{d}y_ i$. Hence we see that

by the properties of $\Theta ^ i$. An elementary computation shows that the trace in the expression above is zero unless $e_1, \ldots , e_ i$ are congruent to $-1$ modulo $p$ and $e_{i + 1}, \ldots , e_ d$ are divisible by $p$. Moreover, in this case we obtain

We conclude that we get zero in characteristic $p$ unless $d = i$ and in this case we get every possible $d$-form. $\square$

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